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I want to modify a Swift string by converting some characters to uppercase, some others to lowercase.
In Obj-c I had following :
- (NSString*) lowercaseDestination:(NSString*) string {
NSUInteger length = string.length;
unichar buf[length+1];
[string getCharacters:buf];
BOOL up = true;
for (int i=0; i< length ; i++) {
unichar chr = buf[i];
if( .... ) {
buf[i] = toupper(chr);
} else {
buf[i] = tolower(chr);
}
}
string = [NSString stringWithCharacters:buf length:length];
return string;
How would you do that in Swift 2 ?
I did no find any Character method to upper or lower the case.
Would be an array of String of 1 character be an option ? (And then use String methods to upper and lower each String
String has a upperCaseString method, but Character doesn't.
The reason is that in exotic languages like German, converting a single
character to upper case can result in multiple characters:
print("ß".uppercaseString) // "SS"
The toupper/tolower functions are not Unicode-safe and not
available in Swift.
So you can enumerate the string characters, convert each character to
a string, convert that to upper/lowercase, and concatenate the results:
func lowercaseDestination(str : String) -> String {
var result = ""
for c in str.characters {
let s = String(c)
if condition {
result += s.lowercaseString
} else {
result += s.uppercaseString
}
}
return result
}
which can be written more compactly as
func lowercaseDestination(str : String) -> String {
return "".join(str.characters.map { c -> String in
let s = String(c)
return condition ? s.lowercaseString : s.uppercaseString
})
}
Re your comment: If the condition needs to check more than one
character then you might want to create an array of all characters
first:
func lowercaseDestination(str : String) -> String {
var result = ""
let characters = Array(str.characters)
for i in 0 ..< characters.count {
let s = String(characters[i])
if condition {
result += s.lowercaseString
} else {
result += s.uppercaseString
}
}
return result
}
Write a method to test if a string meets the preconditions to become a palindrome.
Eg:
Input | Output
mmo | True
yakak | True
travel | False
I'm thinking of this approach:
Make a suffix tree for all permutation of T such that T$Reverse(T)#
Check for all permutation for same node
Am I missing anything?
All you need to do is check that there's at most one character with an odd number of occurrences. Here's a Java example:
private static boolean canMakePalindrom(String s) {
Map<Character, Integer> countChars = new HashMap<>();
// Count the occurrences of each character
for (char c : s.toCharArray()) {
Integer count = countChars.get(c);
if (count == null) {
count = Integer.valueOf(1);
} else {
count = count + 1;
}
countChars.put(c, count);
}
boolean hasOdd = false;
for (int count : countChars.values()) {
if (count % 2 == 1) {
if (hasOdd) {
// Found two chars with odd counts - return false;
return false;
} else {
// Found the first char with odd count
hasOdd = true;
}
}
}
// Haven't found more than one char with an odd count
return true;
}
EDIT4 (yes - these are ordered to make sense, but numbered by chronological order):
The above implementation has a built in inefficiency. I don't think the first iteration over the string can be avoided, but there's no real reason to keep a count of all the occurrences - it's enough to just keep track of those with the an odd count. For this usecase, it's enough to keep track of each character we encounter (e.g., with a Set), and remove it when we encounter it again. In the worst case, where all the characters in the string are different, the performance is comparable, but in the common case, where there are several occurrences of each character, this implementation improves both time and memory complexity of the second loop (which is now reduced to a single condition) dramatically:
private static boolean canMakePalindrom(String s) {
Set<Character> oddChars = new HashSet<>();
// Go over the characters
for (char c : s.toCharArray()) {
// Record the encountered character:
if (!oddChars.add(c)) {
// If the char was already encountered, remove it -
// this is an even time we encounter it
oddChars.remove(c);
}
}
// Check the number of characters with odd counts:
return oddChars.size() <= 1;
}
EDIT3 (yes - these are ordered to make sense, but numbered by chronological order):
Java 8 provides a fluent streaming API which could be used to create an implementation similar to the Python one-liners below:
private static boolean canMakePalindrom(String s) {
return s.chars()
.boxed()
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting()))
.values()
.stream()
.filter(p -> p % 2 == 1)
.count() <= 1;
}
EDIT:
Python built-in functions and comprehension capabilities make this too attractive not to publish this one liner solution. It's probably less efficient than the aforementioned Java one, but is quite elegant:
from collections import Counter
def canMakePalindrom(s):
return len([v for v in Counter(s).values() if v % 2 == 1]) <= 1
EDIT2:
Or, an even cleaner approach as proposed by #DSM in the comments:
from collections import Counter
def canMakePalindrom(s):
return sum(v % 2 == 1 for v in Counter(s).values()) <= 1
Instead of counting how many times each letter occurs, another approach keeps track of whether a letter has occurred an odd or even number of times. If a letter has occurred an even number of times, you don’t need to worry about it, and only need to keep track of the odd occurrences in a set. In Java:
public static boolean canMakePalindrome(String s) {
Set<Character> oddLetters = new HashSet<>();
for ( char c : s.toCharArray() ) {
if ( ! oddLetters.remove(c) ) {
oddLetters.add(c);
}
}
return oddLetters.size() <= 1;
}
Really all you're looking for is if all (or all but one) of the letters are paired off. As long as they are, then they will be able to be turned into a palindrome.
So it would be something like...
bool canBeTurnedIntoAPalindrome(string drome)
{
// If we've found a letter that has no match, the center letter.
bool centerUsed = false;
char center;
char c;
int count = 0;
// TODO: Remove whitespace from the string.
// Check each letter to see if there's an even number of it.
for(int i = 0; i<drome.length(); i++)
{
c = drome[i];
count = 0;
for(int j = 0; j < drome.length(); j++)
if (drome[j] == c)
count++;
// If there was an odd number of those entries
// and the center is already used, then a palindrome
// is impossible, so return false.
if (count % 2 == 1)
{
if (centerUsed == true && center != c)
return false;
else
{
centerused = true;
center = c; // This is so when we encounter it again it
// doesn't count it as another separate center.
}
}
}
// If we made it all the way through that loop without returning false, then
return true;
}
This isn't the most efficient (it's counting letters as many times as it comes across them, even if they've been counted already) but it does work.
If I'm understanding your question correctly, this is how I understand it:
If the input string can be rearranged into a palindrome, output "True", otherwise output "False".
Then you can use these simple rules:
If the length is even, every unique character in the input has to occur a multiple of 2 times.
If the length is odd, every unique character except one has to occur a multiple of 2 times. Only 1 character is allowed to not occur a multiple of 2 times.
So for the 3 given examples:
"mmo", odd length, m occurs twice (multiple of 2), o occurs once (not a multiple of 2), so True.
"yakak", odd length, a occurs twice (multiple of 2), k occurs twice (multiple of 2), y occurs once (not a multiple of 2) , so True.
"travel", more than one character does not occur a multiple of 2, so False.
Additional examples:
"mmorpg", only m occurs a multiple of 2, the rest only once, so False.
"mmom", no characters occur a multiple of 2, more than one character occurs "not a multiple of 2 times", so False.
At this point you should realise that if only 1 character is allowed to occur a non-multiple-of-2 times, then you can disregard the length. A string with an even length will have either 2 or more characters occuring a non-multiple-of-2 times, or none at all.
So the final rule should be this:
If at most 1 unique character occurs a non-multiple-of-2 times in the input, the output is True otherwise the output is False.
def can_permutation_palindrome(s):
counter = {}
for c in s:
counter[c] = counter.get(c, 0) + 1
odd_count = 0
for count in counter.values():
odd_count += count % 2
return odd_count in [0, 1]
def check(string):
bv = 0
for s in string:
bv ^= 1 << ord(s)
return bv == 0 or bv & (bv - 1) == 0
I reached the solution below today (python). I think it's readable, and performance-wise it's really good.
sum(map(lambda x: word.count(x) % 2, set(word))) <= 1
We're basically counting the number of occurrences of each character in the string "word", getting the remainder of the division by 2, summing them all and checking if you have at most 1 of them.
The idea is that you need to have all characters paired, except potentially for one (the middle one).
My idea is, if the number of letters with odd count is one and rest all have even count, a palindrome is possible..Here's my program in Python
string = raw_input()
found = False
char_set = set(string) # Lets find unique letters
d_dict = {}
for c in char_set:
d_dict[c] = string.count(c) # Keep count of each letter
odd_l = [e for e in d_dict.values() if e%2 == 1] # Check how many has odd number of occurrence
if len(odd_l) >1:
pass
else:
found = True
if not found:
print("NO")
else:
print("YES")
Any string can be palindrome only if at most one character occur odd no. of times and all other characters must occur even number of times. The following program can be used to check whether a palindrome can be string or not.
void checkPalindrome(string s)
{
vector<int> vec(256,0); //Vector for all ASCII characters present.
for(int i=0;i<s.length();++i)
{
vec[s[i]-'a']++;
}
int odd_count=0,flag=0;
for(int i=0;i<vec.size();++i)
{
if(vec[i]%2!=0)
odd_count++;
if(odd_count>1)
{
flag=1;
cout<<"Can't be palindrome"<<endl;
break;
}
}
if(flag==0)
cout<<"Yes can be palindrome"<<endl;
}
With O(n) complexity .
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace PallindromePemutation
{
class charcount
{
public char character { get; set; }
public int occurences { get; set; }
}
class Program
{
static void Main(string[] args)
{
List<charcount> list = new List<charcount>();
charcount ch;
int count = 0;
char[] arr = "travel".ToCharArray();
for (int i = 0; i < arr.Length; i++)
{
charcount res = list.Find(x => x.character == arr.ElementAt(i));
if (res == null)
{
ch = new charcount();
ch.character = arr.ElementAt(i);
ch.occurences = 1;
list.Add(ch);
}
else
{
charcount temp= list.Find(x => x.character == arr.ElementAt(i));
temp.occurences++;
}
}
foreach (var item in list)
{
if (!(item.occurences % 2 == 0))
{
count++;
}
}
if (count > 1)
{
Console.WriteLine("false");
}
else
{
Console.WriteLine("true");
}
Console.ReadKey();
}
}
}
If we don't care case sensitivity of characters and spaces within a string, then a sample solution in C# by using Dictionary can be like :
private static bool IsPalindromePermutation(string inputStr)
{
// First, check whether input string is null or whitespace.
// If yes, then return false.
if (string.IsNullOrWhiteSpace(inputStr))
return false;
var inputDict = new Dictionary<char, int>();
// Big/small letter is not important
var lowerInputStr = inputStr.ToLower();
// Fill input dictionary
// If hit a space, then skip it
for (var i = 0; i < lowerInputStr.Length; i++)
{
if (lowerInputStr[i] != ' ')
{
if (inputDict.ContainsKey(lowerInputStr[i]))
inputDict[lowerInputStr[i]] += 1;
else
inputDict.Add(lowerInputStr[i], 1);
}
}
var countOdds = 0;
foreach(var elem in inputDict)
{
if(elem.Value % 2 != 0)
countOdds++;
}
return countOdds <= 1;
}
We can acheive this via collections also
String name = "raa";
List<Character> temp = new ArrayList<>(name.chars()
.mapToObj(e -> (char) e).collect(Collectors.toList()));
for (int i = 0; i < temp.size(); i++) {
for (int j = i + 1; j < temp.size(); j++) {
if (temp.get(i).equals(temp.get(j))) {
temp.remove(j);
temp.remove(i);
i--;
}
}
}
if (temp.size() <= 1) {
System.out.println("Pallindrome");
} else {
System.out.println(temp.size());
System.out.println("Not Pallindrome");
}
}
This is my solution
public static void main(String[] args) {
List<Character> characters = new ArrayList<>();
Scanner scanner = new Scanner(System.in);
String input = scanner.nextLine();
for (int i = 0; i < input.length(); i++){
char val = input.charAt(i);
if (characters.contains(val)){
characters.remove(characters.indexOf(val));
} else{
characters.add(val);
}
}
if (characters.size() == 1 || characters.size() == 0){
System.out.print("Yes");
} else{
System.out.print("No");
}
}
That 's my solution. The string could contain several words with spaces, such as
Input: Tact Coa
Output true
Input: Tact Coa vvu
Output: false
public static boolean checkForPalindrome(String str) {
String strTrimmed = str.replaceAll(" ","");
System.out.println(strTrimmed);
char[] str1 = strTrimmed.toCharArray();
for (int i = 0; i < str1.length; i++) {
str1[i] = Character.toLowerCase(str1[i]);
}
Arrays.sort(str1);
String result = new String(str1);
System.out.println(result);
int count = 0;
for (int j = 0; j < str1.length; j += 2) {
if (j != str1.length-1) {
if (str1[j] != str1[j+1]) {
count++;
j++;
}
} else {
count++;
}
}
if (count > 1) return false;
else return true;
}
Question: Can a String become a palindrome?
Method1: count of characters
IN Java :
public class TEST11 {
public static void main(String[] args) {
String a = "Protijayi";
int[] count = new int[256];
Arrays.fill(count, 0);
for (int i = 0; i < a.length(); i++) {
char ch = a.charAt(i);
count[ch]++;
} // for
// counting of odd letters
int odd = 0;
for (int i = 0; i < count.length; i++) {
if ((count[i] & 1) == 1) {
odd++;
}
} // for
if (odd > 1) {
System.out.println("no");
} else {
System.out.println("yes");
}
}
}
IN Python:
def fix (a):
count = [0] * 256
for i in a: count[ord(i)] += 1
# counting of odd characters
odd = 0
for i in range(256):
if((count[i] & 1) == 1): odd += 1
if(odd > 1):print("no")
else:print("yes")
a = "Protijayi"
fix(a)
Method 2 : Use of HashSet
In Java:
public class TEST11 {
public static void main(String[] args) {
String a = "Protijayi";
Set<Character> set = new HashSet<>();
for (char ch : a.toCharArray()) {
if (set.contains(ch)) {
set.remove(ch);
}
set.add(ch);
} // for
if (set.size() <= 1) {
System.out.println("yes can be a palindrome");
} else {
System.out.println("no");
}
}
}
Swift example for this question.
var str = "mmoosl"
extension String {
func count(of needle: Character) -> Int {
return reduce(0) {
$1 == needle ? $0 + 1 : $0
}
}
}
func canBeTurnedIntoAPalinpolyString(_ polyString: String) -> Bool {
var centerUsed = false
var center = Character("a")
for i in polyString {
let count = polyString.count(of: i)
if count == 1 && !centerUsed {
center = i
centerUsed = true
} else {
if count % 2 != 0 {
return false
}
}
}
return true
}
print(canBeTurnedIntoAPalinpolyString(str))
Java
private static boolean isStringPalindromePermutation(String input) {
if(input == null) return false;
if(input.isEmpty()) return false;
int checker = 0;
for (int i = 0; i < input.length(); i++) {
int character = input.charAt(i) - 'a';
int oneShiftedByNumberInCharacter = 1 << character;
int summaryAnd = checker & oneShiftedByNumberInCharacter;
if ( summaryAnd > 0 ) {
int revertToShiftedByChar = ~oneShiftedByNumberInCharacter;
checker = checker & revertToShiftedByChar;
} else {
checker |= oneShiftedByNumberInCharacter;
}
}
if ( input.length() % 2 == 0 ) {
if ( checker == 0) {
return true;
}
else return false;
} else {
int checkerMinusOne = checker-1;
if((checkerMinusOne & checker) == 0){
return true;
}else{
return false;
}
}
}
Why use a suffix tree or any other data structure?
The basic requirement of a palindromic string is the frequency of all characters must be even or only one character can have odd frequency.
Example :-
Input : aabbaa
Output : frequency of a is 4 and b is 2 (both even)
Input : xxzyzxx
Output : frequency of x is 4, z is 2 and y=1 (only 1 odd)
Sample code for better understanding :
bool ispalin(string str) //function to check
{
int freq[26] = {0}; //to store frequency of character here i am
// considering only lower case letters
for (int i = 0; str.length(); i++)
freq[str[i]]++;
int odd = 0;
for (int i = 0; i < 26; i++) //Count odd occurring characters
{
if (freq[i] & 1) //checking if odd
odd++;
if (odd > 1) //if number of odd freq is greater than 1
return false;
}
return true; //else return true
}
python code to check whether a palindrome can be formed from given string or not:
test_str = input('enter any string = ')
count = 0
for item in set(test_str):
if test_str.count(item)%2 != 0:
count+=1
if (count>1):
print(" palindrome cannot be formed")
else:
print(" palindrome can be formed")
Please try this code if any issue please comments
More efficient implementation - Java
boolean palindromeRearranging(String inputString) {
Map<Character, Integer> charsCount = new HashMap<Character, Integer>();
for(char c : inputString.toCharArray()){
charsCount.compute(c, (key, val) -> val == null ? 1 : val + 1);
}
List<Integer> result = new ArrayList<>();
charsCount.forEach((k, v) -> {
if(v % 2 != 0){
result.add(v);
}
});
return (result.size() == 0 || result.size() == 1);
}
Here is my code :
boolean palindromeRearranging(String inputString) {
HashMap<Character,Integer> stCount=new HashMap<>();
for(int i=0;i<inputString.length();i++){
stCount.put(inputString.charAt(i),0);
}
for(int i=0;i<inputString.length();i++){
int c= stCount.get(inputString.charAt(i));
stCount.put(inputString.charAt(i),++c);
}
int c=0;
for (Map.Entry<Character,Integer> entry : stCount.entrySet()){
if(entry.getValue()%2!=0){
c++;
if(c>1){
return false;
}
}
}
return true;
}
JS solution:
function solution(inputString) {
const arr = inputString.split('');
let hasCoupleList = arr.map( (el) => arr.filter( (el1) => el1 == el).length % 2 == 0).filter( (el) => el == false).length;
return (arr.length % 2 == 0)
? hasCoupleList == 0
: hasCoupleList == 1;
}
With JAVA
import java.util.*;
import java.lang.*;
//Classs
class Permutation {
/*
* We need to have an even number of almost all characters,
* so that half can be on one side and half can be on the other side.
* At most one character (the middle character) can have an odd count.
*/
public static boolean hasPalindrome(String str) {
boolean wasOdd = false;
for (Character c: str.toCharArray()) {
int counter = 0;
for (Character cc: str.toCharArray()) {
if (c == cc) {
counter++;
}
}
if (counter % 2 == 1) {
if (wasOdd) {
return false;
}
wasOdd = true;
}
}
return true;
}
public static void main(String args[]) throws Exception {
//Taking string input
//Scanner
Scanner s = new Scanner(System.in);
String str = s.nextLine();
if (Permutation.hasPalindrome(str)) {
System.out.println("YES"); // Writing output to STDOUT
} else {
System.out.println("NO"); // Writing output to STDOUT
}
}
}
Implementation from Checking if a String is a Permutation of a Palindrome
Time complexity is essentially O(n). This means that the function is linear in the length of the input string
public static boolean isPermutationOfPalindrome(String str) {
// Convert the input string to lower case and remove any non-letter characters
str = str.toLowerCase().replaceAll("[^a-z]", "");
// Create an array to count the frequency of each letter
int[] charCounts = new int[26];
for (int i = 0; i < str.length(); i++) {
charCounts[str.charAt(i) - 'a']++;
}
// Check if there is at most one character with an odd frequency
boolean foundOdd = false;
for (int count : charCounts) {
if (count % 2 == 1) {
if (foundOdd) {
return false;
}
foundOdd = true;
}
}
return true;
}
I have a problem in which I have to SWAP or move characters and integers. Like I have any characters A . now I have some cases, like
NOTE:- Have to use characters A-Z and integers 0-9
A, now I want that when my program run I assign some integer value to this character, If I assign value 3 to this character then A will become D or it just move to 3 places.
Now if I have a character like Y and I add 4 then it will become C means after Z it will again start from character A.
Same condition I have to follow with Integer if i have 9 and we assign 3 to it then it will become 2 because loop start from 0 not from 1. Means we have to use only 0-9 integers.
I know that i am using wrong name to question but i have no idea that what lines i have to use for that kind of question.
Hope you understand my problem.
Thanks in advance.
Try the below extension method, which does the following:
It creates 2 dictionaries in order to speed up the key look up in the alphabet
Will parse the inputString variable, split it in substrings of the length of the moveString variable's length (or the remainder)
On every substring, it will evaluate each character in order to detect if it's a digit
If it's not a digit, it looks up for the value in the swappedAlphabet dictionary, by using the int key
If it's a digit, it applies a modulo operation on the sum of the digit and the corresponding moveint value
It finally aggregates all the characters in the final result string
Here's the code:
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static void Main(string[] args)
{
string
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string inputString = "ABC123D", moveString = "12";
var result = inputString.Swap(alphabet, moveString);
Console.WriteLine(result);
}
}
static class ExtensionMethods
{
public static Dictionary<TValue, TKey>
SwapKeysValues<TKey, TValue>(this Dictionary<TKey, TValue> input)
{
var result = new Dictionary<TValue, TKey>();
input.ToList().ForEach((keyValuePair) =>
{
result.Add(keyValuePair.Value, keyValuePair.Key);
});
return result;
}
public static string Swap(
this string input,
string alphabet,
string move)
{
Dictionary<char, int>
alphabetDictionary = new Dictionary<char, int>();
for (int i = 0; i < alphabet.Length; i++)
{
alphabetDictionary.Add(alphabet[i], i);
}
var swapedAlphabet = alphabetDictionary.SwapKeysValues();
return Enumerable
.Range(0, (int)Math.Ceiling(input.Length / (move.Length * 1M)))
.ToList()
.Aggregate<int, string>("", (s, i) =>
{
var l = i * move.Length + move.Length;
var cInput = input.Substring(i * move.Length,
(l > input.Length)
? input.Length - i * move.Length : move.Length);
return s + cInput
.Select((c, index) =>
{
int intCandidate;
if (!Int32.TryParse(c.ToString(), out intCandidate))
{
var length = (alphabetDictionary[c] +
Int32.Parse(move[index].ToString()));
return
swapedAlphabet[(alphabet.Length > length)
? length : length % alphabet.Length];
}
else
{
var moveInt = Int32.Parse(move[index].ToString());
return Char.Parse(((intCandidate + moveInt) % 10)
.ToString());
}
})
.Aggregate<char, string>("", (a, b) => a + b);
});
}
}
Another alternative you have is relying on the in-built character/integer types which follow the order you want; with an additional consideration: if you account for caps, it would deliver caps ("B" after "A" and "b" after "a"). The only thing you need to worry about is making sure that the iterations will be limited to the A-Z/0-9 boundaries. Sample code:
public string moveChar(string inputChar, int noPos)
{
string outChar = checkBoundaries(inputChar, noPos);
if (outChar == "")
{
outChar = basicConversion(inputChar, noPos);
}
return outChar;
}
public string basicConversion(string inputChar, int noPos)
{
return Convert.ToString(Convert.ToChar(Convert.ToInt32(Convert.ToChar(inputChar)) + noPos));
}
public string checkBoundaries(string inputChar, int noPos)
{
string outString = "";
int count1 = 0;
do
{
count1 = count1 + 1;
string curTemp = basicConversion(inputChar, 1);
if (inputChar.ToLower() == "z" || curTemp.ToLower() == "z")
{
if (inputChar.ToLower() != "z")
{
noPos = noPos - count1;
}
inputChar = "a";
outString = "a";
if (inputChar == "Z" || curTemp == "Z")
{
inputChar = "A";
outString = "A";
}
count1 = 1;
}
else if (inputChar == "9" || curTemp == "9")
{
if (inputChar != "9")
{
noPos = noPos - count1;
}
inputChar = "0";
outString = "0";
count1 = 1;
}
else
{
inputChar = curTemp;
outString = inputChar;
}
} while (count1 < noPos);
return outString;
}
It expects strings (just one character (letter or number) per call) and you can call it simply by using: moveChar("current letter or number", no_of_pos_to_move). This version accounts just for "positive"/"forwards" movements but it might easily be edited to account for the inverse situation.
Here's a very simple way to implement a Caesar Cipher with the restrictions you defined.
var shift = 3;
var input = "HELLO WORLD 678";
var classAlphabets = new Dictionary<UnicodeCategory, string>
{
{ UnicodeCategory.SpaceSeparator, " " },
{ UnicodeCategory.UppercaseLetter, "ABCDEFGHIJKLMNOPQRSTUVWXYZ" },
{ UnicodeCategory.DecimalDigitNumber, "0123456789" }
};
var encoded = input.ToUpperInvariant()
.Select(c => new { Alphabet = classAlphabets[Char.GetUnicodeCategory(c)], Character = c })
.Select(x => new { x.Alphabet, Index = x.Alphabet.IndexOf(x.Character) })
.Select(x => new { x.Alphabet, Index = x.Index + shift })
.Select(x => new { x.Alphabet, Index = x.Index % x.Alphabet.Length })
.Select(x => x.Alphabet.ElementAt(x.Index))
.Aggregate(new StringBuilder(), (builder, character) => builder.Append(character))
.ToString();
Console.Write(encoded);
// encoded = "KHOOR ZRUOG 901"
Decoding is simply a case of inverting the shift.
Caesar cipher can be easier like this:
static char Encrypt(char ch, int code)
{
if (!char.IsLetter(ch))
{
return ch;
}
char offset = char.IsUpper(ch) ? 'A' : 'a';
return (char)(((ch + code - offset) % 26) + offset);
}
static string Encrypt(string input, int code)
{
return new string(input.ToCharArray().Select(ch => Encrypt(ch, code)).ToArray());
}
static string Decrypt(string input, int code)
{
return Encrypt(input, 26 - code);
}
const string TestCase = "Pack my box with five dozen liquor jugs.";
static void Main()
{
string str = TestCase;
Console.WriteLine(str);
str = Encrypt(str, 5);
Console.WriteLine("Encrypted: {0}", str);
str = Decrypt(str, 5);
Console.WriteLine("Decrypted: {0}", str);
Console.ReadKey();
}
Given a string S of length N find longest substring without repeating characters.
Example:
Input: "stackoverflow"
Output: "stackoverfl"
If there are two such candidates, return first from left. I need linear time and constant space algorithm.
You are going to need a start and an end locator(/pointer) for the
string and an array where you store information for each character:
did it occour at least once?
Start at the beginning of the string, both locators point to the
start of the string.
Move the end locator to the right till you find
a repetition (or reach the end of the string). For each processed character, store it in the array.
When stopped store the position if this is the largest substring. Also remember the repeated character.
Now do the same thing with the start locator, when processing
each character, remove its flags from the array. Move the locator till
you find the earlier occurrence of the repeated character.
Go back to step 3 if you haven't reached the end of string.
Overall: O(N)
import java.util.HashSet;
public class SubString {
public static String subString(String input){
HashSet<Character> set = new HashSet<Character>();
String longestOverAll = "";
String longestTillNow = "";
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (set.contains(c)) {
longestTillNow = "";
set.clear();
}
longestTillNow += c;
set.add(c);
if (longestTillNow.length() > longestOverAll.length()) {
longestOverAll = longestTillNow;
}
}
return longestOverAll;
}
public static void main(String[] args) {
String input = "substringfindout";
System.out.println(subString(input));
}
}
You keep an array indicating the position at which a certain character occurred last. For convenience all characters occurred at position -1. You iterate on the string keeping a window, if a character is repeated in that window, you chop off the prefix that ends with the first occurrence of this character. Throughout, you maintain the longest length. Here's a python implementation:
def longest_unique_substr(S):
# This should be replaced by an array (size = alphabet size).
last_occurrence = {}
longest_len_so_far = 0
longest_pos_so_far = 0
curr_starting_pos = 0
curr_length = 0
for k, c in enumerate(S):
l = last_occurrence.get(c, -1)
# If no repetition within window, no problems.
if l < curr_starting_pos:
curr_length += 1
else:
# Check if it is the longest so far
if curr_length > longest_len_so_far:
longest_pos_so_far = curr_starting_pos
longest_len_so_far = curr_length
# Cut the prefix that has repetition
curr_length -= l - curr_starting_pos
curr_starting_pos = l + 1
# In any case, update last_occurrence
last_occurrence[c] = k
# Maybe the longest substring is a suffix
if curr_length > longest_len_so_far:
longest_pos_so_far = curr_starting_pos
longest_len_so_far = curr_length
return S[longest_pos_so_far:longest_pos_so_far + longest_len_so_far]
EDITED:
following is an implementation of the concesus. It occured to me after my original publication. so as not to delete original, it is presented following:
public static String longestUniqueString(String S) {
int start = 0, end = 0, length = 0;
boolean bits[] = new boolean[256];
int x = 0, y = 0;
for (; x < S.length() && y < S.length() && length < S.length() - x; x++) {
bits[S.charAt(x)] = true;
for (y++; y < S.length() && !bits[S.charAt(y)]; y++) {
bits[S.charAt(y)] = true;
}
if (length < y - x) {
start = x;
end = y;
length = y - x;
}
while(y<S.length() && x<y && S.charAt(x) != S.charAt(y))
bits[S.charAt(x++)]=false;
}
return S.substring(start, end);
}//
ORIGINAL POST:
Here is my two cents. Test strings included. boolean bits[] = new boolean[256] may be larger to encompass some larger charset.
public static String longestUniqueString(String S) {
int start=0, end=0, length=0;
boolean bits[] = new boolean[256];
int x=0, y=0;
for(;x<S.length() && y<S.length() && length < S.length()-x;x++) {
Arrays.fill(bits, false);
bits[S.charAt(x)]=true;
for(y=x+1;y<S.length() && !bits[S.charAt(y)];y++) {
bits[S.charAt(y)]=true;
}
if(length<y-x) {
start=x;
end=y;
length=y-x;
}
}
return S.substring(start,end);
}//
public static void main(String... args) {
String input[][] = { { "" }, { "a" }, { "ab" }, { "aab" }, { "abb" },
{ "aabc" }, { "abbc" }, { "aabbccdefgbc" },
{ "abcdeafghicabcdefghijklmnop" },
{ "abcdeafghicabcdefghijklmnopqrabcdx" },
{ "zxxaabcdeafghicabcdefghijklmnopqrabcdx" },
{"aaabcdefgaaa"}};
for (String[] a : input) {
System.out.format("%s *** GIVES *** {%s}%n", Arrays.toString(a),
longestUniqueString(a[0]));
}
}
Here is one more solution with only 2 string variables:
public static String getLongestNonRepeatingString(String inputStr){
if(inputStr == null){
return null;
}
String maxStr = "";
String tempStr = "";
for(int i=0; i < inputStr.length(); i++){
// 1. if tempStr contains new character, then change tempStr
if(tempStr.contains("" + inputStr.charAt(i))){
tempStr = tempStr.substring(tempStr.lastIndexOf(inputStr.charAt(i)) + 1);
}
// 2. add new character
tempStr = tempStr + inputStr.charAt(i);
// 3. replace maxStr with tempStr if tempStr is longer
if(maxStr.length() < tempStr.length()){
maxStr = tempStr;
}
}
return maxStr;
}
Algorithm in JavaScript (w/ lots of comments)..
/**
Given a string S find longest substring without repeating characters.
Example:
Input: "stackoverflow"
Output: "stackoverfl"
Input: "stackoverflowabcdefghijklmn"
Output: "owabcdefghijklmn"
*/
function findLongestNonRepeatingSubStr(input) {
var chars = input.split('');
var currChar;
var str = "";
var longestStr = "";
var hash = {};
for (var i = 0; i < chars.length; i++) {
currChar = chars[i];
if (!hash[chars[i]]) { // if hash doesn't have the char,
str += currChar; //add it to str
hash[chars[i]] = {index:i};//store the index of the char
} else {// if a duplicate char found..
//store the current longest non-repeating chars. until now
//In case of equal-length, <= right-most str, < will result in left most str
if(longestStr.length <= str.length) {
longestStr = str;
}
//Get the previous duplicate char's index
var prevDupeIndex = hash[currChar].index;
//Find all the chars AFTER previous duplicate char and current one
var strFromPrevDupe = input.substring(prevDupeIndex + 1, i);
//*NEW* longest string will be chars AFTER prevDupe till current char
str = strFromPrevDupe + currChar;
//console.log(str);
//Also, Reset hash to letters AFTER duplicate letter till current char
hash = {};
for (var j = prevDupeIndex + 1; j <= i; j++) {
hash[input.charAt(j)] = {index:j};
}
}
}
return longestStr.length > str.length ? longestStr : str;
}
//console.log("stackoverflow => " + findLongestNonRepeatingSubStr("stackoverflow"));
//returns stackoverfl
//console.log("stackoverflowabcdefghijklmn => " +
findLongestNonRepeatingSubStr("stackoverflowabcdefghijklmn")); //returns owabcdefghijklmn
//console.log("1230123450101 => " + findLongestNonRepeatingSubStr("1230123450101")); //
returns 234501
We can consider all substrings one by one and check for each substring whether it contains all unique characters or not.
There will be n*(n+1)/2 substrings. Whether a substirng contains all unique characters or not can be checked in linear time by
scanning it from left to right and keeping a map of visited characters. Time complexity of this solution would be O(n^3).`
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
public class LengthOfLongestSubstringWithOutRepeatingChar {
public static void main(String[] args)
{
String s="stackoverflow";
//allSubString(s);
System.out.println("result of find"+find(s));
}
public static String find(String s)
{
List<String> allSubsring=allSubString(s);
Set<String> main =new LinkedHashSet<String>();
for(String temp:allSubsring)
{
boolean a = false;
for(int i=0;i<temp.length();i++)
{
for(int k=temp.length()-1;k>i;k--)
{
if(temp.charAt(k)==temp.charAt(i))
a=true;
}
}
if(!a)
{
main.add(temp);
}
}
/*for(String x:main)
{
System.out.println(x);
}*/
String res=null;
int min=0,max=s.length();
for(String temp:main)
{
if(temp.length()>min&&temp.length()<max)
{
min=temp.length();
res=temp;
}
}
System.out.println(min+"ha ha ha"+res+"he he he");
return res;
}
//substrings left to right ban rahi hai
private static List<String> allSubString(String str) {
List<String> all=new ArrayList<String>();
int c=0;
for (int i = 0; i < str.length(); i++) {
for (int j = 0; j <= i; j++) {
if (!all.contains(str.substring(j, i + 1)))
{
c++;
all.add(str.substring(j, i + 1));
}
}
}
for(String temp:all)
{
System.out.println("substring :-"+temp);
}
System.out.println("count"+c);
return all;
}
}
Another O(n) JavaScript solution. It does not alter strings during the looping; it just keeps track of the offset and length of the longest sub string so far:
function longest(str) {
var hash = {}, start, end, bestStart, best;
start = end = bestStart = best = 0;
while (end < str.length) {
while (hash[str[end]]) hash[str[start++]] = 0;
hash[str[end]] = 1;
if (++end - start > best) bestStart = start, best = end - start;
}
return str.substr(bestStart, best);
}
// I/O for snippet
document.querySelector('input').addEventListener('input', function () {
document.querySelector('span').textContent = longest(this.value);
});
Enter word:<input><br>
Longest: <span></span>
simple python snippet
l=length p=position
maxl=maxlength maxp=maxposition
Tested and working. For easy understanding, I suppose there's a drawer to put the letters.
Function:
public int lengthOfLongestSubstring(String s) {
int maxlen = 0;
int start = 0;
int end = 0;
HashSet<Character> drawer = new HashSet<Character>();
for (int i=0; i<s.length(); i++) {
char ch = s.charAt(i);
if (drawer.contains(ch)) {
//search for ch between start and end
while (s.charAt(start)!=ch) {
//drop letter from drawer
drawer.remove(s.charAt(start));
start++;
}
//Do not remove from drawer actual char (it's the new recently found)
start++;
end++;
}
else {
drawer.add(ch);
end++;
int _maxlen = end-start;
if (_maxlen>maxlen) {
maxlen=_maxlen;
}
}
}
return maxlen;
}
Longest substring without repeating character in python
public int lengthOfLongestSubstring(String s) {
if(s.equals(""))
return 0;
String[] arr = s.split("");
HashMap<String,Integer> map = new HashMap<>();
Queue<String> q = new LinkedList<>();
int l_till = 1;
int l_all = 1;
map.put(arr[0],0);
q.add(arr[0]);
for(int i = 1; i < s.length(); i++){
if (map.containsKey(arr[i])) {
if(l_till > l_all){
l_all = l_till;
}
while(!q.isEmpty() && !q.peek().equals(arr[i])){
map.remove(q.remove());
}
if(!q.isEmpty())
map.remove(q.remove());
q.add(arr[i]);
map.put(arr[i],i);
//System.out.println(q);
//System.out.println(map);
l_till = q.size();
}
else {
l_till = l_till + 1;
map.put(arr[i],i);
q.add(arr[i]);
}
}
if(l_till > l_all){
l_all = l_till;
}
return l_all;
}
I was asked the same question in an interview.
I have written Python3 code, to find the first occurrence of the substring with all distinct chars. In my implementations, I start with index = 0 and iterate over the input string. While iterating used a Python dict seems to store indexes of chars in input-string those has been visited in the iteration.
In iteration, if char c, does not find in current substring – raise KeyError exception
if c is found to be a duplicate char in the current substring (as c previously appeared during iteration – named that index last_seen) start a new substring
def lds(string: str) -> str:
""" returns first longest distinct substring in input `string` """
seens = {}
start, end, curt_start = 0, 0, 0
for curt_end, c in enumerate(string):
try:
last_seen = seens[c]
if last_seen < curt_start:
raise KeyError(f"{c!r} not found in {string[curt_start: curt_end]!r}")
if end - start < curt_end - curt_start:
start, end = curt_start, curt_end
curt_start = last_seen + 1
except KeyError:
pass
seens[c] = curt_end
else:
# case when the longest substring is suffix of the string, here curt_end
# do not point to a repeating char hance included in the substring
if string and end - start < curt_end - curt_start + 1:
start, end = curt_start, curt_end + 1
return string[start: end]
private static string LongestSubstring(string word)
{
var set = new HashSet<char>();
string longestOverAll = "";
string longestTillNow = "";
foreach (char c in word)
{
if (!set.Contains(c))
{
longestTillNow += c;
set.Add(c);
}
else
{
longestTillNow = string.Empty;
}
if (longestTillNow.Length > longestOverAll.Length)
{
longestOverAll = longestTillNow;
}
}
return longestOverAll;
}
import java.util.ArrayList;
import java.util.HashSet;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.Set;
import java.util.TreeMap;
public class LongestSubString2 {
public static void main(String[] args) {
String input = "stackoverflowabcdefghijklmn";
List<String> allOutPuts = new ArrayList<String>();
TreeMap<Integer, Set> map = new TreeMap<Integer, Set>();
for (int k = 0; k < input.length(); k++) {
String input1 = input.substring(k);
String longestSubString = getLongestSubString(input1);
allOutPuts.add(longestSubString);
}
for (String str : allOutPuts) {
int strLen = str.length();
if (map.containsKey(strLen)) {
Set set2 = (HashSet) map.get(strLen);
set2.add(str);
map.put(strLen, set2);
} else {
Set set1 = new HashSet();
set1.add(str);
map.put(strLen, set1);
}
}
System.out.println(map.lastKey());
System.out.println(map.get(map.lastKey()));
}
private static void printArray(Object[] currentObjArr) {
for (Object obj : currentObjArr) {
char str = (char) obj;
System.out.println(str);
}
}
private static String getLongestSubString(String input) {
Set<Character> set = new LinkedHashSet<Character>();
String longestString = "";
int len = input.length();
for (int i = 0; i < len; i++) {
char currentChar = input.charAt(i);
boolean isCharAdded = set.add(currentChar);
if (isCharAdded) {
if (i == len - 1) {
String currentStr = getStringFromSet(set);
if (currentStr.length() > longestString.length()) {
longestString = currentStr;
}
}
continue;
} else {
String currentStr = getStringFromSet(set);
if (currentStr.length() > longestString.length()) {
longestString = currentStr;
}
set = new LinkedHashSet<Character>(input.charAt(i));
}
}
return longestString;
}
private static String getStringFromSet(Set<Character> set) {
Object[] charArr = set.toArray();
StringBuffer strBuff = new StringBuffer();
for (Object obj : charArr) {
strBuff.append(obj);
}
return strBuff.toString();
}
}
This is my solution, and it was accepted by leetcode. However, after I saw the stats, I saw whole lot solutions has much faster result....meaning, my solution is around 600ms for all their test cases, and most of the js solutions are around 200 -300 ms bracket.. who can tell me why my solution is slowwww??
var lengthOfLongestSubstring = function(s) {
var arr = s.split("");
if (s.length === 0 || s.length === 1) {
return s.length;
}
var head = 0,
tail = 1;
var str = arr[head];
var maxL = 0;
while (tail < arr.length) {
if (str.indexOf(arr[tail]) == -1) {
str += arr[tail];
maxL = Math.max(maxL, str.length);
tail++;
} else {
maxL = Math.max(maxL, str.length);
head = head + str.indexOf(arr[tail]) + 1;
str = arr[head];
tail = head + 1;
}
}
return maxL;
};
I am posting O(n^2) in python . I just want to know whether the technique mentioned by Karoly Horvath has any steps that are similar to existing search/sort algorithms ?
My code :
def main():
test='stackoverflow'
tempstr=''
maxlen,index=0,0
indexsubstring=''
print 'Original string is =%s\n\n' %test
while(index!=len(test)):
for char in test[index:]:
if char not in tempstr:
tempstr+=char
if len(tempstr)> len(indexsubstring):
indexsubstring=tempstr
elif (len(tempstr)>=maxlen):
maxlen=len(tempstr)
indexsubstring=tempstr
break
tempstr=''
print 'max substring length till iteration with starting index =%s is %s'%(test[index],indexsubstring)
index+=1
if __name__=='__main__':
main()
Simple and Easy
import java.util.Scanner;
public class longestsub {
static Scanner sn = new Scanner(System.in);
static String word = sn.nextLine();
public static void main(String[] args) {
System.out.println("The Length is " +check(word));
}
private static int check(String word) {
String store="";
for (int i = 0; i < word.length(); i++) {
if (store.indexOf(word.charAt(i))<0) {
store = store+word.charAt(i);
}
}
System.out.println("Result word " +store);
return store.length();
}
}
Not quite optimized but simple answer in Python
def lengthOfLongestSubstring(s):
temp,maxlen,newstart = {},0,0
for i,x in enumerate(s):
if x in temp:
newstart = max(newstart,s[:i].rfind(x)+1)
else:
temp[x] = 1
maxlen = max(maxlen, len(s[newstart:i + 1]))
return maxlen
I think the costly affair is rfind which is why it's not quite optimized.
This is my solution. Hope it helps.
function longestSubstringWithoutDuplication(str) {
var max = 0;
//if empty string
if (str.length === 0){
return 0;
} else if (str.length === 1){ //case if the string's length is 1
return 1;
}
//loop over all the chars in the strings
var currentChar,
map = {},
counter = 0; //count the number of char in each substring without duplications
for (var i=0; i< str.length ; i++){
currentChar = str.charAt(i);
//if the current char is not in the map
if (map[currentChar] == undefined){
//push the currentChar to the map
map[currentChar] = i;
if (Object.keys(map).length > max){
max = Object.keys(map).length;
}
} else { //there is duplacation
//update the max
if (Object.keys(map).length > max){
max = Object.keys(map).length;
}
counter = 0; //initilize the counter to count next substring
i = map[currentChar]; //start from the duplicated char
map = {}; // clean the map
}
}
return max;
}
here is my javascript and cpp implementations with great details: https://algorithm.pingzhang.io/String/longest_substring_without_repeating_characters.html
We want to find the longest substring without repeating characters. The first thing comes to my mind is that we need a hash table to store every character in a substring so that when a new character comes in, we can easily know whether this character is already in the substring or not. I call it as valueIdxHash. Then, a substring has a startIdx and endIdx. So we need a variable to keep track of the starting index of a substring and I call it as startIdx. Let's assume we are at index i and we already have a substring (startIdx, i - 1). Now, we want to check whether this substring can keep growing or not.
If the valueIdxHash contains str[i], it means it is a repeated character. But we still need to check whether this repeated character is in the substring (startIdx, i - 1). So we need to retrieve the index of str[i] that is appeared last time and then compare this index with startIdx.
If startIdx is larger, it means the last appeared str[i] is outside of the substring. Thus the subtring can keep growing.
If startIdx is smaller, it means the last appeared str[i] is within of the substring. Thus, the substring cannot grow any more. startIdx will be updated as valueIdxHash[str[i]] + 1 and the new substring (valueIdxHash[str[i]] + 1, i) has potential to keep growing.
If the valueIdxHash does not contain str[i], the substring can keep growing.
I modified my solution to "find the length of the longest substring without repeating characters".
public string LengthOfLongestSubstring(string s) {
var res = 0;
var dict = new Dictionary<char, int>();
var start = 0;
for(int i =0; i< s.Length; i++)
{
if(dict.ContainsKey(s[i]))
{
start = Math.Max(start, dict[s[i]] + 1); //update start index
dict[s[i]] = i;
}
else
{
dict.Add(s[i], i);
}
res = Math.Max(res, i - start + 1); //track max length
}
return s.Substring(start,res);
}
import java.util.HashMap;
import java.util.HashSet;
public class SubString {
public static String subString(String input) {
String longesTillNOw = "";
String longestOverAll = "";
HashMap<Character,Integer> chars = new HashMap<>();
char[] array=input.toCharArray();
int start=0;
for (int i = 0; i < array.length; i++) {
char charactor = array[i];
if (chars.containsKey(charactor) ) {
start=chars.get(charactor)+1;
i=start;
chars.clear();
longesTillNOw = "";
} else {
chars.put(charactor,i);
longesTillNOw = longesTillNOw + charactor;
if (longesTillNOw.length() > longestOverAll.length()) {
longestOverAll = longesTillNOw;
}
}
}
return longestOverAll;
}
public static void main(String[] args) {
String input = "stackoverflowabcdefghijklmn";
System.out.println(subString(input));
}
}
Here are two ways to approach this problem in JavaScript.
A Brute Force approach is to loop through the string twice, checking every substring against every other substring and finding the maximum length where the substring is unique. We'll need two functions: one to check if a substring is unique and a second function to perform our double loop.
// O(n) time
const allUnique = str => {
const set = [...new Set(str)];
return (set.length == str.length) ? true: false;
}
// O(n^3) time, O(k) size where k is the size of the set
const lengthOfLongestSubstring = str => {
let result = 0,
maxResult = 0;
for (let i=0; i<str.length-1; i++) {
for (let j=i+1; j<str.length; j++) {
if (allUnique(str.substring(i, j))) {
result = str.substring(i, j).length;
if (result > maxResult) {
maxResult = result;
}
}
}
return maxResult;
}
}
This has a time complexity of O(n^3) since we perform a double loop O(n^2) and then another loop on top of that O(n) for our unique function. The space is the size of our set which can be generalized to O(n) or more accurately O(k) where k is the size of the set.
A Greedy Approach is to loop through only once and keep track of the maximum unique substring length as we go. We can use either an array or a hash map, but I think the new .includes() array method is cool, so let's use that.
const lengthOfLongestSubstring = str => {
let result = [],
maxResult = 0;
for (let i=0; i<str.length; i++) {
if (!result.includes(str[i])) {
result.push(str[i]);
} else {
maxResult = i;
}
}
return maxResult;
}
This has a time complexity of O(n) and a space complexity of O(1).
This problem can be solved in O(n) time complexity.
Initialize three variables
Start (index pointing to the start of the non repeating substring, Initialize it as 0 ).
End (index pointing to the end of the non repeating substring, Initialize it as 0 )
Hasmap (Object containing the last visited index positions of characters. Ex : {'a':0, 'b':1} for string "ab")
Steps :
Iterate over the string and perform following actions.
If the current character is not present in hashmap (), add it as to
hashmap, character as key and its index as value.
If current character is present in hashmap, then
a) Check whether the start index is less than or equal to the value present in the hashmap against the character (last index of same character earlier visited),
b) it is less then assign start variables value as the hashmaps' value + 1 (last index of same character earlier visited + 1);
c) Update hashmap by overriding the hashmap's current character's value as current index of character.
d) Calculate the end-start as the longest substring value and update if it's greater than earlier longest non-repeating substring.
Following is the Javascript Solution for this problem.
var lengthOfLongestSubstring = function(s) {
let length = s.length;
let ans = 0;
let start = 0,
end = 0;
let hashMap = {};
for (var i = 0; i < length; i++) {
if (!hashMap.hasOwnProperty(s[i])) {
hashMap[s[i]] = i;
} else {
if (start <= hashMap[s[i]]) {
start = hashMap[s[i]] + 1;
}
hashMap[s[i]] = i;
}
end++;
ans = ans > (end - start) ? ans : (end - start);
}
return ans;
};
Question: Find the longest substring without repeating characters.
Example 1 :
import java.util.LinkedHashMap;
import java.util.Map;
public class example1 {
public static void main(String[] args) {
String a = "abcabcbb";
// output => 3
System.out.println( lengthOfLongestSubstring(a));
}
private static int lengthOfLongestSubstring(String a) {
if(a == null || a.length() == 0) {return 0 ;}
int res = 0 ;
Map<Character , Integer> map = new LinkedHashMap<>();
for (int i = 0; i < a.length(); i++) {
char ch = a.charAt(i);
if (!map.containsKey(ch)) {
//If ch is not present in map, adding ch into map along with its position
map.put(ch, i);
}else {
/*
If char ch is present in Map, reposition the cursor i to the position of ch and clear the Map.
*/
i = map.put(ch, i);// updation of index
map.clear();
}//else
res = Math.max(res, map.size());
}
return res;
}
}
if you want the longest string without the repeating characters as output then do this inside the for loop:
String res ="";// global
int len = 0 ;//global
if(len < map.size()) {
len = map.size();
res = map.keySet().toString();
}
System.out.println("len -> " + len);
System.out.println("res => " + res);
def max_substring(string):
last_substring = ''
max_substring = ''
for x in string:
k = find_index(x,last_substring)
last_substring = last_substring[(k+1):]+x
if len(last_substring) > len(max_substring):
max_substring = last_substring
return max_substring
def find_index(x, lst):
k = 0
while k <len(lst):
if lst[k] == x:
return k
k +=1
return -1
can we use something like this .
def longestpalindrome(str1):
arr1=list(str1)
s=set(arr1)
arr2=list(s)
return len(arr2)
str1='abadef'
a=longestpalindrome(str1)
print(a)
if only length of the substring is to be returned
Algorithm: 1) Initialise an empty dictionary dct to check if any character already exists in the string. 2) cnt - to keep the count of substring without repeating characters. 3)l and r are the two pointers initialised to first index of the string. 4)loop through each char of the string. 5) If the character not present in the dct add itand increse the cnt. 6)If its already present then check if cnt is greater then resStrLen.7)Remove the char from dct and shift the left pointer by 1 and decrease the count.8)Repeat 5,6,7 till l,r greater or equal to length of the input string. 9)Have one more check at the end to handle cases like input string with non-repeating characters.Here is the simple python program to Find longest substring without repeating characters
a="stackoverflow"
strLength = len(a)
dct={}
resStrLen=0
cnt=0
l=0
r=0
strb=l
stre=l
while(l<strLength and r<strLength):
if a[l] in dct:
if cnt>resStrLen:
resStrLen=cnt
strb=r
stre=l
dct.pop(a[r])
cnt=cnt-1
r+=1
else:
cnt+=1
dct[a[l]]=1
l+=1
if cnt>resStrLen:
resStrLen=cnt
strb=r
stre=l
print "Result String Length : "+str(resStrLen)
print "Result String : " + a[strb:stre]
The solution in C.
#include<stdio.h>
#include <string.h>
void longstr(char* a, int *start, int *last)
{
*start = *last = 0;
int visited[256];
for (int i = 0; i < 256; i++)
{
visited[i] = -1;
}
int max_len = 0;
int cur_len = 0;
int prev_index;
visited[a[0]] = 0;
for (int i = 1; i < strlen(a); i++)
{
prev_index = visited[a[i]];
if (prev_index == -1 || i - cur_len > prev_index)
{
cur_len++;
*last = i;
}
else
{
if (max_len < cur_len)
{
*start = *last - cur_len;
max_len = cur_len;
}
cur_len = i - prev_index;
}
visited[a[i]] = i;
}
if (max_len < cur_len)
{
*start = *last - cur_len;
max_len = cur_len;
}
}
int main()
{
char str[] = "ABDEFGABEF";
printf("The input string is %s \n", str);
int start, last;
longstr(str, &start, &last);
//printf("\n %d %d \n", start, last);
memmove(str, (str + start), last - start);
str[last] = '\0';
printf("the longest non-repeating character substring is %s", str);
return 0;
}
public int lengthOfLongestSubstring(String s) {
int startIndex = 0;
int maxLength = 0;
//since we have 256 ascii chars
int[] lst = new int[256];
Arrays.fill(lst,-1);
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
//to get ascii value of c
int ic = (int) c;
int value = lst[ic];
//this will say to move start index to next index of the repeating char
//we only do this if the repeating char index is greater than start index
if (value >= startIndex) {
maxLength = Math.max(maxLength, i - startIndex);
startIndex = value + 1;
}
lst[ic] = i;
}
//when we came to an end of string
return Math.max(maxLength,s.length()-startIndex);
}
This is the fastest and it is linear time and constant space
I am developing a J2ME application.
I want to split the following string at "<br>" & comma:
3,toothpaste,2<br>4,toothbrush,3
How can I do this?
private String[] split(String original,String separator) {
Vector nodes = new Vector();
// Parse nodes into vector
int index = original.indexOf(separator);
while(index >= 0) {
nodes.addElement( original.substring(0, index) );
original = original.substring(index+separator.length());
index = original.indexOf(separator);
}
// Get the last node
nodes.addElement( original );
// Create split string array
String[] result = new String[ nodes.size() ];
if( nodes.size() > 0 ) {
for(int loop = 0; loop < nodes.size(); loop++)
{
result[loop] = (String)nodes.elementAt(loop);
System.out.println(result[loop]);
}
}
return result;
}
The above method will let you split a string about the passed separator, much like J2EE's String.split(). So first split the string on the line break tag, and then do it at each offset of the returned array for the "," comma.
e.g.
String[] lines = this.split(myString,"<br>");
for(int i = 0; i < lines.length; i++)
{
String[] splitStr = this.split(lines[i],",");
System.out.println(splitStr[0] + " " + splitStr[1] + " " + splitStr[2]);
}
private String[] split(String original, String separator, int count)
{
String[] result;
int index = original.indexOf(separator);
if(index >= 0)
result = split(original.substring(index + separator.length()), separator, count + 1);
else
{
result = new String[count + 1];
index = original.length();
}
result[count] = original.substring(0, index);
return result;
}
String[] lines = this.split(myString,"<br>",0);