Related
How to read empty cell in excel file using POI?
When I upload Excel file, the empty cell is filled with previous cell.
like this:
this is the sample of excel file
this is the DB that the empty cell be filled with previous cell
POI VERSION
org.apache.poi
poi
4.1.2
org.apache.poi
poi-ooxml
4.1.2
code
#RequestMapping("/mng/mngSaupExcelUpload14.do")
public ModelAndView mngBizApplyExcelUpload14(MultipartHttpServletRequest request, HttpServletResponse response,
HttpSession session, INDEYearVO attach) throws IOException, ParseException {
Iterator<String> iterator = request.getFileNames();
int actResult = 0;
ModelAndView mav = new ModelAndView();
response.setContentType("text/html");
response.setCharacterEncoding("UTF-8");
PrintWriter out = response.getWriter();
MultipartFile mFile = null;
while (iterator.hasNext()) {
String uploadFileName = iterator.next();
mFile = request.getFile(uploadFileName);
// String originFileName = mFile.getOriginalFilename();
// String saveFileName = originFileName;
}
XSSFWorkbook workbook = new XSSFWorkbook(mFile.getInputStream());
for (int sheetNum = 0; sheetNum < workbook.getNumberOfSheets(); sheetNum++) {
XSSFSheet sheet = workbook.getSheetAt(sheetNum);
int rows = sheet.getPhysicalNumberOfRows();
int rowindex = 0;
int columnindex = 0;
if (sheetNum == 0) {
for (rowindex = 1; rowindex < rows; rowindex++) {
sheet = workbook.getSheetAt(0);
XSSFRow row = sheet.getRow(rowindex);
if (row != null) {
int cells = row.getPhysicalNumberOfCells();
for (columnindex = 0; columnindex <= cells; columnindex++) {
XSSFCell cell = row.getCell(columnindex);
String value = "";
if (cell == null) {
continue;
} else {
switch (cell.getCellType()) {
case FORMULA:
value = cell.getCellFormula();
break;
case NUMERIC:
//value = (int)cell.getNumericCellValue() + "";
double cellValue = cell.getNumericCellValue();
if (cellValue == Math.rint(cellValue)) {
value = String.valueOf((int) cellValue);
} else {
value = String.valueOf(cellValue);
}
break;
case STRING:
value = cell.getStringCellValue() + "";
break;
case BLANK:
value = null;
// value = cell.getBooleanCellValue() + "";
break;
case ERROR:
value = cell.getErrorCellValue() + "";
break;
default:
}
if (columnindex == 0) {
attach.setSn(Integer.parseInt(value));
}else if (columnindex == 1) {
attach.setYear(Integer.parseInt(value));
}else if (columnindex == 2) {
attach.setSigngu(value);
} else if (columnindex == 3) {
attach.setJan(Integer.parseInt(value));
} else if (columnindex == 4) {
attach.setFeb(Integer.parseInt(value));
} else if (columnindex == 5) {
attach.setMar(Integer.parseInt(value));
} else if (columnindex == 6) {
attach.setApr(Integer.parseInt(value));
} else if (columnindex == 7) {
attach.setMay(Integer.parseInt(value));
} else if (columnindex == 8) {
attach.setJne(Integer.parseInt(value));
} else if (columnindex == 9) {
attach.setJly(Integer.parseInt(value));
} else if (columnindex == 10) {
attach.setAug(Integer.parseInt(value));
} else if (columnindex == 11) {
attach.setSep(Integer.parseInt(value));
} else if (columnindex == 12) {
attach.setOct(Integer.parseInt(value));
} else if (columnindex == 13) {
attach.setNov(Integer.parseInt(value));
} else if (columnindex == 14) {
if ("false".equals(value) || "".equals(value) || "0".equals(value) || value == null ) {
attach.setDec(Integer.parseInt(""));
} else {
attach.setDec(Integer.parseInt(value));
}
}
}
}
actResult = attachFileService.fileInsertActForExcel14(attach);
}
}
}
}
if (actResult > 0) {
out.println("<script>");
out.println("alert('등록에 성공 하였습니다.');");
out.println("location.replace('" + request.getContextPath() + "attachFile.do');");
out.println("</script>");
out.flush();
out.close();
return null;
} else {
out.println("<script>");
out.println("alert('등록 중 오류가 발생했습니다.');");
out.println("location.replace('" + request.getContextPath() + "attachFile.do');");
out.println("</script>");
out.flush();
out.close();
return null;
}
}
I don't really know why empty cell is filled..
I don't think it's a problem with poi. Also check the db insert code.
And I'm developing a library that might be helpful, so I'd appreciate it if you could watch it.
https://github.com/scndry/jackson-dataformat-spreadsheet
i am running a test but i am receiving null after the test and i am receiving a AssertionFailedError
assertEquals("", this.myCustomString.remove(""));
this.myCustomString.setString(null);
assertEquals("", this.myCustomString.remove(""));
this.myCustomString.setString("my lucky numbers are 6, 8, and 19.");
assertEquals("my lucky numbes e 6, 8, nd 19.", this.myCustomString.remove("ra6"));
public String remove(String arg){//removes specified characters from the string
if (myString == null || myString == "") {
return this.myString;
}
if (myString != null) {
this.myString = myString.replaceAll(arg,"");
return myString;
}
return myString;
}
This should help
String result = "";
if (myString == null || myString == "") {
return "";
}
//places the String arg into a character array
char[] c = arg.toCharArray();
// then use a for loop to check if there is only letters in the string arg
// after we have checked to see if there is only letters when remove anything that is not letters
for(int i= 0; i< c.length; i++){
if (Character.isLetter(c[i])) {
result = result + c[i];
}
}
// when then use the result from above and place it into its own character array, we then use the characters
// from the array to be replaced by blank
for (char ch : result.toCharArray()) {
myString = myString.replace(String.valueOf(ch), "");
}
return myString;
}
I have a string which contains binary digits. How to separate it in to pairs of digits?
Suppose the string is:
let x = "11231245"
I want to add a separator such as ":" (i.e., a colon) after each 2 characters.
I would like the output to be:
"11:23:12:45"
How could I do this in Swift ?
Swift 5.2 • Xcode 11.4 or later
extension Collection {
func unfoldSubSequences(limitedTo maxLength: Int) -> UnfoldSequence<SubSequence,Index> {
sequence(state: startIndex) { start in
guard start < endIndex else { return nil }
let end = index(start, offsetBy: maxLength, limitedBy: endIndex) ?? endIndex
defer { start = end }
return self[start..<end]
}
}
func every(n: Int) -> UnfoldSequence<Element,Index> {
sequence(state: startIndex) { index in
guard index < endIndex else { return nil }
defer { let _ = formIndex(&index, offsetBy: n, limitedBy: endIndex) }
return self[index]
}
}
var pairs: [SubSequence] { .init(unfoldSubSequences(limitedTo: 2)) }
}
extension StringProtocol where Self: RangeReplaceableCollection {
mutating func insert<S: StringProtocol>(separator: S, every n: Int) {
for index in indices.every(n: n).dropFirst().reversed() {
insert(contentsOf: separator, at: index)
}
}
func inserting<S: StringProtocol>(separator: S, every n: Int) -> Self {
.init(unfoldSubSequences(limitedTo: n).joined(separator: separator))
}
}
Testing
let str = "112312451"
let final0 = str.unfoldSubSequences(limitedTo: 2).joined(separator: ":")
print(final0) // "11:23:12:45:1"
let final1 = str.pairs.joined(separator: ":")
print(final1) // "11:23:12:45:1"
let final2 = str.inserting(separator: ":", every: 2)
print(final2) // "11:23:12:45:1\n"
var str2 = "112312451"
str2.insert(separator: ":", every: 2)
print(str2) // "11:23:12:45:1\n"
var str3 = "112312451"
str3.insert(separator: ":", every: 3)
print(str3) // "112:312:451\n"
var str4 = "112312451"
str4.insert(separator: ":", every: 4)
print(str4) // "1123:1245:1\n"
I'll go for this compact solution (in Swift 4) :
let s = "11231245"
let r = String(s.enumerated().map { $0 > 0 && $0 % 2 == 0 ? [":", $1] : [$1]}.joined())
You can make an extension and parameterize the stride and the separator so that you can use it for every value you want (In my case, I use it to dump 32-bit space-operated hexadecimal data):
extension String {
func separate(every stride: Int = 4, with separator: Character = " ") -> String {
return String(enumerated().map { $0 > 0 && $0 % stride == 0 ? [separator, $1] : [$1]}.joined())
}
}
In your case this gives the following results:
let x = "11231245"
print (x.separate(every:2, with: ":")
$ 11:23:12:45
Swift 5.3
/// Adds a separator at every N characters
/// - Parameters:
/// - separator: the String value to be inserted, to separate the groups. Default is " " - one space.
/// - stride: the number of characters in the group, before a separator is inserted. Default is 4.
/// - Returns: Returns a String which includes a `separator` String at every `stride` number of characters.
func separated(by separator: String = " ", stride: Int = 4) -> String {
return enumerated().map { $0.isMultiple(of: stride) && ($0 != 0) ? "\(separator)\($1)" : String($1) }.joined()
}
Short and simple, add a let or two if you want
extension String {
func separate(every: Int, with separator: String) -> String {
return String(stride(from: 0, to: Array(self).count, by: every).map {
Array(Array(self)[$0..<min($0 + every, Array(self).count)])
}.joined(separator: separator))
}
}
let a = "separatemepleaseandthankyou".separate(every: 4, with: " ")
a is
sepa rate mepl ease andt hank you
Its my code in swift 4
let x = "11231245"
var newText = String()
for (index, character) in x.enumerated() {
if index != 0 && index % 2 == 0 {
newText.append(":")
}
newText.append(String(character))
}
print(newText)
Outputs 11:23:12:45
My attempt at that code would be:
func insert(seperator: String, afterEveryXChars: Int, intoString: String) -> String {
var output = ""
intoString.characters.enumerate().forEach { index, c in
if index % afterEveryXChars == 0 && index > 0 {
output += seperator
}
output.append(c)
}
return output
}
insert(":", afterEveryXChars: 2, intoString: "11231245")
Which outputs
11:23:12:45
let y = String(
x.characters.enumerate().map() {
$0.index % 2 == 0 ? [$0.element] : [$0.element, ":"]
}.flatten()
)
A simple One line of code for inserting separater ( Swift 4.2 ):-
let testString = "123456789"
let ansTest = testString.enumerated().compactMap({ ($0 > 0) && ($0 % 2 == 0) ? ":\($1)" : "\($1)" }).joined() ?? ""
print(ansTest) // 12:34:56:78:9
Swift 4.2.1 - Xcode 10.1
extension String {
func insertSeparator(_ separatorString: String, atEvery n: Int) -> String {
guard 0 < n else { return self }
return self.enumerated().map({String($0.element) + (($0.offset != self.count - 1 && $0.offset % n == n - 1) ? "\(separatorString)" : "")}).joined()
}
mutating func insertedSeparator(_ separatorString: String, atEvery n: Int) {
self = insertSeparator(separatorString, atEvery: n)
}
}
Usage
let testString = "11231245"
let test1 = testString.insertSeparator(":", atEvery: 2)
print(test1) // 11:23:12:45
var test2 = testString
test2.insertedSeparator(",", atEvery: 3)
print(test2) // 112,312,45
I'm little late here, but i like to use regex like in this:
extension String {
func separating(every: Int, separator: String) -> String {
let regex = #"(.{\#(every)})(?=.)"#
return self.replacingOccurrences(of: regex, with: "$1\(separator)", options: [.regularExpression])
}
}
"111222333".separating(every: 3, separator: " ")
the output:
"111 222 333"
extension String{
func separate(every: Int) -> [String] {
return stride(from: 0, to: count, by: every).map {
let ix0 = index(startIndex, offsetBy: $0);
let ix1 = index(after:ix0);
if ix1 < endIndex {
return String(self[ix0...ix1]);
}else{
return String(self[ix0..<endIndex]);
}
}
}
/// or O(1) implementation (without count)
func separate(every: Int) -> [String] {
var parts:[String] = [];
var ix1 = startIndex;
while ix1 < endIndex {
let ix0 = ix1;
var n = 0;
while ix1 < endIndex && n < every {
ix1 = index(after: ix1);
n += 1;
}
parts.append(String(self[ix0..<ix1]));
}
return parts;
}
"asdf234sdf".separate(every: 2).joined(separator: ":");
A simple String extension that doesn't require the original string to be a multiple of the step size (increment):
extension String {
func inserted(_ newElement: Character,atEach increment:Int)->String {
var newStr = self
for indx in stride(from: increment, to: newStr.count, by: increment).reversed() {
let index = String.Index(encodedOffset: indx)
newStr.insert(newElement, at: index)
}
return newStr
}
}
Write a method to test if a string meets the preconditions to become a palindrome.
Eg:
Input | Output
mmo | True
yakak | True
travel | False
I'm thinking of this approach:
Make a suffix tree for all permutation of T such that T$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();
}