I am trying to increment an integer by 1 every-time the letter in a string isn't equal to the specific character (e.g. a),
For example a string of dfla, would count 3. Because the loop will break at 'a'.
How would I able to do this?
private int countToFirstCharacter(String name, String character) {
int count = 0;
for (int i = 0; i < materialName.length(); i++) {
//Increment count if it isn't equal to a, then break loop.
//Stuck here.
}
return count;
}
No Need to a loop:
string s = "dfla";
int x = s.IndexOf("a"); // It will show you 3
Another solution can be:
private static int countToFirstCharacter(String name, char character)
{
int count = 0;
for (int i = 0; i < name.Length; i++)
{
if (name[i].Equals(character))
break;
else
count++;
}
return count;
}
Replace all occurences of the character with an empty character, and then check the length of the string
http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#replace(char, char)
http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#length()
Related
Good day,
I have to write a program that prints all permutations of the String "abcd" with the following restrictions:
the strings must always have 4 characters;
one character can be used more then once in a string;
"b" must always be followed by "a";
a string cannot have both "d" and "a"; and
the program must also print at the end the number of strings printed.
here is the working code i came up with so far.
Now i have to add code for :
"b" must always be followed by "a";
a string cannot have both "d" and "a";
Can someone help me with this ?
public class Combination2 {
public static void main(String[] args) {
String s = "abcd";
int count = 0;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
for (int k = 0; k < 4; k++) {
for (int d = 0; d < 4; d++) {
String res = "" + s.charAt(i) + s.charAt(j) + s.charAt(k) + s.charAt(d);
count++;
System.out.println("" + res);
}
}
}
}
System.out.println("There are " + count + " combinations");
}
}
The simplest place to start would probably be to generate all permutations of the String "abcd" while ignoring the restrictions. Then once you have that figured out try and work on the restrictions.
We are given a string and we have to find out the minimum number of swaps to convert it into a palindrome.
Ex-
Given string: ntiin
Palindrome: nitin
Minimum number of swaps: 1
If it is not possible to convert it into a palindrome, return -1.
I am unable to think of any approach except brute force. We can check on the first and last characters, if they are equal, we check for the smaller substring, and then apply brute force on it. But this will be of a very high complexity, and I feel this question can be solved in another way. Maybe dynamic programming. How to approach it?
First you could check if the string can be converted to a palindrome.
Just have an array of letters (26 chars if all letters are latin lowercase), and count the number of each letter in the input string.
If string length is even, all letters counts should be even.
If string length is odd, all letters counts should be even except one.
This first pass in O(n) will already treat all -1 cases.
If the string length is odd, start by moving the element with odd count to the middle.
Then you can apply following procedure:
Build a weighted graph with the following logic for an input string S of length N:
For every element from index 0 to N/2-1:
- If symmetric element S[N-index-1] is same continue
- If different, create edge between the 2 characters (alphabetic order), or increment weight of an existing one
The idea is that when a weight is even you can do a 'good swap' by forming two pairs in one swap.
When weight is odd, you cannot place two pairs in one swap, your swaps need to form a cycle
1. For instance "a b a b"
One edge between a,b of weight 2:
a - b (2)
Return 1
2. For instance: "a b c b a c"
a - c (1)
b - a (1)
c - b (1)
See the cycle: a - b, b - c, c - a
After a swap of a,c you get:
a - a (1)
b - c (1)
c - b (1)
Which is after ignoring first one and merge 2 & 3:
c - b (2)
Which is even, you get to the result in one swap
Return 2
3. For instance: "a b c a b c"
a - c (2)
One swap and you are good
So basically after your graph is generated, add to the result the weight/2 (integer division e.g. 7/3 = 3) of each edge
Plus find the cycles and add to the result length-1 of each cycle
there is the same question as asked!
https://www.codechef.com/problems/ENCD12
I got ac for this solution
https://www.ideone.com/8wF9DT
//minimum adjacent swaps to make a string to its palindrome
#include<bits/stdc++.h>
using namespace std;
bool check(string s)
{
int n=s.length();
map<char,int> m;
for(auto i:s)
{
m[i]++;
}
int cnt=0;
for(auto i=m.begin();i!=m.end();i++)
{
if(i->second%2)
{
cnt++;
}
}
if(n%2&&cnt==1){return true;}
if(!(n%2)&&cnt==0){return true;}
return false;
}
int main()
{
string a;
while(cin>>a)
{
if(a[0]=='0')
{
break;
}
string s;s=a;
int n=s.length();
//first check if
int cnt=0;
bool ini=false;
if(n%2){ini=true;}
if(check(s))
{
for(int i=0;i<n/2;i++)
{
bool fl=false;
int j=0;
for(j=n-1-i;j>i;j--)
{
if(s[j]==s[i])
{
fl=true;
for(int k=j;k<n-1-i;k++)
{
swap(s[k],s[k+1]);
cnt++;
// cout<<cnt<<endl<<flush;
}
// cout<<" "<<i<<" "<<cnt<<endl<<flush;
break;
}
}
if(!fl&&ini)
{
for(int k=i;k<n/2;k++)
{
swap(s[k],s[k+1]);
cnt++;
}
// cout<<cnt<<" "<<i<<" "<<endl<<flush;
}
}
cout<<cnt<<endl;
}
else{
cout<<"Impossible"<<endl;
}
}
}
Hope it helps!
Technique behind my code is Greedy
first check if palindrome string can exist for the the string and if it can
there would be two cases one is when the string length would be odd then only count of one char has be odd
and if even then no count should be odd
then
from index 0 to n/2-1 do the following
fix this character and search for this char from n-i-1 to i+1
if found then swap from that position (lets say j) to its new position n-i-1
if the string length is odd then every time you encounter a char with no other occurence shift it to n/2th position..
My solution revolves around the palindrome property that first element and last element should match and if their adjacent elements also do not match then its not a palindrome. Keep comparing and swapping till both reach the same element or adjacent elements.
Written solution in java as below:
public static void main(String args[]){
String input = "natinat";
char[] arr = input.toCharArray();
int swap = 0;
int i = 0;
int j = arr.length-1;
char temp;
while(i<j){
if(arr[i] != arr[j]){
if(arr[i+1] == arr[j]){
//swap i and i+1 and increment i, decrement j, swap++
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
i++;j--;
swap++;
} else if(arr[i] == arr[j-1]){
//swap j and j-1 and increment i, decrement j, swap++
temp = arr[j];
arr[j] = arr[j-1];
arr[j-1] = temp;
i++;j--;
swap++;
} else if(arr[i+1] == arr[j-1] && i+1 != j-1){
//swap i and i+1, swap j and j-1 and increment i, decrement j, swap+2
temp = arr[j];
arr[j] = arr[j-1];
arr[j-1] = temp;
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
i++;j--;
swap = swap+2;
}else{
swap = -1;break;
}
} else{
//increment i, decrement j
i++;j--;
}
}
System.out.println("No Of Swaps: "+swap);
}
My solution in java for any type of string i.e Binary String, Numbers
public int countSwapInPalindrome(String s){
int length = s.length();
if (length == 0 || length == 1) return -1;
char[] str = s.toCharArray();
int start = 0, end = length - 1;
int count = 0;
while (start < end) {
if (str[start] != str[end]){
boolean isSwapped = false;
for (int i = start + 1; i < end; i++){
if (str[start] == str[i]){
char temp = str[i];
str[i] = str[end];
str[end] = temp;
count++;
isSwapped = true;
break;
}else if (str[end] == str[i]){
char temp = str[i];
str[i] = str[start];
str[start] = temp;
count++;
isSwapped = true;
break;
}
}
if (!isSwapped) return -1;
}
start++;
end--;
}
return (s.equals(String.valueOf(str))) ? -1 : count;
}
I hope it helps
string s;
cin>>s;
int n = s.size(),odd=0;
vi cnt(26,0);
unordered_map<int,set<int>>mp;
for(int i=0;i<n;i++){
cnt[s[i]-'a']++;
mp[s[i]-'a'].insert(i);
}
for(int i=0;i<26;i++){
if(cnt[i]&1) odd++;
}
int ans=0;
if((n&1 && odd == 1)|| ((n&1) == 0 && odd == 0)){
int left=0,right=n-1;
while(left < right){
if(s[left] == s[right]){
cnt[left]--;
cnt[right]--;
mp[s[left]-'a'].erase(left);
mp[s[right]-'a'].erase(right);
left++;
right--;
}else{
if(cnt[left]&1 == 0){
ans++;
int index = *mp[s[left]-'a'].rbegin();
mp[s[left]-'a'].erase(index);
mp[s[right]-'a'].erase(right);
mp[s[right]-'a'].insert(index);
swap(s[right],s[index]);
cnt[left]-=2;
}else{
ans++;
int index = *mp[s[right]-'a'].begin();
mp[s[right]-'a'].erase(index);
mp[s[left]-'a'].erase(left);
mp[s[left]-'a'].insert(index);
swap(s[left],s[index]);
cnt[right]-=2;
}
left++;
right--;
}
}
}else{
// cout<<odd<<" ";
cout<<"-1\n";
return;
}
cout<<ans<<"\n";
java program to return the characters which occurs more than 3 times in a string using only string functions
use only charAt and length functions to get solution
import java.util.*;
public class StringExample {
public static void main(String args[]) {
String str = "sofiiiffjjjh";
Map < Character, Integer > charFreq = new HashMap < Character, Integer > ();
if (str != null) {
for (Character c: str.toCharArray()) {
Integer count = charFreq.get(c);
int newCount = (count == null ? 1 : count + 1);
charFreq.put(c, newCount);
if (newCount >= 3) {
System.out.println(c);
}
}
}
}
}
This solution first sorts the character in the input string. It loops through the characters in the string, comparing the current character with the previous character
1. If they are the same, it increments the count
2. If not, it checks the number of times previousChar has occurred and prints it if it is more than 3 and continues.
char [] strAsArray = s.toCharArray();
Arrays.sort(strAsArray);
String sorted = new String(strAsArray);
System.out.println(sorted);
if (sorted.length() > 0) {
char previousChar = s.charAt(0);
int count = 1;
for (int i = 1; i < sorted.length(); i++) {
if (sorted.charAt(i) == previousChar) {
count++;
} else { // encountered new character
if (count >= 3) {
//if the previous character has appeared more than 3 times
System.out.println(sorted.charAt(i - 1));
}
count = 1;
previousChar = sorted.charAt(i);
}
}
//To handle case where the String has only one character any number of times
if (count >= 3) {
System.out.println(sorted.charAt(sorted.length() - 1));
}
}
Shil has a string S , consisting of N lowercase English letters. In one operation, he can delete any pair of adjacent letters with same value. For example, string "aabcc" would become either "aab" or "bcc" after operation.
Shil wants to reduce S as much as possible. To do this, he will repeat the above operation as many times as it can be performed. Help Shil out by finding and printing 's non-reducible form!
If the final string is empty, print Empty String; otherwise, print the final non-reducible string.
Sample Input 0
aaabccddd
Sample Output 0
abd
Sample Input 1
baab
Sample Output 1
Empty String
Sample Input 2
aa
Sample Output 2
Empty String
Explanation
Sample Case 0: Shil can perform the following sequence of operations to get the final string:
Thus, we print .
Sample Case 1: Shil can perform the following sequence of operations to get the final string: aaabccddd -> abccddd
abccddd -> abddd
abddd -> abd
Thus we print abd
Sample case 1: baab -> bb
bb -> Empty String.
in my code in the while loop when i assign s[i] to str[i].the value of s[i] is not getting assigned to str[i].the str[i] has a garbage value.
my code :
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
string s;
cin>>s;
int len = s.length();
int len1 = 0;
string str;
for(int i = 0;i < len-1;i++){
if(s[i]!= '*'){
for(int j=i+1;j < len;j++){
if(s[j] != '*'){
if(s[i] == s[j]){
s[i] = s[j] = '*';
}
}
}
}
}
int i = 0;
while(i<len){
if(s[i] != '*'){
str[len1] = s[i];
len1++;
}
i++;
}
if(len1 != 0){
cout<<str;
}
else{
cout<<"Empty String";
}
return 0;
}
#include <iostream>
using namespace std;
int main(){
string s,tempS;
bool condition = false;
cin >> s;
tempS = s;
while(condition==false){
for(int i=1; i<s.size(); i++){
if(s[i]==s[i-1]){
s.erase(s.begin()+i-1, s.begin()+i+1);
}
}
if(tempS == s){
condition = true;
} else{
tempS = s;
}
}
if(s.size()==0){
cout << "Empty String" ;
} else{
cout << s;
}
return 0;
}
the first while loop keeps on modifying the string until and unless it becomes equal to temp (which is equal to the string pre-modification)
It is comparing the adjacent elements if they are equal or not and then deleting them both.
As soon as string becomes equal to temp after modifications, string has reached it's most reduced state !
I have a string:
string mystring="part1, part2, part3, part4, part5";
How can I just return the first 3 elements without splitting them up first?
so like this:
string newstring="part1, part2, part3";
You could get the first three using:
RegEx r = new RegEx(#"(\S+, \S+, \S+), \S+");
I'm sure there is a better way to write the regex, but I think that would do it for basic inputs.
Try to find Index of 3rd Comma, and then get the substring.
Example
void Main()
{
string mystring="part1, part2, part3, part4, part5";
int thirdCommaIndex = IndexOf(mystring, ',', 3);
var substring = mystring.Substring(0,thirdCommaIndex-1);
Console.WriteLine(substring);
}
int IndexOf(string s, char c, int n)
{
int index = 0;
int count = 0;
foreach(char ch in s)
{
index++;
if (ch == c)
count++;
if (count == n )
break;
}
if (count == 0) index = -1;
return index;
}
This will parse the string trying to find the third comma and throwing it and everything after it away.
string mystring = "part1, part2, part3, part4, part5";
UInt16 CommasFound = 0;
UInt16 Location = 0;
for (Location = 0; (CommasFound < 3) &&
(Location < mystring.Count()); Location++)
if (mystring[Location].Equals(','))
CommasFound++;
if (CommasFound == 3)
{
string newstring = mystring.Substring(0, Location-1);
}
else { // Handle the case where there isn't a third item
}