Related Articles
Maximum length substring having all same characters after k changes
• Difficulty Level : Hard
• Last Updated : 29 Apr, 2021

We have a string of length n, which consist only UPPER and LOWER CASE characters and we have a number k (always less than n and greater than 0). We can make at most k changes in our string such that we can get a sub-string of maximum length which have all same characters.
Examples:

```n = length of string
k = changes you can make
Input : n = 5 k = 2
str = ABABA
Output : maximum length = 5
which will be (AAAAA)

Input : n = 6 k = 4
str = HHHHHH
Output : maximum length=6
which will be(HHHHHH) ```

We check for each character of English alphabet (both upper and lower cases one by one). We are basically looking for maximum length of sub-string that can be formed by each character and whichever character will form the sub-string of maximum length then that length will be our answer.

1. We check for maximum length of sub-string that can be formed by every character in a set of 52 characters (From ‘A’ to ‘Z’ and from ‘a’ to ‘z’).
2. For doing this we traverse whole string and whenever we find a different character, we increase the count.
3. If count becomes greater than k (at right index), we again start from 0th index and if we found different character we will decrease the count.
4. When count will be equal to k (at left index) then at that point the length will be rightIndex-leftIndex+1.
5. We repeat this process until we reach at the end of string and at that point we will return the maximum length.
6. We do this for all characters and finally return the maximum length.

## C++

 `// C++ program to find maximum length equal``// character string with k changes``#include ``using` `namespace` `std;` `// function to find the maximum length of``// substring having character ch``int` `findLen(string& A, ``int` `n, ``int` `k, ``char` `ch)``{``    ``int` `maxlen = 1;``    ``int` `cnt = 0;``    ``int` `l = 0, r = 0;``    ` `    ``// traverse the whole string``    ``while` `(r < n) {``    ` `        ``/* if character is not same as ch``           ``increase count */``        ``if` `(A[r] != ch)``            ``++cnt;` `        ``/* While  count > k  traverse the string``           ``again until count becomes less than k``           ``and decrease the count when characters``           ``are not same */``        ``while` `(cnt > k) {``            ``if` `(A[l] != ch)``                ``--cnt;``            ``++l;``        ``}` `        ``/* length of substring will be rightIndex -``           ``leftIndex + 1. Compare this with the maximum``           ``length and return maximum length */``        ``maxlen = max(maxlen, r - l + 1);``        ``++r;``    ``}``    ``return` `maxlen;``}` `// function which returns maximum length of substring``int` `answer(string& A, ``int` `n, ``int` `k)``{``    ``int` `maxlen = 1;``    ``for` `(``int` `i = 0; i < 26; ++i) {``        ``maxlen = max(maxlen, findLen(A, n, k, i+``'A'``));``        ``maxlen = max(maxlen, findLen(A, n, k, i+``'a'``));``    ``}``    ``return` `maxlen;``}` `// Driver code``int` `main()``{``    ``int` `n = 5, k = 2;``    ``string A = ``"ABABA"``;``    ``cout << ``"Maximum length = "` `<< answer(A, n, k) << endl;` `    ``n = 6, k = 4;``    ``string B = ``"HHHHHH"``;``    ``cout << ``"Maximum length = "` `<< answer(B, n, k) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find maximum length equal``// character string with k changes` `public` `class` `GFG``{``    ``// method to find the maximum length of``    ``// substring having character ch``    ``static` `int` `findLen(String A, ``int` `n, ``int` `k, ``char` `ch)``    ``{``        ``int` `maxlen = ``1``;``        ``int` `cnt = ``0``;``        ``int` `l = ``0``, r = ``0``;``         ` `        ``// traverse the whole string``        ``while` `(r < n) {``         ` `            ``/* if character is not same as ch``               ``increase count */``            ``if` `(A.charAt(r) != ch)``                ``++cnt;``     ` `            ``/* While  count > k  traverse the string``               ``again until count becomes less than k``               ``and decrease the count when characters``               ``are not same */``            ``while` `(cnt > k) {``                ``if` `(A.charAt(l) != ch)``                    ``--cnt;``                ``++l;``            ``}``     ` `            ``/* length of substring will be rightIndex -``               ``leftIndex + 1. Compare this with the maximum``               ``length and return maximum length */``            ``maxlen = Math.max(maxlen, r - l + ``1``);``            ``++r;``        ``}``        ``return` `maxlen;``    ``}``     ` `    ``// method which returns maximum length of substring``    ``static` `int` `answer(String A, ``int` `n, ``int` `k)``    ``{``        ``int` `maxlen = ``1``;``        ``for` `(``int` `i = ``0``; i < ``26``; ++i) {``            ``maxlen = Math.max(maxlen, findLen(A, n, k, (``char``) (i+``'A'``)));``            ``maxlen = Math.max(maxlen, findLen(A, n, k, (``char``) (i+``'a'``)));``        ``}``        ``return` `maxlen;``    ``}``    ` `    ``// Driver Method``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``5``, k = ``2``;``        ``String A = ``"ABABA"``;``        ``System.out.println(``"Maximum length = "` `+ answer(A, n, k));``     ` `        ``n = ``6``; k = ``4``;``        ``String B = ``"HHHHHH"``;``        ``System.out.println(``"Maximum length = "` `+ answer(B, n, k));``    ``}``}`

## Python3

 `# Python3 program to find maximum length``# equal character string with k changes` `# function to find the maximum length of``# substring having character ch``def` `findLen(A, n, k, ch):``    ``maxlen ``=` `1``    ``cnt ``=` `0``    ``l ``=` `0``    ``r ``=` `0` `    ``# traverse the whole string``    ``while` `r < n:` `        ``# if character is not same as ch``        ``# increase count``        ``if` `A[r] !``=` `ch:``            ``cnt ``+``=` `1` `        ``# While count > k traverse the string``        ``# again until count becomes less than k``        ``# and decrease the count when characters``        ``# are not same``        ``while` `cnt > k:``            ``if` `A[l] !``=` `ch:``                ``cnt ``-``=` `1``            ``l ``+``=` `1` `        ``# length of substring will be rightIndex -``        ``# leftIndex + 1. Compare this with the``        ``# maximum length and return maximum length``        ``maxlen ``=` `max``(maxlen, r ``-` `l ``+` `1``)``        ``r ``+``=` `1` `    ``return` `maxlen` `# function which returns``# maximum length of substring``def` `answer(A, n, k):``    ``maxlen ``=` `1``    ``for` `i ``in` `range``(``26``):``        ``maxlen ``=` `max``(maxlen, findLen(A, n, k,``                             ``chr``(i ``+` `ord``(``'A'``))))``        ``maxlen ``=` `max``(maxlen, findLen(A, n, k,``                             ``chr``(i ``+` `ord``(``'a'``))))` `    ``return` `maxlen` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``n ``=` `5``    ``k ``=` `2``    ``A ``=` `"ABABA"``    ``print``(``"Maximum length ="``,``             ``answer(A, n, k))` `    ``n ``=` `6``    ``k ``=` `4``    ``B ``=` `"HHHHHH"``    ``print``(``"Maximum length ="``,``             ``answer(B, n, k))` `# This code is contributed by``# sanjeev2552`

## C#

 `// C# program to find maximum length equal``// character string with k changes``using` `System;` `class` `GFG``{``    ` `    ``// method to find the maximum length of``    ``// substring having character ch``    ``public` `static` `int` `findLen(``string` `A, ``int` `n,``                               ``int` `k, ``char` `ch)``    ``{``        ``int` `maxlen = 1;``        ``int` `cnt = 0;``        ``int` `l = 0, r = 0;` `        ``// traverse the whole string``        ``while` `(r < n)``        ``{` `            ``// if character is``            ``// not same as ch``            ``// increase count``            ``if` `(A[r] != ch)``            ``{``                ``++cnt;``            ``}` `            ``// While count > k traverse``            ``// the string again until``            ``// count becomes less than``            ``// k and decrease the``            ``// count when characters``            ``// are not same``            ``while` `(cnt > k)``            ``{``                ``if` `(A[l] != ch)``                ``{``                    ``--cnt;``                ``}``                ``++l;``            ``}` `            ``// length of substring``            ``// will be rightIndex -``            ``// leftIndex + 1.``            ``// Compare this with the maximum``            ``// length and return maximum length``            ``maxlen = Math.Max(maxlen, r - l + 1);``            ``++r;``        ``}``        ``return` `maxlen;``    ``}` `    ``// method which returns``    ``// maximum length of substring``    ``public` `static` `int` `answer(``string` `A, ``int` `n, ``int` `k)``    ``{``        ``int` `maxlen = 1;``        ``for` `(``int` `i = 0; i < 26; ++i)``        ``{``            ``maxlen = Math.Max(maxlen,``              ``findLen(A, n, k, (``char``)(i + ``'A'``)));``              ` `            ``maxlen = Math.Max(maxlen,``              ``findLen(A, n, k, (``char``)(i + ``'a'``)));``        ``}``        ``return` `maxlen;``    ``}` `// Driver Method``public` `static` `void` `Main(``string``[] args)``{``    ``int` `n = 5, k = 2;``    ``string` `A = ``"ABABA"``;``    ``Console.WriteLine(``"Maximum length = "``                      ``+ answer(A, n, k));` `    ``n = 6;``    ``k = 4;``    ``string` `B = ``"HHHHHH"``;``    ``Console.WriteLine(``"Maximum length = "``                      ``+ answer(B, n, k));``}``}` `// This code is contributed by Shrikant13`

## PHP

 ` k traverse the string``        ``again until count becomes less than k``        ``and decrease the count when characters``        ``are not same */``        ``while` `(``\$cnt` `> ``\$k``)``        ``{``            ``if` `(``\$A``[``\$l``] != ``\$ch``)``                ``--``\$cnt``;``            ``++``\$l``;``        ``}` `        ``/* length of substring will be rightIndex -``        ``leftIndex + 1. Compare this with the maximum``        ``length and return maximum length */``        ``\$maxlen` `= max(``\$maxlen``, ``\$r` `- ``\$l` `+ 1);``        ``++``\$r``;``    ``}``    ``return` `\$maxlen``;``}` `// function which returns maximum``// length of substring``function` `answer(``\$A``, ``\$n``, ``\$k``)``{``    ``\$maxlen` `= 1;``    ``for` `(``\$i` `= 0; ``\$i` `< 26; ++``\$i``)``    ``{``        ``\$maxlen` `= max(``\$maxlen``,``                      ``findLen(``\$A``, ``\$n``, ``\$k``, ``\$i` `+ ``'A'``));``        ``\$maxlen` `= max(``\$maxlen``,``                      ``findLen(``\$A``, ``\$n``, ``\$k``, ``\$i` `+ ``'a'``));``    ``}``    ``return` `\$maxlen``;``}` `// Driver code``\$n` `= 5; ``\$k` `= 2; ``\$A` `= ``"ABABA"``;``echo` `"Maximum length = "` `. answer(``\$A``, ``\$n``, ``\$k``) . ``"\n"``;` `\$n` `= 6; ``\$k` `= 4; ``\$B` `= ``"HHHHHH"``;``echo` `"Maximum length = "` `. answer(``\$B``, ``\$n``, ``\$k``) . ``"\n"``;`` ` `// This code is contributed by ita_c``?>`

## Javascript

 ``

Output:

```Maximum length = 5
Maximum length = 6```

This article is contributed by Niteesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.