# Longest equal substring with cost less than K

Given two string X and Y of the same length which consists of lowercase letters and also an integer K. The task is to find the maximum length up to which X can be changed to Y within the given cost K.
The cost of changing a character is given by the absolute difference between the ASCII value of the characters. That is, to change a character at index i, cost = |x[i] – Y[i]|

Examples:

Input: X = abcd, Y = bcde, K = 3
Output: 3
Explanation: A maximum of 3 characters can be changed because the cost to change each article is 1.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to maintain a prefix array of length N to store the absolute sum of the strings. That is, the cost to change the string X to Y. The following steps can be followed to compute the result:

• Maintain two pointers say i and j.
• In a while loop check if the difference between ith index and jth index of prefix array is greater than given cost or not.
• If the difference is greater than given cost then increase the j pointer to compensate for the cost else increase the i pointer.

Below is the implementation of the above approach:

## CPP

 `// C++ program to find the ` `// maximum length of equal substring ` `// within a given cost ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum length ` `int` `solve(string X, string Y, ``int` `N, ``int` `K) ` `{ ` ` `  `    ``int` `count[N + 1] = { 0 }; ` `    ``int` `sol = 0; ` `    ``count = 0; ` ` `  `    ``// Fill the prefix array with ` `    ``// the difference of letters ` `    ``for` `(``int` `i = 1; i <= N; i++) { ` ` `  `        ``count[i] = count[i - 1] + ``abs``(X[i - 1] - Y[i - 1]); ` `    ``} ` ` `  `    ``int` `j = 0; ` ` `  `    ``for` `(``int` `i = 1; i <= N; i++) { ` `        ``while` `((count[i] - count[j]) > K) { ` ` `  `            ``j++; ` `        ``} ` ` `  `        ``// Update the maximum length ` `        ``sol = max(sol, i - j); ` `    ``} ` ` `  `    ``return` `sol; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `N = 4; ` ` `  `    ``string X = ``"abcd"``, Y = ``"bcde"``; ` ` `  `    ``int` `K = 3; ` ` `  `    ``cout << solve(X, Y, N, K) << ``"\n"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the ` `// maximum length of equal subString ` `// within a given cost ` `class` `GFG ` `{ ` ` `  `// Function to find the maximum length ` `static` `int` `solve(String X, String Y, ` `                ``int` `N, ``int` `K) ` `{ ` ` `  `    ``int` `[]count = ``new` `int``[N + ``1``]; ` `    ``int` `sol = ``0``; ` `    ``count[``0``] = ``0``; ` ` `  `    ``// Fill the prefix array with ` `    ``// the difference of letters ` `    ``for` `(``int` `i = ``1``; i <= N; i++)  ` `    ``{ ` ` `  `        ``count[i] = count[i - ``1``] +  ` `                ``Math.abs(X.charAt(i - ``1``) -  ` `                ``Y.charAt(i - ``1``)); ` `    ``} ` ` `  `    ``int` `j = ``0``; ` ` `  `    ``for` `(``int` `i = ``1``; i <= N; i++) ` `    ``{ ` `        ``while` `((count[i] - count[j]) > K) ` `        ``{ ` `            ``j++; ` `        ``} ` ` `  `        ``// Update the maximum length ` `        ``sol = Math.max(sol, i - j); ` `    ``} ` ` `  `    ``return` `sol; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` `  `    ``int` `N = ``4``; ` `    ``String X = ``"abcd"``, Y = ``"bcde"``; ` `    ``int` `K = ``3``; ` ` `  `    ``System.out.print(solve(X, Y, N, K) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 program to find the ` `# maximum length of equal subString ` `# within a given cost ` ` `  `# Function to find the maximum length ` `def` `solve(X, Y, N, K): ` `    ``count ``=` `[``0``] ``*` `(N ``+` `1``); ` `    ``sol ``=` `0``; ` `    ``count[``0``] ``=` `0``; ` ` `  `    ``# Fill the prefix array with ` `    ``# the difference of letters ` `    ``for` `i ``in` `range``(``1``, N ``+` `1``): ` `        ``count[i] ``=` `(count[i ``-` `1``] ``+` `                    ``abs``(``ord``(X[i ``-` `1``]) ``-`  `                    ``ord``(Y[i ``-` `1``]))); ` ` `  `    ``j ``=` `0``; ` ` `  `    ``for` `i ``in` `range``(``1``, N ``+` `1``): ` `        ``while` `((count[i] ``-` `count[j]) > K): ` `            ``j ``+``=` `1``; ` ` `  `        ``# Update the maximum length ` `        ``sol ``=` `max``(sol, i ``-` `j); ` ` `  `    ``return` `sol; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``N ``=` `4``; ` `    ``X ``=` `"abcd"``; ` `    ``Y ``=` `"bcde"``; ` `    ``K ``=` `3``; ` ` `  `    ``print``(solve(X, Y, N, K)); ` ` `  `# This code is contributed by PrinciRaj1992 `

## C#

 `// C# program to find the  ` `// maximum length of equal subString  ` `// within a given cost  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to find the maximum length  ` `    ``static` `int` `solve(``string` `X, ``string` `Y,  ` `                    ``int` `N, ``int` `K)  ` `    ``{  ` `     `  `        ``int` `[]count = ``new` `int``[N + 1];  ` `        ``int` `sol = 0;  ` `        ``count = 0;  ` `     `  `        ``// Fill the prefix array with  ` `        ``// the difference of letters  ` `        ``for` `(``int` `i = 1; i <= N; i++)  ` `        ``{  ` `     `  `            ``count[i] = count[i - 1] +  ` `                    ``Math.Abs(X[i - 1] -  ` `                    ``Y[i - 1]);  ` `        ``}  ` `     `  `        ``int` `j = 0;  ` `     `  `        ``for` `(``int` `i = 1; i <= N; i++)  ` `        ``{  ` `            ``while` `((count[i] - count[j]) > K)  ` `            ``{  ` `                ``j++;  ` `            ``}  ` `     `  `            ``// Update the maximum length  ` `            ``sol = Math.Max(sol, i - j);  ` `        ``}  ` `     `  `        ``return` `sol;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `     `  `        ``int` `N = 4;  ` `        ``string` `X = ``"abcd"``, Y = ``"bcde"``;  ` `        ``int` `K = 3;  ` `     `  `        ``Console.WriteLine(solve(X, Y, N, K) + ``"\n"``);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

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