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# Longest equal substring with cost less than K

• Last Updated : 10 Jun, 2021

Given two strings X and Y of the same length, which consist 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:
Explanation: A maximum of 3 characters can be changed because the cost to change each article is 1.

Approach: The idea is to maintain a prefix array of length N to store the absolute sum of the strings. That is, the cost of changing 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 the 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:

## C++

 `// 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`

## Javascript

 ``
Output:
`3`

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