# Longest Common Subsequence with at most k changes allowed

Given two sequence **P** and **Q** of numbers. The task is to find Longest Common Subsequence of two sequence if we are allowed to change at most **k** element in first sequence to any value.

Examples:

Input : P = { 8, 3 } Q = { 1, 3 } K = 1 Output : 2 If we change first element of first sequence from 8 to 1, both sequences become same. Input : P = { 1, 2, 3, 4, 5 } Q = { 5, 3, 1, 4, 2 } K = 1 Output : 3 By changing first element of first sequence to 5 to get the LCS ( 5, 3, 4 }.

The idea is to use Dynamic Programming. Define a 3D matrix dp[][][], where dp[i][j][k] defines the Longest Common Subsequence for the first i numbers of first array, first j number of second array when we are allowed to change at max k number in the first array.

Therefore, recursion will look like

If P[i] != Q[j], dp[i][j][k] = max(dp[i - 1][j][k], dp[i][j - 1][k], dp[i - 1][j - 1][k - 1] + 1) If P[i] == Q[j], dp[i][j][k] = max(dp[i - 1][j][k], dp[i][j - 1][k], dp[i - 1][j - 1][k] + 1)

Below is The implementation of this approach:

## C++

`// CPP program to find LCS of two arrays with ` `// k changes allowed in first array. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `#define MAX 10 ` ` ` `// Return LCS with at most k changes allowed. ` `int` `lcs(` `int` `dp[MAX][MAX][MAX], ` `int` `arr1[], ` `int` `n, ` ` ` `int` `arr2[], ` `int` `m, ` `int` `k) ` `{ ` ` ` `// If at most changes is less than 0. ` ` ` `if` `(k < 0) ` ` ` `return` `-1e7; ` ` ` ` ` `// If any of two array is over. ` ` ` `if` `(n < 0 || m < 0) ` ` ` `return` `0; ` ` ` ` ` `// Making a reference variable to dp[n][m][k] ` ` ` `int` `& ans = dp[n][m][k]; ` ` ` ` ` `// If value is already calculated, return ` ` ` `// that value. ` ` ` `if` `(ans != -1) ` ` ` `return` `ans; ` ` ` ` ` `// calculating LCS with no changes made. ` ` ` `ans = max(lcs(dp, arr1, n - 1, arr2, m, k), ` ` ` `lcs(dp, arr1, n, arr2, m - 1, k)); ` ` ` ` ` `// calculating LCS when array element are same. ` ` ` `if` `(arr1[n] == arr2[m]) ` ` ` `ans = max(ans, 1 + lcs(dp, arr1, n - 1, ` ` ` `arr2, m - 1, k)); ` ` ` ` ` `// calculating LCS with changes made. ` ` ` `ans = max(ans, 1 + lcs(dp, arr1, n - 1, ` ` ` `arr2, m - 1, k - 1)); ` ` ` ` ` `return` `ans; ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ` `int` `k = 1; ` ` ` `int` `arr1[] = { 1, 2, 3, 4, 5 }; ` ` ` `int` `arr2[] = { 5, 3, 1, 4, 2 }; ` ` ` `int` `n = ` `sizeof` `(arr1) / ` `sizeof` `(arr1[0]); ` ` ` `int` `m = ` `sizeof` `(arr2) / ` `sizeof` `(arr2[0]); ` ` ` ` ` `int` `dp[MAX][MAX][MAX]; ` ` ` `memset` `(dp, -1, ` `sizeof` `(dp)); ` ` ` ` ` `cout << lcs(dp, arr1, n, arr2, m, k) << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Python3

`# Python3 program to find LCS of two arrays ` `# with k changes allowed in the first array. ` `MAX` `=` `10` ` ` `# Return LCS with at most k changes allowed. ` `def` `lcs(dp, arr1, n, arr2, m, k): ` ` ` ` ` `# If at most changes is less than 0. ` ` ` `if` `k < ` `0` `: ` ` ` `return` `-` `(` `10` `*` `*` `7` `) ` ` ` ` ` `# If any of two array is over. ` ` ` `if` `n < ` `0` `or` `m < ` `0` `: ` ` ` `return` `0` ` ` ` ` `# Making a reference variable to dp[n][m][k] ` ` ` `ans ` `=` `dp[n][m][k] ` ` ` ` ` `# If value is already calculated, ` ` ` `# return that value. ` ` ` `if` `ans !` `=` `-` `1` `: ` ` ` `return` `ans ` ` ` ` ` `# calculating LCS with no changes made. ` ` ` `ans ` `=` `max` `(lcs(dp, arr1, n ` `-` `1` `, arr2, m, k), ` ` ` `lcs(dp, arr1, n, arr2, m ` `-` `1` `, k)) ` ` ` ` ` `# calculating LCS when array element are same. ` ` ` `if` `arr1[n` `-` `1` `] ` `=` `=` `arr2[m` `-` `1` `]: ` ` ` `ans ` `=` `max` `(ans, ` `1` `+` `lcs(dp, arr1, n ` `-` `1` `, ` ` ` `arr2, m ` `-` `1` `, k)) ` ` ` ` ` `# calculating LCS with changes made. ` ` ` `ans ` `=` `max` `(ans, lcs(dp, arr1, n ` `-` `1` `, ` ` ` `arr2, m ` `-` `1` `, k ` `-` `1` `)) ` ` ` ` ` `return` `ans ` ` ` `# Driven Program ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `k ` `=` `1` ` ` `arr1 ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `] ` ` ` `arr2 ` `=` `[` `5` `, ` `3` `, ` `1` `, ` `4` `, ` `2` `] ` ` ` `n ` `=` `len` `(arr1) ` ` ` `m ` `=` `len` `(arr2) ` ` ` ` ` `dp ` `=` `[[[` `-` `1` `for` `i ` `in` `range` `(` `MAX` `)] ` `for` `j ` `in` `range` `(` `MAX` `)] ` `for` `k ` `in` `range` `(` `MAX` `)] ` ` ` ` ` `print` `(lcs(dp, arr1, n, arr2, m, k)) ` ` ` `# This code is contributed by Rituraj Jain ` |

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**Output:**

3

**Time Complexity: **O(N*M*K).

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

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