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 C++ implementation of this approach:

// 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;
}

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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