Make Longest Common Subsequence at least K

Given two strings A of length N and B of length M. These strings contain lowercase English alphabets. You are also given an integer K. You can change the character of x of string A to any other character y. The cost of this conversion is abs(ASCII(x)- ASCII(y)). Find the minimum cost required such that the length of the Longest Common Subsequence (LCS) of A and B is at least K.

Constraints:

• 1 <= N <= 100
• 1 <= M <= 100

Examples:

Input: N = 5, M = 4, K = 3, A = "abcba", B = "acyx"
Output: 22
Explanation: As the minimum length of the subsequences is 3, we have to match any 3 characters of A with B. We can convert A[3] to "x" with cost = 22, then we can get the Longest Common Subsequence as "acx" of length 3.

Input: N = 3, M = 3, K = 3, A = "abc", B = "abc"
Output: 0
Explanation: As the minimum length is 3 both the string is of length 3 and every character is matches no change is needed and the minimum cost will be 0.

Approach: To solve the problem follow the below idea

The problem can be solved using Dynamic Programming. We can make a recursive function rec(i1, i2, K) which returns the minimum cost to have a Longest Common Subsequence of length K if we are at index i1 in string A and index i2 in string B. For any function rec(i1, i2, K) we have 3 choices:

• Skip the i1th character of A, cost = rec(i1 - 1, i2, K)
• Skip the i2th character of B, cost = rec(i1, i2 - 1, K)
• Include the i1th character of A and i2th character of B for LCS with cost = abs(A[i1] - B[i2]) + rec(i1 - 1, i2 - 1, K - 1)

Use a 3D array dp[N][M][K] for memoization. Iterate for each N, M and K get the minimum cost.

Below are the steps to solve this problem:

• As maximum length can be 100, Initialize a 3D dp of size [105] [105] [105] filled with -1.
• If characters present at index M -1 and N -1 are equal, we need 0 cost.
• Otherwise, get the absolute difference of characters A[n] and B[m] and add it as cost.
• Look for the minimum cost from { rec(A, B, n - 1, m, k), x + rec(A, B, n - 1, m - 1, k - 1), rec(A, B, n, m - 1, k) }. And proceed further with it.
• Return the dp[n][m][k] as the minimum cost.

Below is the implementation of the code:

// C++ Implementation
#include <bits/stdc++.h>
using namespace std;

// Declare a dp
int dp[105][105][105];

// Function to iterate in dp
int rec(string A, string B, int i1, int i2, int k)
{
    // If length of lcs string is reached
    if (k == 0)
        return 0;

    // if any of the string is completed
    if (i1 < 0 || i2 < 0) {

        if (k == 0) {
            return 0;
        }
        return 1e9;
    }

    // If value of that is already found
    if (dp[i1][i2][k] != -1) {
        return dp[i1][i2][k];
    }

    // If values are equal
    if (A[i1] == B[i2]) {
        dp[i1][i2][k] = rec(A, B, i1 - 1, i2 - 1, k - 1);
    }

    // Otherwise check by adding cost
    else {
        int cost = abs((int)A[i1] - (int)B[i2]);
        dp[i1][i2][k]
            = min({ rec(A, B, i1 - 1, i2, k),
                    cost + rec(A, B, i1 - 1, i2 - 1, k - 1),
                    rec(A, B, i1, i2 - 1, k) });
    }

    // Return the value
    return dp[i1][i2][k];
}

// Function to find minimum cost to get lcs of length
// k
int generateSub(string A, string B, int N, int M, int K)
{

    // Fill the dp with -1
    memset(dp, -1, sizeof(dp));

    // Iterate in dp
    return rec(A, B, N - 1, M - 1, K);
}

// Driver code
int main()
{
    int K = 3;
    string A = "abcba";
    string B = "acyx";
    int N = A.length();
    int M = B.length();

    // Function call
    cout << generateSub(A, B, N, M, K);

    return 0;
}

import java.util.Arrays;

public class LCS {

    // Declare a dp
    static int[][][] dp;

    // Function to iterate in dp
    static int rec(String A, String B, int i1, int i2, int k) {
        // If length of LCS string is reached
        if (k == 0)
            return 0;

        // if any of the string is completed
        if (i1 < 0 || i2 < 0) {
            if (k == 0) {
                return 0;
            }
            return (int) 1e9;
        }

        // If value of that is already found
        if (dp[i1][i2][k] != -1) {
            return dp[i1][i2][k];
        }

        // If values are equal
        if (A.charAt(i1) == B.charAt(i2)) {
            dp[i1][i2][k] = rec(A, B, i1 - 1, i2 - 1, k - 1);
        }

        // Otherwise check by adding cost
        else {
            int cost = Math.abs((int) A.charAt(i1) - (int) B.charAt(i2));
            dp[i1][i2][k] = Math.min(
                    Math.min(rec(A, B, i1 - 1, i2, k), cost + rec(A, B, i1 - 1, i2 - 1, k - 1)),
                    rec(A, B, i1, i2 - 1, k));
        }

        // Return the value
        return dp[i1][i2][k];
    }

    // Function to find minimum cost to get LCS of length k
    static int generateSub(String A, String B, int N, int M, int K) {
        // Fill the dp with -1
        dp = new int[N][M][K + 1];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < M; j++) {
                Arrays.fill(dp[i][j], -1);
            }
        }

        // Iterate in dp
        return rec(A, B, N - 1, M - 1, K);
    }

    // Driver code
    public static void main(String[] args) {
        int K = 3;
        String A = "abcba";
        String B = "acyx";
        int N = A.length();
        int M = B.length();

        // Function call
        System.out.println(generateSub(A, B, N, M, K));
    }
}

// This code is contributed by akshitaguprzj3

def generate_sub(A, B, K):
    N, M = len(A), len(B)
    dp = [[[-1 for _ in range(K + 1)] for _ in range(M + 1)] for _ in range(N + 1)]

    def rec(i1, i2, k):
        if k == 0:
            return 0

        if i1 < 0 or i2 < 0:
            if k == 0:
                return 0
            return float('inf')

        if dp[i1][i2][k] != -1:
            return dp[i1][i2][k]

        if A[i1] == B[i2]:
            dp[i1][i2][k] = rec(i1 - 1, i2 - 1, k - 1)
        else:
            cost = abs(ord(A[i1]) - ord(B[i2]))
            dp[i1][i2][k] = min(rec(i1 - 1, i2, k),
                                 cost + rec(i1 - 1, i2 - 1, k - 1),
                                 rec(i1, i2 - 1, k))

        return dp[i1][i2][k]

    return rec(N - 1, M - 1, K)

# Driver code
K = 3
A = "abcba"
B = "acyx"

result = generate_sub(A, B, K)
print(result)

using System;

class Program
{
    // Declare a dp
    static int[,,] dp = new int[105, 105, 105];

    // Function to iterate in dp
    static int Rec(string A, string B, int i1, int i2, int k)
    {
        // If length of lcs string is reached
        if (k == 0)
            return 0;

        // if any of the string is completed
        if (i1 < 0 || i2 < 0)
        {
            if (k == 0)
                return 0;
            return int.MaxValue;
        }

        // If value of that is already found
        if (dp[i1, i2, k] != -1)
            return dp[i1, i2, k];

        // If values are equal
        if (A[i1] == B[i2])
            dp[i1, i2, k] = Rec(A, B, i1 - 1, i2 - 1, k - 1);
        else
        {
            // Otherwise check by adding cost
            int cost = Math.Abs((int)A[i1] - (int)B[i2]);
            dp[i1, i2, k] = Math.Min(Rec(A, B, i1 - 1, i2, k),
                            Math.Min(cost + Rec(A, B, i1 - 1, i2 - 1, k - 1),
                                     Rec(A, B, i1, i2 - 1, k)));
        }

        // Return the value
        return dp[i1, i2, k];
    }

    // Function to find minimum cost to get lcs of length
    // k
    static int GenerateSub(string A, string B, int N, int M, int K)
    {
        // Fill the dp with -1
        for (int i = 0; i < 105; i++)
            for (int j = 0; j < 105; j++)
                for (int l = 0; l < 105; l++)
                    dp[i, j, l] = -1;

        // Iterate in dp
        return Rec(A, B, N - 1, M - 1, K);
    }

    // Driver code
    static void Main()
    {
        int K = 3;
        string A = "abcba";
        string B = "acyx";
        int N = A.Length;
        int M = B.Length;

        // Function call
        Console.WriteLine(GenerateSub(A, B, N, M, K));
    }
}

function generateSub(A, B, K) {
    const N = A.length;
    const M = B.length;

    // Initialize a 3D array for memoization
    const dp = Array.from({ length: N + 1 }, () =>
        Array.from({ length: M + 1 }, () =>
            Array(K + 1).fill(-1)
        )
    );

    // Recursive function to calculate the minimum cost
    function rec(i1, i2, k) {
        // If k becomes 0, no more substitutions are allowed
        if (k === 0) {
            return 0;
        }

        // Base cases
        if (i1 < 0 || i2 < 0) {
            return (k === 0) ? 0 : Infinity;
        }

        // Check if the result is already memoized
        if (dp[i1][i2][k] !== -1) {
            return dp[i1][i2][k];
        }

        // If the characters at the current positions are equal
        if (A[i1] === B[i2]) {
            dp[i1][i2][k] = rec(i1 - 1, i2 - 1, k - 1);
        } else {
            // Calculate cost for substitution
            const cost = Math.abs(A[i1].charCodeAt(0) - B[i2].charCodeAt(0));

            // Update the memoization table with the minimum cost
            dp[i1][i2][k] = Math.min(
                rec(i1 - 1, i2, k),
                cost + rec(i1 - 1, i2 - 1, k - 1),
                rec(i1, i2 - 1, k)
            );
        }

        return dp[i1][i2][k];
    }

    // Start the recursion from the last characters of both strings
    return rec(N - 1, M - 1, K);
}

// Driver code
const K = 3;
const A = "abcba";
const B = "acyx";

const result = generateSub(A, B, K);
console.log(result);

22

Time Complexity: O(N * M * K), where N is the length of string A and M is the length of string B.
Auxiliary Space: O(N * M * K)

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Article Tags :

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• DSA
• Algorithms-Dynamic Programming
• Memoization
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