Minimum cost to make Longest Common Subsequence of length k

Given two string X, Y and an integer k. Now the task is to convert string X with the minimum cost such that the Longest Common Subsequence of X and Y after conversion is of length k. The cost of conversion is calculated as XOR of old character value and new character value. The character value of ‘a’ is 0, ‘b’ is 1, and so on.

Examples:

Input : X = "abble", 
        Y = "pie",
        k = 2
Output : 25

If you changed 'a' to 'z', it will cost 0 XOR 25.

The problem can be solved by slight change in Dynamic Programming problem of Longest Increasing Subsequence. Instead of two states, we maintain three states. 
Note, that if k > min(n, m) then it’s impossible to attain LCS of atleast k length, else it’s always possible. 
Let dp[i][j][p] stores the minimum cost to achieve LCS of length p in x[0…i] and y[0….j]. 
With base step as dp[i][j][0] = 0 because we can achieve LCS of 0 length without any cost and for i < 0 or j 0 in such case). 
Else there are 3 cases: 
1. Convert x[i] to y[j]. 
2. Skip i^th character from x. 
3. Skip j^th character from y.

If we convert x[i] to y[j], then cost = f(x[i]) XOR f(y[j]) will be added and LCS will decrease by 1. f(x) will return the character value of x. 
Note that the minimum cost to convert a character ‘a’ to any character ‘c’ is always f(a) XOR f(c) because f(a) XOR f(c) <= (f(a) XOR f(b) + f(b) XOR f(c)) for all a, b, c. 
If you skip i^th character from x then i will be decreased by 1, no cost will be added and LCS will remain the same. 
If you skip j^th character from x then j will be decreased by 1, no cost will be added and LCS will remain the same.

Therefore,

dp[i][j][k] = min(cost + dp[i - 1][j - 1][k - 1], 
                  dp[i - 1][j][k], 
                  dp[i][j - 1][k])
The minimum cost to make the length of their
LCS atleast k is dp[n - 1][m - 1][k]

C++

#include <bits/stdc++.h>
using namespace std;
const int N = 30;

// Return Minimum cost to make LCS of length k
int solve(char X[], char Y[], int l, int r,
                     int k, int dp[][N][N])
{
    // If k is 0.
    if (!k)
        return 0;

    // If length become less than 0, return
    // big number.
    if (l < 0 | r < 0)
        return 1e9;

    // If state already calculated.
    if (dp[l][r][k] != -1)
        return dp[l][r][k];

    // Finding the cost
    int cost = (X[l] - 'a') ^ (Y[r] - 'a');

    // Finding minimum cost and saving the state value
    return dp[l][r][k] = min({cost +
                      solve(X, Y, l - 1, r - 1, k - 1, dp),
                             solve(X, Y, l - 1, r, k, dp),
                             solve(X, Y, l, r - 1, k, dp)});
}

// Driven Program
int main()
{
    char X[] = "abble";
    char Y[] = "pie";
    int n = strlen(X);
    int m = strlen(Y);
    int k = 2;

    int dp[N][N][N];
    memset(dp, -1, sizeof dp);
    int ans = solve(X, Y, n - 1, m - 1, k, dp);

    cout << (ans == 1e9 ? -1 : ans) << endl;
    return 0;
}

Java

class GFG
{

    static int N = 30;

    // Return Minimum cost to make LCS of length k
    static int solve(char X[], char Y[], int l, int r,
                                    int k, int dp[][][])
    {
        // If k is 0.
        if (k == 0)
        {
            return 0;
        }

        // If length become less than 0, return
        // big number.
        if (l < 0 | r < 0)
        {
            return (int) 1e9;
        }

        // If state already calculated.
        if (dp[l][r][k] != -1)
        {
            return dp[l][r][k];
        }

        // Finding the cost
        int cost = (X[l] - 'a') ^ (Y[r] - 'a');

        // Finding minimum cost and saving the state value
        return dp[l][r][k] = Math.min(Math.min(cost +
                solve(X, Y, l - 1, r - 1, k - 1, dp),
                solve(X, Y, l - 1, r, k, dp)),
                solve(X, Y, l, r - 1, k, dp));
    }

    // Driver code
    public static void main(String[] args)
    {
        char X[] = "abble".toCharArray();
        char Y[] = "pie".toCharArray();
        int n = X.length;
        int m = Y.length;
        int k = 2;

        int[][][] dp = new int[N][N][N];
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                for (int l = 0; l < N; l++)
                {
                    dp[i][j][l] = -1;
                }
            }
        }
        int ans = solve(X, Y, n - 1, m - 1, k, dp);

        System.out.println(ans == 1e9 ? -1 : ans);
    }
}

// This code contributed by Rajput-Ji

Python3

# Python3 program to calculate Minimum cost
# to make Longest Common Subsequence of length k
N = 30

# Return Minimum cost to make LCS of length k
def solve(X, Y, l, r, k, dp):

    # If k is 0
    if k == 0:
        return 0

    # If length become less than 0,
    # return big number
    if l < 0 or r < 0:
        return 1000000000

    # If state already calculated
    if dp[l][r][k] != -1:
        return dp[l][r][k]

    # Finding cost
    cost = ((ord(X[l]) - ord('a')) ^
            (ord(Y[r]) - ord('a')))

    dp[l][r][k] = min([cost + solve(X, Y, l - 1,
                                          r - 1, k - 1, dp),
                              solve(X, Y, l - 1, r, k, dp),
                              solve(X, Y, l, r - 1, k, dp)])

    return dp[l][r][k]

# Driver Code
if __name__ == "__main__":
    X = "abble"
    Y = "pie"
    n = len(X)
    m = len(Y)
    k = 2
    dp = [[[-1] * N for __ in range(N)]
                    for ___ in range(N)]
    ans = solve(X, Y, n - 1, m - 1, k, dp)

    print(-1 if ans == 1000000000 else ans)

# This code is contributed
# by vibhu4agarwal

C#

// C# program to find subarray with
// sum closest to 0
using System;

class GFG
{

    static int N = 30;

    // Return Minimum cost to make LCS of length k
    static int solve(char []X, char []Y, int l, int r,
                                    int k, int [,,]dp)
    {
        // If k is 0.
        if (k == 0)
        {
            return 0;
        }

        // If length become less than 0, return
        // big number.
        if (l < 0 | r < 0)
        {
            return (int) 1e9;
        }

        // If state already calculated.
        if (dp[l,r,k] != -1)
        {
            return dp[l,r,k];
        }

        // Finding the cost
        int cost = (X[l] - 'a') ^ (Y[r] - 'a');

        // Finding minimum cost and saving the state value
        return dp[l,r,k] = Math.Min(Math.Min(cost +
                solve(X, Y, l - 1, r - 1, k - 1, dp),
                solve(X, Y, l - 1, r, k, dp)),
                solve(X, Y, l, r - 1, k, dp));
    }

    // Driver code
    public static void Main(String[] args)
    {
        char []X = "abble".ToCharArray();
        char []Y = "pie".ToCharArray();
        int n = X.Length;
        int m = Y.Length;
        int k = 2;

        int[,,] dp = new int[N, N, N];
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                for (int l = 0; l < N; l++)
                {
                    dp[i,j,l] = -1;
                }
            }
        }
        int ans = solve(X, Y, n - 1, m - 1, k, dp);

        Console.WriteLine(ans == 1e9 ? -1 : ans);
    }
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript program to calculate Minimum cost
// to make Longest Common Subsequence of length k
const N = 30

// Return Minimum cost to make LCS of length k
function solve(X, Y, l, r, k, dp){

    // If k is 0
    if(k == 0)
        return 0

    // If length become less than 0,
    // return big number
    if(l < 0 || r < 0)
        return 1000000000

    // If state already calculated
    if(dp[l][r][k] != -1)
        return dp[l][r][k]

    // Finding cost
    let cost = ((X[l].charCodeAt(0) - 'a'.charCodeAt(0)) ^ (Y[r].charCodeAt(0) - 'a'.charCodeAt(0)))

    dp[l][r][k] = Math.min(Math.min(cost + solve(X, Y, l - 1, r - 1, k - 1, dp),solve(X, Y, l - 1, r, k, dp)),
                              solve(X, Y, l, r - 1, k, dp))

    return dp[l][r][k]
}

// Driver Code

let X = "abble"
let Y = "pie"
let n = X.length
let m = Y.length
let k = 2
let dp = new Array(N);
for(let i = 0; i < N; i++)
{
    dp[i] = new Array(N);
    for(let j = 0; j < N; j++)
    {
        dp[i][j] = new Array(N).fill(-1);
    }
}
let ans = solve(X, Y, n - 1, m - 1, k, dp)

document.write((ans == 1000000000)? -1:ans)

// This code is contributed by shinjanpatra

</script>

Output:

3