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 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. 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 legth 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 ith character from x.
3. Skip jth 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 charcater ‘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 ith character from x then i will be decreased by 1, no cost will be added and LCS will remain the same.
If you skip jth 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]

.

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#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;
}

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Output:

3


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