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 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 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 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]

.

#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; }

Output:

3

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.