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 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 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].
C++
#include <bits/stdc++.h>using namespace std;const int N = 30; // Return Minimum cost to make LCS of length kint 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 Programint 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 kN = 30 # Return Minimum cost to make LCS of length kdef 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 Codeif __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 0using 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 |
Output:
3
