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 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
import java.util.*;import java.io.*;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 |
Javascript
<script>// JavaScript program to calculate Minimum cost // to make Longest Common Subsequence of length kconst N = 30// Return Minimum cost to make LCS of length kfunction 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 Codelet X = "abble"let Y = "pie"let n = X.lengthlet m = Y.lengthlet k = 2let 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
Time Complexity: O(n3)
Auxiliary Space: O(n3)
Efficient approach: Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a 3d DP to store the solution of the subproblems.
- Initialize the DP with base cases when k= 0 dp value is 0 and when n=0 or m=0 dp value is 1e9 and dp[0][0][0] = 0 .
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in dp[n][m][k].
Implementation :
C++
// C++ code for above approach#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 n, int m, int k){ // Create 3D DP int dp[n + 1][m + 1][k + 1]; // initialize DP with base cases for (int i = 0; i <= n; i++) { for (int j = 0; j <= m; j++) { for (int p = 0; p <= k; p++) { if (k == 0) { dp[i][j][p] = 0; } else if (i == 0 || j == 0) { dp[i][j][p] = 1e9; } } } } dp[0][0][0] = 0; // iterate over subproblem to get the current // value from previous computation store in DP for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { for (int p = 1; p <= k; p++) { int cost = (X[i - 1] - 'a') ^ (Y[j - 1] - 'a'); // update DP dp[i][j][p] = min( { cost + dp[i - 1][j - 1][p - 1], dp[i - 1][j][p], dp[i][j - 1][p] }); } } } // return final answer return dp[n][m][k];}// Driven Programint main(){ char X[] = "abble"; char Y[] = "pie"; int n = strlen(X); int m = strlen(Y); int k = 2; // function call int ans = solve(X, Y, n, m, k); cout << (ans == 1e9 ? -1 : ans) << endl; return 0;} |
Python3
INF = float('inf')# Return Minimum cost to make LCS of length kdef solve(X, Y, n, m, k): # Create 3D DP dp = [[[0 for p in range(k+1)] for j in range(m+1)] for i in range(n+1)] # initialize DP with base cases for i in range(n+1): for j in range(m+1): for p in range(k+1): if k == 0: dp[i][j][p] = 0 elif i == 0 or j == 0: dp[i][j][p] = INF dp[0][0][0] = 0 # iterate over subproblem to get the current # value from previous computation store in DP for i in range(1, n+1): for j in range(1, m+1): for p in range(1, k+1): cost = ord(X[i-1]) - ord('a') ^ ord(Y[j-1]) - ord('a') # update DP dp[i][j][p] = min(cost + dp[i-1][j-1][p-1], dp[i-1][j][p], dp[i][j-1][p]) # return final answer return -1 if dp[n][m][k] == INF else dp[n][m][k]# Driven Programif __name__ == '__main__': X = "abble" Y = "pie" n = len(X) m = len(Y) k = 2 # function call ans = solve(X, Y, n, m, k) print(ans) |
Javascript
const N = 30;// Return Minimum cost to make LCS of length kfunction solve(X, Y, n, m, k) { // Create 3D DP const dp = new Array(n + 1).fill(null) .map(() => new Array(m + 1).fill(null) .map(() => new Array(k + 1).fill(0))); // initialize DP with base cases for (let i = 0; i <= n; i++) { for (let j = 0; j <= m; j++) { for (let p = 0; p <= k; p++) { if (k === 0) { dp[i][j][p] = 0; } else if (i === 0 || j === 0) { dp[i][j][p] = Infinity; } } } } dp[0][0][0] = 0; // iterate over subproblem to get the current // value from previous computation store in DP for (let i = 1; i <= n; i++) { for (let j = 1; j <= m; j++) { for (let p = 1; p <= k; p++) { const cost = (X[i - 1].charCodeAt() - 'a'.charCodeAt()) ^ (Y[j - 1].charCodeAt() - 'a'.charCodeAt()); // update DP dp[i][j][p] = Math.min( cost + dp[i - 1][j - 1][p - 1], dp[i - 1][j][p], dp[i][j - 1][p]); } } } // return final answer return dp[n][m][k];}// Driven Programconst X = "abble";const Y = "pie";const n = X.length;const m = Y.length;const k = 2;// function callconst ans = solve(X, Y, n, m, k);console.log(ans === Infinity ? -1 : ans); |
Output
3
Time Complexity: O(n*m*k)
Auxiliary Space: O(n*m*k)
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