Given two sequence P and Q of numbers. The task is to find Longest Common Subsequence of two sequences if we are allowed to change at most k element in first sequence to any value.
Examples:
Input : P = { 8, 3 }
Q = { 1, 3 }
K = 1
Output : 2
If we change first element of first
sequence from 8 to 1, both sequences
become same.
Input : P = { 1, 2, 3, 4, 5 }
Q = { 5, 3, 1, 4, 2 }
K = 1
Output : 3
By changing first element of first
sequence to 5 to get the LCS ( 5, 3, 4 }.
The idea is to use Dynamic Programming. Define a 3D matrix dp[][][], where dp[i][j][k] defines the Longest Common Subsequence for the first i numbers of first array, first j number of second array when we are allowed to change at max k number in the first array.
Therefore, recursion will look like
If P[i] != Q[j],
dp[i][j][k] = max(dp[i - 1][j][k],
dp[i][j - 1][k],
dp[i - 1][j - 1][k - 1] + 1)
If P[i] == Q[j],
dp[i][j][k] = max(dp[i - 1][j][k],
dp[i][j - 1][k],
dp[i - 1][j - 1][k] + 1)
Below is the implementation of this approach:
C++
// CPP program to find LCS of two arrays with // k changes allowed in first array. #include <bits/stdc++.h> using namespace std; #define MAX 10 // Return LCS with at most k changes allowed. int lcs(int dp[MAX][MAX][MAX], int arr1[], int n, int arr2[], int m, int k) { // If at most changes is less than 0. if (k < 0) return -1e7; // If any of two array is over. if (n < 0 || m < 0) return 0; // Making a reference variable to dp[n][m][k] int& ans = dp[n][m][k]; // If value is already calculated, return // that value. if (ans != -1) return ans; // calculating LCS with no changes made. ans = max(lcs(dp, arr1, n - 1, arr2, m, k), lcs(dp, arr1, n, arr2, m - 1, k)); // calculating LCS when array element are same. if (arr1[n-1] == arr2[m-1]) ans = max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k)); // calculating LCS with changes made. ans = max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k - 1)); return ans; } // Driven Program int main() { int k = 1; int arr1[] = { 1, 2, 3, 4, 5 }; int arr2[] = { 5, 3, 1, 4, 2 }; int n = sizeof(arr1) / sizeof(arr1[0]); int m = sizeof(arr2) / sizeof(arr2[0]); int dp[MAX][MAX][MAX]; memset(dp, -1, sizeof(dp)); cout << lcs(dp, arr1, n, arr2, m, k) << endl; return 0; } |
Java
// Java program to find LCS of two arrays with // k changes allowed in first array. import java.util.*; import java.io.*; class GFG { static int MAX = 10; // Return LCS with at most k changes allowed. static int lcs(int[][][] dp, int[] arr1, int n, int[] arr2, int m, int k) { // If at most changes is less than 0. if (k < 0) return -10000000; // If any of two array is over. if (n < 0 || m < 0) return 0; // Making a reference variable to dp[n][m][k] int ans = dp[n][m][k]; // If value is already calculated, return // that value. if (ans != -1) return ans; try { // calculating LCS with no changes made. ans = Math.max(lcs(dp, arr1, n - 1, arr2, m, k), lcs(dp, arr1, n, arr2, m - 1, k)); // calculating LCS when array element are same. if (arr1[n - 1] == arr2[m - 1]) ans = Math.max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k)); // calculating LCS with changes made. ans = Math.max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k - 1)); } catch (Exception e) { } // Storing the value in dp. dp[n][m][k] = ans; return ans; } // Driver Code public static void main(String[] args) { int k = 1; int[] arr1 = { 1, 2, 3, 4, 5 }; int[] arr2 = { 5, 3, 1, 4, 2 }; int n = arr1.length; int m = arr2.length; int[][][] dp = new int[MAX][MAX][MAX]; for (int i = 0; i < MAX; i++) for (int j = 0; j < MAX; j++) for (int l = 0; l < MAX; l++) dp[i][j][l] = -1; System.out.println(lcs(dp, arr1, n, arr2, m, k)); } } // This code is contributed by // krishnat208026 |
Python3
# Python3 program to find LCS of two arrays # with k changes allowed in the first array. MAX = 10 # Return LCS with at most k changes allowed. def lcs(dp, arr1, n, arr2, m, k): # If at most changes is less than 0. if k < 0: return -(10 ** 7) # If any of two array is over. if n < 0 or m < 0: return 0 # Making a reference variable to dp[n][m][k] ans = dp[n][m][k] # If value is already calculated, # return that value. if ans != -1: return ans # calculating LCS with no changes made. ans = max(lcs(dp, arr1, n - 1, arr2, m, k), lcs(dp, arr1, n, arr2, m - 1, k)) # calculating LCS when array element are same. if arr1[n-1] == arr2[m-1]: ans = max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k)) # calculating LCS with changes made. ans = max(ans, lcs(dp, arr1, n - 1, arr2, m - 1, k - 1)) dp[n][m][k]=ans return dp[n][m][k] # Driven Program if __name__ == "__main__": k = 1 arr1 = [1, 2, 3, 4, 5] arr2 = [5, 3, 1, 4, 2] n = len(arr1) m = len(arr2) dp = [[[-1 for i in range(MAX)] for j in range(MAX)] for k in range(MAX)] print(lcs(dp, arr1, n, arr2, m, k)) # This code is contributed by Rituraj Jain |
C#
// C# program to find LCS of two arrays with // k changes allowed in first array. using System; class GFG { static int MAX = 10; // Return LCS with at most // k changes allowed. static int lcs(int[,,] dp, int[] arr1, int n, int[] arr2, int m, int k) { // If at most changes is less than 0. if (k < 0) return -10000000; // If any of two array is over. if (n < 0 || m < 0) return 0; // Making a reference variable // to dp[n,m,k] int ans = dp[n, m, k]; // If value is already calculated, // return that value. if (ans != -1) return ans; try { // calculating LCS with no changes made. ans = Math.Max(lcs(dp, arr1, n - 1, arr2, m, k), lcs(dp, arr1, n, arr2, m - 1, k)); // calculating LCS when // array element are same. if (arr1[n - 1] == arr2[m - 1]) ans = Math.Max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k)); // calculating LCS with changes made. ans = Math.Max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k - 1)); } catch (Exception e) { } return ans; } // Driver Code public static void Main(String[] args) { int k = 1; int[] arr1 = { 1, 2, 3, 4, 5 }; int[] arr2 = { 5, 3, 1, 4, 2 }; int n = arr1.Length; int m = arr2.Length; int[,,] dp = new int[MAX, MAX, MAX]; for (int i = 0; i < MAX; i++) for (int j = 0; j < MAX; j++) for (int l = 0; l < MAX; l++) dp[i, j, l] = -1; Console.WriteLine(lcs(dp, arr1, n, arr2, m, k)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript program to find LCS of two // arrays with k changes allowed in // first array. let MAX = 10; // Return LCS with at most k changes allowed. function lcs(dp, arr1, n, arr2, m, k) { // If at most changes is less than 0. if (k < 0) return -10000000; // If any of two array is over. if (n < 0 || m < 0) return 0; // Making a reference variable // to dp[n][m][k] let ans = dp; // If value is already calculated, // return that value. if (ans != -1) return ans; try { // Calculating LCS with no changes made. ans = Math.max(lcs(dp, arr1, n - 1, arr2, m, k), lcs(dp, arr1, n, arr2, m - 1, k)); // Calculating LCS when array element are same. if (arr1[n - 1] == arr2[m - 1]) ans = Math.max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k)); // Calculating LCS with changes made. ans = Math.max(ans, 1 + lcs(dp, arr1, n - 1, arr2, m - 1, k - 1)); } catch (e) { } return ans; } // Driver Code let k = 0; let arr1 = [ 1, 2, 3, 4, 5 ]; let arr2 = [ 5, 3, 1, 4, 2 ]; let n = arr1.length; let m = arr2.length; let dp = new Array(MAX); for(let i = 0; i < MAX; i++) for(let j = 0; j < MAX; j++) for(let l = 0; l < MAX; l++) dp = -1; document.write(lcs(dp, arr1, n, arr2, m, k)); // This code is contributed by shivanisinghss2110 </script> |
Output
4
Time Complexity: O(N*M*K).
Auxiliary Space: O(MAX3)
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