Given two strings A and B of lengths N and M respectively, the task is to find the length of the longest common subsequence that can be two strings if any character from string A can be swapped with any other character from B any number of times.
Examples:
Input: A = “abdeff”, B = “abbet”
Output: 4
Explanation: Swapping A[5] and B[4] modifies A to “abdeft” and B to “abbef”. LCS of the given strings is “abef”. Therefore, length is 4.Input: A = “abcd”, B = “ab”
Output: 2
Explanation: LCS of the given strings is “ab”. Therefore, length is 2.
Approach: The idea is based on the observation that if any character from string A can be swapped with any other character from string B, then it is also possible to swap characters within string A and also within string B.
Proof: If characters A[i] and A[j] are required to be swapped, then take a temporary element at any index k in string B. Follow the steps below to solve the problem:
- Swap A[i] with B[k].
- Swap B[k] with A[j].
- Swap B[k] with A[i].
In this way, the characters within a string can be swapped. Now, the elements can be arranged in any order. Therefore, the idea is to find the frequencies of all the characters present in both the strings and divide them equally.
Follow the steps below to solve the problem:
- Initialize an array, say freq, of size 26, to store the frequency of each character present in the strings.
- Traverse the strings A and B and update the frequency of each character in the array freq[].
- Initialize a variable, say cnt, to store the required length.
- Traverse the array freq[] and increase the value of cnt by freq[i] / 2.
- Store the minimum of cnt, N, and M in a variable, say ans.
- Print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the length of LCS// possible by swapping any character// of a string with that of another stringvoid lcsBySwapping(string A, string B){ // Store the size of the strings int N = A.size(); int M = B.size(); // Stores frequency of characters int freq[26]; memset(freq, 0, sizeof(freq)); // Iterate over characters of the string A for (int i = 0; i < A.size(); i++) { // Update frequency of character A[i] freq[A[i] - 'a'] += 1; } // Iterate over characters of the string B for (int i = 0; i < B.size(); i++) { // Update frequency of character B[i] freq[B[i] - 'a'] += 1; } // Store the count of all pairs // of similar characters int cnt = 0; // Traverse the array freq[] for (int i = 0; i < 26; i++) { // Update cnt cnt += freq[i] / 2; } // Print the minimum of cnt, N and M cout << min(cnt, min(N, M));}// Driver Codeint main(){ // Given strings string A = "abdeff"; string B = "abbet"; lcsBySwapping(A, B); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG{// Function to find the length of LCS// possible by swapping any character// of a string with that of another stringstatic void lcsBySwapping(String A, String B){ // Store the size of the strings int N = A.length(); int M = B.length(); // Stores frequency of characters int freq[] = new int[26]; // Iterate over characters of the string A for(int i = 0; i < N; i++) { // Update frequency of character A[i] freq[A.charAt(i) - 'a'] += 1; } // Iterate over characters of the string B for(int i = 0; i < M; i++) { // Update frequency of character B[i] freq[B.charAt(i) - 'a'] += 1; } // Store the count of all pairs // of similar characters int cnt = 0; // Traverse the array freq[] for(int i = 0; i < 26; i++) { // Update cnt cnt += freq[i] / 2; } // Print the minimum of cnt, N and M System.out.println(Math.min(cnt, Math.min(N, M)));}// Driver Codepublic static void main(String[] args){ // Given strings String A = "abdeff"; String B = "abbet"; lcsBySwapping(A, B);}}// This code is contributed by Kingash |
Python3
# Python3 program for the above approach# Function to find the length of LCS# possible by swapping any character# of a with that of another stringdef lcsBySwapping(A, B): # Store the size of the strings N = len(A) M = len(B) # Stores frequency of characters freq = [0] * 26 # Iterate over characters of the A for i in range(len(A)): # Update frequency of character A[i] freq[ord(A[i]) - ord('a')] += 1 # Iterate over characters of the B for i in range(len(B)): # Update frequency of character B[i] freq[ord(B[i]) - ord('a')] += 1 # Store the count of all pairs # of similar characters cnt = 0 # Traverse the array freq[] for i in range(26): # Update cnt cnt += freq[i] // 2 # Print the minimum of cnt, N and M print (min(cnt, min(N, M)))# Driver Codeif __name__ == '__main__': # Given strings A = "abdeff" B = "abbet" lcsBySwapping(A, B)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to find the length of LCS// possible by swapping any character// of a string with that of another stringstatic void lcsBySwapping(string A, string B){ // Store the size of the strings int N = A.Length; int M = B.Length; // Stores frequency of characters int[] freq = new int[26]; // Iterate over characters of the string A for(int i = 0; i < N; i++) { // Update frequency of character A[i] freq[A[i] - 'a'] += 1; } // Iterate over characters of the string B for(int i = 0; i < M; i++) { // Update frequency of character B[i] freq[B[i] - 'a'] += 1; } // Store the count of all pairs // of similar characters int cnt = 0; // Traverse the array freq[] for(int i = 0; i < 26; i++) { // Update cnt cnt += freq[i] / 2; } // Print the minimum of cnt, N and M Console.WriteLine(Math.Min(cnt, Math.Min(N, M)));}// Driver Codepublic static void Main(string[] args){ // Given strings string A = "abdeff"; string B = "abbet"; lcsBySwapping(A, B);}}// This code is contributed by ukasp |
Javascript
<script>//Javascript program for the above approach// Function to find the length of LCS// possible by swapping any character// of a string with that of another stringfunction lcsBySwapping(A, B){ // Store the size of the strings var N = A.length; var M = B.length; // Stores frequency of characters var freq = new Array(26); freq.fill(0); // Iterate over characters of the string A for (var i = 0; i < A.length; i++) { // Update frequency of character A[i] freq[A[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1; } // Iterate over characters of the string B for (var i = 0; i < B.length; i++) { // Update frequency of character B[i] freq[B[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1; } // Store the count of all pairs // of similar characters var cnt = 0; // Traverse the array freq[] for (var i = 0; i < 26; i++) { // Update cnt cnt += parseInt(freq[i] / 2); } // Print the minimum of cnt, N and M document.write( Math.min(cnt, Math.min(N, M)));}var A = "abdeff";var B = "abbet";lcsBySwapping(A, B);//This code is contributed by SoumikMondal</script> |
Output:
4
Time Complexity: O(N + M)
Auxiliary Space: O(1)
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