Check if given strings can be made same by swapping two characters of same or different strings
Given an array of equal-length strings, arr[] of size N, the task is to check if all the strings can be made equal by repeatedly swapping any pair of characters of same or different strings from the given array. If found to be true, then print “YES”. Otherwise, print “NO”.
Examples:
Input: arr[] = { “acbdd”, “abcee” }
Output: YES
Explanation:
Swapping arr[0][1] and arr[0][2] modifies arr[] to { “abcdd”, “abcee” }
Swapping arr[0][3] and arr[1][4] modifies arr[] to { “abced”, “abced” }
Therefore, the required output is “YES”.Input: arr[] = { “abb”, “acc”, “abc” }
Output: YES
Explanation:
Swapping arr[0][2] with arr[1][1] modifies arr[] to { “abc”, “abc”, “abc” }.
Therefore, the required output is “YES”.
Approach: The idea is to count the frequency of each distinct character of all the strings and check if the frequency of each distinct character is divisible by N or not. If found to be true, then print “YES”. Otherwise, print “NO”. Follow the steps below to solve the problem:
- Initialize an array, say cntFreq[], to store the frequency of each distinct character of all the strings.
- Traverse the array and for every string encountered, count the frequency of each distinct character of the string.
- Traverse the array cntFreq[] using variable i. For every ith character, check if cntFreq[i] is divisible by N or not. If found to be false, then print “NO”.
- Otherwise, print “YES”.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to check if all strings can be // made equal by swapping any pair of // characters from the same or different strings bool isEqualStrings(string arr[], int N) { // Stores length of string int M = arr[0].length(); // Store frequency of each distinct // character of the strings int cntFreq[256] = { 0 }; // Traverse the array for ( int i = 0; i < N; i++) { // Iterate over the characters for ( int j = 0; j < M; j++) { // Update frequency // of arr[i][j] cntFreq[arr[i][j] - 'a' ]++; } } // Traverse the array cntFreq[] for ( int i = 0; i < 256; i++) { // If cntFreq[i] is // divisible by N or not if (cntFreq[i] % N != 0) { return false ; } } return true ; } // Driver Code int main() { string arr[] = { "aab" , "bbc" , "cca" }; int N = sizeof (arr) / sizeof (arr[0]); // Function Call if (isEqualStrings(arr, N)) { cout << "YES" ; } else { cout << "NO" ; } return 0; } |
Java
// Java program to implement // the above approach class GFG{ // Function to check if all strings can be // made equal by swapping any pair of // characters from the same or different strings static boolean isEqualStrings(String[] arr, int N) { // Stores length of string int M = arr[ 0 ].length(); // Store frequency of each distinct // character of the strings int [] cntFreq = new int [ 256 ]; for ( int i = 0 ; i < N; i++) { cntFreq[i] = 0 ; } // Traverse the array for ( int i = 0 ; i < N; i++) { // Iterate over the characters for ( int j = 0 ; j < M; j++) { // Update frequency // of arr[i].charAt(j) cntFreq[arr[i].charAt(j) - 'a' ] += 1 ; } } // Traverse the array cntFreq[] for ( int i = 0 ; i < 256 ; i++) { // If cntFreq[i] is // divisible by N or not if (cntFreq[i] % N != 0 ) { return false ; } } return true ; } // Driver Code public static void main(String[] args) { String[] arr = { "aab" , "bbc" , "cca" }; int N = arr.length; // Function Call if (isEqualStrings(arr, N)) { System.out.println( "YES" ); } else { System.out.println( "NO" ); } } } // This code is contributed by AnkThon |
Python3
# Python3 program to implement # the above approach # Function to check if all strings can # be made equal by swapping any pair of # characters from the same or different # strings def isEqualStrings(arr, N): # Stores length of string M = len (arr[ 0 ]) # Store frequency of each distinct # character of the strings cntFreq = [ 0 ] * 256 # Traverse the array for i in range (N): # Iterate over the characters for j in range (M): # Update frequency # of arr[i][j] cntFreq[ ord (arr[i][j]) - ord ( 'a' )] + = 1 # Traverse the array cntFreq[] for i in range ( 256 ): # If cntFreq[i] is # divisible by N or not if (cntFreq[i] % N ! = 0 ): return False return True # Driver Code if __name__ = = "__main__" : arr = [ "aab" , "bbc" , "cca" ] N = len (arr) # Function Call if isEqualStrings(arr, N): print ( "YES" ) else : print ( "NO" ) # This code is contributed by jana_sayantan |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to check if all strings can be // made equal by swapping any pair of // characters from the same or different strings static bool isEqualStrings( string [] arr, int N) { // Stores length of string int M = arr[0].Length; // Store frequency of each distinct // character of the strings int [] cntFreq = new int [256]; for ( int i = 0; i < N; i++) { cntFreq[i] = 0; } // Traverse the array for ( int i = 0; i < N; i++) { // Iterate over the characters for ( int j = 0; j < M; j++) { // Update frequency // of arr[i][j] cntFreq[arr[i][j] - 'a' ] += 1; } } // Traverse the array cntFreq[] for ( int i = 0; i < 256; i++) { // If cntFreq[i] is // divisible by N or not if (cntFreq[i] % N != 0) { return false ; } } return true ; } // Driver Code public static void Main() { string [] arr = { "aab" , "bbc" , "cca" }; int N = arr.Length; // Function Call if (isEqualStrings(arr, N)) { Console.WriteLine( "YES" ); } else { Console.WriteLine( "NO" ); } } } // This code is contributed by susmitakundugoaldanga |
Javascript
<script> // javascript program of the above approach // Function to check if all strings can be // made equal by swapping any pair of // characters from the same or different strings function isEqualStrings(arr, N) { // Stores length of string let M = arr[0].length; // Store frequency of each distinct // character of the strings let cntFreq = new Array(256).fill(0); for (let i = 0; i < N; i++) { cntFreq[i] = 0; } // Traverse the array for (let i = 0; i < N; i++) { // Iterate over the characters for (let j = 0; j < M; j++) { // Update frequency // of arr[i].charAt(j) cntFreq[arr[i][j] - 'a' ] += 1; } } // Traverse the array cntFreq[] for (let i = 0; i < 256; i++) { // If cntFreq[i] is // divisible by N or not if (cntFreq[i] % N != 0) { return false ; } } return true ; } // Driver Code let arr = [ "aab" , "bbc" , "cca" ]; let N = arr.length; // Function Call if (isEqualStrings(arr, N)) { document.write( "YES" ); } else { document.write( "NO" ); } // This code is contributed by target_2. </script> |
YES
Time Complexity: O(N * M + 256), where M is the length of the string
Auxiliary Space: O(256)
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