Given two strings str1 and str2, the task is to check whether both of the string can be made equal by copying any character of the string with its adjacent character. Note that this operation can be performed any number of times.
Examples:
Input: str1 = “abc”, str2 = “def”
Output: No
As all the characters in both the string are different.
So, there is no way they can be made equal.
Input: str1 = “abc”, str2 = “fac”
Output: Yes
str1 = “abc” -> “aac”
str2 = “fac” -> “aac”
Approach: In order for the strings to be made equal with the given operation, they have to be of equal lengths and there has to be at least one character which is common in both the strings. To check that, create a frequency array freq[] which will store the frequency of all the characters of str1 and then for every character of str2 if its frequency in str1 is greater than 0 then it is possible to make both the strings equal.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 26// Function that returns true if both// the strings can be made equal// with the given operationbool canBeMadeEqual(string str1, string str2){ int len1 = str1.length(); int len2 = str2.length(); // Lengths of both the strings // have to be equal if (len1 == len2) { // To store the frequency of the // characters of str1 int freq[MAX]; for (int i = 0; i < len1; i++) { freq[str1[i] - 'a']++; } // For every character of str2 for (int i = 0; i < len2; i++) { // If current character of str2 // also appears in str1 if (freq[str2[i] - 'a'] > 0) return true; } } return false;}// Driver codeint main(){ string str1 = "abc", str2 = "defa"; if (canBeMadeEqual(str1, str2)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the above approach class GFG { static int MAX = 26; // Function that returns true if both // the strings can be made equal // with the given operation static boolean canBeMadeEqual(String str1, String str2) { int len1 = str1.length(); int len2 = str2.length(); // Lengths of both the strings // have to be equal if (len1 == len2) { // To store the frequency of the // characters of str1 int freq[] = new int[MAX]; for (int i = 0; i < len1; i++) { freq[str1.charAt(i) - 'a']++; } // For every character of str2 for (int i = 0; i < len2; i++) { // If current character of str2 // also appears in str1 if (freq[str2.charAt(i) - 'a'] > 0) return true; } } return false; } // Driver code public static void main (String[] args) { String str1 = "abc", str2 = "defa"; if (canBeMadeEqual(str1, str2)) System.out.println("Yes"); else System.out.println("No"); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approachMAX = 26# Function that returns true if both# the strings can be made equal# with the given operationdef canBeMadeEqual(str1, str2): len1 = len(str1) len2 = len(str2) # Lengths of both the strings # have to be equal if (len1 == len2): # To store the frequency of the # characters of str1 freq = [0 for i in range(MAX)] for i in range(len1): freq[ord(str1[i]) - ord('a')] += 1 # For every character of str2 for i in range(len2): # If current character of str2 # also appears in str1 if (freq[ord(str2[i]) - ord('a')] > 0): return True return False# Driver codestr1 = "abc"str2 = "defa"if (canBeMadeEqual(str1, str2)): print("Yes")else: print("No")# This code is contributed by Mohit Kumar |
C#
// C# implementation of the above approach using System;class GFG { static int MAX = 26; // Function that returns true if both // the strings can be made equal // with the given operation static Boolean canBeMadeEqual(String str1, String str2) { int len1 = str1.Length; int len2 = str2.Length; // Lengths of both the strings // have to be equal if (len1 == len2) { // To store the frequency of the // characters of str1 int []freq = new int[MAX]; for (int i = 0; i < len1; i++) { freq[str1[i] - 'a']++; } // For every character of str2 for (int i = 0; i < len2; i++) { // If current character of str2 // also appears in str1 if (freq[str2[i] - 'a'] > 0) return true; } } return false; } // Driver code public static void Main (String []args) { String str1 = "abc", str2 = "defa"; if (canBeMadeEqual(str1, str2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } }// This code is contributed by Arnab Kundu |
Javascript
<script> // JavaScript implementation of the above approach let MAX = 26; // Function that returns true if both // the strings can be made equal // with the given operation function canBeMadeEqual(str1, str2) { let len1 = str1.length; let len2 = str2.length; // Lengths of both the strings // have to be equal if (len1 == len2) { // To store the frequency of the // characters of str1 let freq = new Array(MAX); freq.fill(0); for (let i = 0; i < len1; i++) { freq[str1[i].charCodeAt() - 'a'.charCodeAt()]++; } // For every character of str2 for (let i = 0; i < len2; i++) { // If current character of str2 // also appears in str1 if (freq[str2[i].charCodeAt() - 'a'.charCodeAt()] > 0) return true; } } return false; } let str1 = "abc", str2 = "defa"; if (canBeMadeEqual(str1, str2)) document.write("Yes"); else document.write("No"); </script> |
Output:
No
Time Complexity: O(N)
Auxiliary Space: O(26)
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