# Check if two strings can be made equal by swapping one character among each other

• Difficulty Level : Basic
• Last Updated : 05 Dec, 2022

Given two strings A and B of length N, the task is to check whether the two strings can be made equal by swapping any character of A with any other character of B only once.
Examples:

Input: A = “SEEKSFORGEEKS”, B = “GEEKSFORGEEKG”
Output: Yes
SEEKSFORGEEKS” and “GEEKSFORGEEKG”

After removing  the elements which are the same and have the same index in both the strings. We have string A=”SS” and B=”GG”.

The new strings are of length two and both the elements in each string are same.Hence swapping is possible.
Input: A = “GEEKSFORGEEKS”, B = “THESUPERBSITE”
Output: No

Approach: First omit the elements which are the same and have the same index in both the strings. Then if the new strings are of length two and both the elements in each string are the same then only the swap is possible.
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;   // Function that returns true if the string // can be made equal after one swap bool canBeEqual(string a, string b, int n) {     // A and B are new a and b     // after we omit the same elements     vector A, B;       // Take only the characters which are     // different in both the strings     // for every pair of indices     for (int i = 0; i < n; i++)     {           // If the current characters differ         if (a[i]!= b[i])         {             A.push_back(a[i]);             B.push_back(b[i]);         }     }           // The strings were already equal     if (A.size() == B.size() and         B.size() == 0)         return true;       // If the lengths of the     // strings are two     if (A.size() == B.size() and         B.size() == 2)     {           // If swapping these characters         // can make the strings equal         if (A[0] == A[1] and B[0] == B[1])             return true;     }     return false; }   // Driver code int main() {     string A = "SEEKSFORGEEKS";     string B = "GEEKSFORGEEKG";           if (canBeEqual(A, B, A.size()))         printf("Yes");     else         printf("No"); }   // This code is contributed by Mohit Kumar

## Java

 // Java implementation of the approach import java.util.*; class GFG {       // Function that returns true if the string // can be made equal after one swap static boolean canBeEqual(char []a,                           char []b, int n) {     // A and B are new a and b     // after we omit the same elements     Vector A = new Vector<>();     Vector B = new Vector<>();       // Take only the characters which are     // different in both the strings     // for every pair of indices     for (int i = 0; i < n; i++)     {           // If the current characters differ         if (a[i] != b[i])         {             A.add(a[i]);             B.add(b[i]);         }     }           // The strings were already equal     if (A.size() == B.size() &&         B.size() == 0)         return true;       // If the lengths of the     // strings are two     if (A.size() == B.size() &&         B.size() == 2)     {           // If swapping these characters         // can make the strings equal         if (A.get(0) == A.get(1) &&             B.get(0) == B.get(1))             return true;     }     return false; }   // Driver code public static void main(String[] args) {     char []A = "SEEKSFORGEEKS".toCharArray();     char []B = "GEEKSFORGEEKG".toCharArray();           if (canBeEqual(A, B, A.length))         System.out.printf("Yes");     else         System.out.printf("No"); } }   // This code is contributed by Rajput-Ji

## Python3

 # Python3 implementation of the approach   # Function that returns true if the string # can be made equal after one swap def canBeEqual(a, b, n):     # A and B are new a and b     # after we omit the same elements     A =[]     B =[]           # Take only the characters which are     # different in both the strings     # for every pair of indices     for i in range(n):               # If the current characters differ         if a[i]!= b[i]:             A.append(a[i])             B.append(b[i])                   # The strings were already equal     if len(A)== len(B)== 0:         return True           # If the lengths of the     # strings are two     if len(A)== len(B)== 2:               # If swapping these characters         # can make the strings equal         if A[0]== A[1] and B[0]== B[1]:             return True           return False   # Driver code A = 'SEEKSFORGEEKS' B = 'GEEKSFORGEEKG'   if (canBeEqual(A, B, len(A))):     print("Yes") else:     print("No")

## C#

 // C# implementation of the above approach using System; using System.Collections.Generic;   class GFG {       // Function that returns true if the string // can be made equal after one swap static Boolean canBeEqual(char []a,                           char []b, int n) {     // A and B are new a and b     // after we omit the same elements     List A = new List();     List B = new List();       // Take only the characters which are     // different in both the strings     // for every pair of indices     for (int i = 0; i < n; i++)     {           // If the current characters differ         if (a[i] != b[i])         {             A.Add(a[i]);             B.Add(b[i]);         }     }           // The strings were already equal     if (A.Count == B.Count &&         B.Count == 0)         return true;       // If the lengths of the     // strings are two     if (A.Count == B.Count &&         B.Count == 2)     {           // If swapping these characters         // can make the strings equal         if (A[0] == A[1] &&             B[0] == B[1])             return true;     }     return false; }   // Driver code public static void Main(String[] args) {     char []A = "SEEKSFORGEEKS".ToCharArray();     char []B = "GEEKSFORGEEKG".ToCharArray();           if (canBeEqual(A, B, A.Length))         Console.WriteLine("Yes");     else         Console.WriteLine("No"); } }   // This code is contributed by PrinciRaj1992

## Javascript



Output:

Yes

Time Complexity: O(N), where N is the size of the given strings.

Auxiliary Space: O(N), where N is the size of the given strings.

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