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# Check if a string can be made equal to another string by swapping or replacement of characters

• Last Updated : 19 May, 2021

Given two strings S1 and S2, the task is to check if string S1 can be made equal to string S2 by swapping any pair of characters replacing any character in the string S1. If it is possible to make the string S1 equal to S2, then print “Yes”. Otherwise, print “No”.

Examples:

Input: S1 = “abc”, S2 = “bca”
Output: Yes
Explanation:
Operation 1: “abc” -> “acb”
Operation 2: “acb” -> “bca”

Input: S1 = “a”, S2 = “aa”
Output: No
Explanation: It is impossible to make the two strings same.

Approach: The idea to solve this problem is to find the frequencies of the characters in the strings and check that the same character is being used in both strings or not. Follow the steps below to solve the problem:

Below is the implementation of this approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find if given strings``// are same or not``bool` `sameStrings(string str1,``                 ``string str2)``{``    ``int` `N = str1.length();``    ``int` `M = str2.length();` `    ``// Base Condition``    ``if` `(N != M) {``        ``return` `false``;``    ``}` `    ``// Stores frequency of characters``    ``// of the string str1 and str2``    ``int` `a = { 0 }, b = { 0 };` `    ``// Traverse strings str1 & str2 and``    ``// store frequencies in a[] and b[]``    ``for` `(``int` `i = 0; i < N; i++) {``        ``a[str1[i] - ``'a'``]++;``        ``b[str2[i] - ``'a'``]++;``    ``}` `    ``// Check if both strings have``    ``// same characters or not``    ``int` `i = 0;` `    ``while` `(i < 256) {` `        ``if` `((a[i] == 0 && b[i] == 0)``            ``|| (a[i] != 0 && b[i] != 0)) {``            ``i++;``        ``}` `        ``// If a character is present``        ``// in one string and is not in``        ``// another string, return false``        ``else` `{``            ``return` `false``;``        ``}``    ``}` `    ``// Sort the array a[] and b[]``    ``sort(a, a + 256);``    ``sort(b, b + 256);` `    ``// Check arrays a and b contain``    ``// the same frequency or not``    ``for` `(``int` `i = 0; i < 256; i++) {` `        ``// If the frequencies are not``        ``// the same after sorting``        ``if` `(a[i] != b[i])``            ``return` `false``;``    ``}` `    ``// At this point, str1 can``    ``// be converted to str2``    ``return` `true``;``}``// Driver Code``int` `main()``{``    ``string S1 = ``"cabbba"``, S2 = ``"abbccc"``;``    ``if` `(sameStrings(S1, S2))``        ``cout << ``"YES"` `<< endl;``    ``else``        ``cout << ``" NO"` `<< endl;` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG``{` `// Function to find if given Strings``// are same or not``static` `boolean` `sameStrings(String str1,``                 ``String str2)``{``    ``int` `N = str1.length();``    ``int` `M = str2.length();` `    ``// Base Condition``    ``if` `(N != M)``    ``{``        ``return` `false``;``    ``}` `    ``// Stores frequency of characters``    ``// of the String str1 and str2``    ``int` `[]a = ``new` `int``[``256``];``    ``int` `[]b = ``new` `int``[``256``];` `    ``// Traverse Strings str1 & str2 and``    ``// store frequencies in a[] and b[]``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``a[str1.charAt(i) - ``'a'``]++;``        ``b[str2.charAt(i) - ``'a'``]++;``    ``}` `    ``// Check if both Strings have``    ``// same characters or not``    ``int` `i = ``0``;``    ``while` `(i < ``256``)``    ``{``        ``if` `((a[i] == ``0` `&& b[i] == ``0``)``            ``|| (a[i] != ``0` `&& b[i] != ``0``))``        ``{``            ``i++;``        ``}` `        ``// If a character is present``        ``// in one String and is not in``        ``// another String, return false``        ``else``        ``{``            ``return` `false``;``        ``}``    ``}` `    ``// Sort the array a[] and b[]``    ``Arrays.sort(a);``    ``Arrays.sort(b);` `    ``// Check arrays a and b contain``    ``// the same frequency or not``    ``for` `(i = ``0``; i < ``256``; i++)``    ``{` `        ``// If the frequencies are not``        ``// the same after sorting``        ``if` `(a[i] != b[i])``            ``return` `false``;``    ``}` `    ``// At this point, str1 can``    ``// be converted to str2``    ``return` `true``;``}``  ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``String S1 = ``"cabbba"``, S2 = ``"abbccc"``;``    ``if` `(sameStrings(S1, S2))``        ``System.out.print(``"YES"` `+``"\n"``);``    ``else``        ``System.out.print(``" NO"` `+``"\n"``);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program for the above approach` `# Function to find if given strings``# are same or not``def` `sameStrings(str1, str2):``    ``N ``=` `len``(str1)``    ``M ``=` `len``(str2)` `    ``# Base Condition``    ``if` `(N !``=` `M):``        ``return` `False` `    ``# Stores frequency of characters``    ``# of the str1 and str2``    ``a, b ``=` `[``0``]``*``256``, [``0``]``*``256` `    ``# Traverse strings str1 & str2 and``    ``# store frequencies in a[] and b[]``    ``for` `i ``in` `range``(N):``        ``a[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``+``=` `1``        ``b[``ord``(str2[i]) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``# Check if both strings have``    ``# same characters or not``    ``i ``=` `0``    ``while` `(i < ``256``):` `        ``if` `((a[i] ``=``=` `0` `and` `b[i] ``=``=` `0``) ``or` `(a[i] !``=` `0` `and` `b[i] !``=` `0``)):``            ``i ``+``=` `1` `        ``# If a character is present``        ``# in one and is not in``        ``# another string, return false``        ``else``:``            ``return` `False` `    ``# Sort the array a[] and b[]``    ``a ``=` `sorted``(a)``    ``b ``=` `sorted``(b)` `    ``# Check arrays a and b contain``    ``# the same frequency or not``    ``for` `i ``in` `range``(``256``):` `        ``# If the frequencies are not``        ``# the same after sorting``        ``if` `(a[i] !``=` `b[i]):``            ``return` `False` `    ``# At this point, str1 can``    ``# be converted to str2``    ``return` `True` `  ``# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``S1, S2 ``=` `"cabbba"``, ``"abbccc"``    ``if` `(sameStrings(S1, S2)):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{` `// Function to find if given Strings``// are same or not``static` `bool` `sameStrings(``string` `str1,``                 ``string` `str2)``{``    ``int` `N = str1.Length;``    ``int` `M = str2.Length;` `    ``// Base Condition``    ``if` `(N != M)``    ``{``        ``return` `false``;``    ``}` `    ``// Stores frequency of characters``    ``// of the String str1 and str2``    ``int` `[]a = ``new` `int``;``    ``int` `[]b = ``new` `int``;` `    ``// Traverse Strings str1 & str2 and``    ``// store frequencies in a[] and b[]``    ``for` `(``int` `j = 0; j < N; j++)``    ``{``        ``a[str1[j] - ``'a'``]++;``        ``b[str2[j] - ``'a'``]++;``    ``}` `    ``// Check if both Strings have``    ``// same characters or not``    ``int` `i = 0 ;``    ``while` `(i < 256)``    ``{``        ``if` `((a[i] == 0 && b[i] == 0)``            ``|| (a[i] != 0 && b[i] != 0))``        ``{``            ``i++;``        ``}` `        ``// If a character is present``        ``// in one String and is not in``        ``// another String, return false``        ``else``        ``{``            ``return` `false``;``        ``}``    ``}` `    ``// Sort the array a[] and b[]``    ``Array.Sort(a);``    ``Array.Sort(b);` `    ``// Check arrays a and b contain``    ``// the same frequency or not``    ``for` `(``int` `j = 0; j < 256; j++)``    ``{` `        ``// If the frequencies are not``        ``// the same after sorting``        ``if` `(a[j] != b[j])``            ``return` `false``;``    ``}` `    ``// At this point, str1 can``    ``// be converted to str2``    ``return` `true``;``}` `  ``// Driver Code``  ``static` `public` `void` `Main()``  ``{``    ``string` `S1 = ``"cabbba"``, S2 = ``"abbccc"``;``    ``if` `(sameStrings(S1, S2))``        ``Console.Write(``"YES"` `+``"\n"``);``    ``else``        ``Console.Write(``" NO"` `+``"\n"``);``  ``}``}` `// This code is contributed by code_hunt.`

## Javascript

 ``
Output:
`YES`

Time Complexity: O(N)
Auxiliary Space: O(256)

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