GeeksforGeeks App
Open App
Browser
Continue

# Minimum Number of Manipulations required to make two Strings Anagram Without Deletion of Character

Given two strings s1 and s2, we need to find the minimum number of manipulations required to make two strings anagram without deleting any character.

Note:- The anagram strings have same set of characters, sequence of characters can be different.

If deletion of character is allowed and cost is given, refer to Minimum Cost To Make Two Strings Identical
Question Source: Yatra.com Interview Experience | Set 7

Examples:

```Input :
s1 = "aba"
s2 = "baa"
Output : 0
Explanation: Both String contains identical characters

Input :
s1 = "ddcf"
s2 = "cedk"
Output : 2
Explanation : Here, we need to change two characters
in either of the strings to make them identical. We
can change 'd' and 'f' in s1 or 'e' and 'k' in s2.```

Assumption: Length of both the Strings is considered similar

Implementation:

## C++

 `// C++ Program to find minimum number``// of manipulations required to make``// two strings identical``#include ``using` `namespace` `std;` `    ``// Counts the no of manipulations``    ``// required``    ``int` `countManipulations(string s1, string s2)``    ``{``        ` `        ``int` `count = 0;` `        ``// store the count of character``        ``int` `char_count[26];``        ` `        ``for` `(``int` `i = 0; i < 26; i++)``        ``{``            ``char_count[i] = 0;``        ``}` `        ``// iterate though the first String``        ``// and update count``        ``for` `(``int` `i = 0; i < s1.length(); i++)``            ``char_count[s1[i] - ``'a'``]++;` `        ``// iterate through the second string``        ``// update char_count.``        ``// if character is not found in``        ``// char_count then increase count``        ``for` `(``int` `i = 0; i < s2.length(); i++)``        ``{``            ``char_count[s2[i] - ``'a'``]--;      ``        ``}``      ` `        ``for``(``int` `i = 0; i < 26; ++i)``        ``{``          ``if``(char_count[i] != 0)``          ``{``            ``count+=``abs``(char_count[i]);``          ``}``        ``}``        ``return` `count / 2;``    ``}` `    ``// Driver code``    ``int` `main()``    ``{` `        ``string s1 = ``"ddcf"``;``        ``string s2 = ``"cedk"``;``        ` `        ``cout<

## Java

 `// Java Program to find minimum number of manipulations``// required to make two strings identical``public` `class` `Similar_strings {` `    ``// Counts the no of manipulations required``    ``static` `int` `countManipulations(String s1, String s2)``    ``{``        ``int` `count = ``0``;` `        ``// store the count of character``        ``int` `char_count[] = ``new` `int``[``26``];` `        ``// iterate though the first String and update``        ``// count``        ``for` `(``int` `i = ``0``; i < s1.length(); i++)``            ``char_count[s1.charAt(i) - ``'a'``]++;       ` `        ``// iterate through the second string``        ``// update char_count.``        ``// if character is not found in char_count``        ``// then increase count``        ``for` `(``int` `i = ``0``; i < s2.length(); i++)``        ``{``            ``char_count[s2.charAt(i) - ``'a'``]--;``        ``}``      ` `        ``for``(``int` `i = ``0``; i < ``26``; ++i)``        ``{``          ``if``(char_count[i] != ``0``)``          ``{``            ``count+= Math.abs(char_count[i]);``          ``}``        ``}``        ` `        ``return` `count / ``2``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``String s1 = ``"ddcf"``;``        ``String s2 = ``"cedk"``;``        ``System.out.println(countManipulations(s1, s2));``    ``}``}`

## Python3

 `# Python3 Program to find minimum number``# of manipulations required to make``# two strings identical` `# Counts the no of manipulations``# required``def` `countManipulations(s1, s2):``    ` `    ``count ``=` `0` `    ``# store the count of character``    ``char_count ``=` `[``0``] ``*` `26``    ` `    ``for` `i ``in` `range``(``26``):``        ``char_count[i] ``=` `0` `    ``# iterate though the first String``    ``# and update count``    ``for` `i ``in` `range``(``len``( s1)):``        ``char_count[``ord``(s1[i]) ``-``                   ``ord``(``'a'``)] ``+``=` `1` `    ``# iterate through the second string``    ``# update char_count.``    ``# if character is not found in``    ``# char_count then increase count``    ``for` `i ``in` `range``(``len``(s2)):``        ``char_count[``ord``(s2[i]) ``-` `ord``(``'a'``)] ``-``=` `1``        ` `    ``for` `i ``in` `range``(``26``):``        ``if` `char_count[i] !``=` `0``:``            ``count ``+``=` `abs``(char_count[i])``        `  `    ``return` `count ``/` `2` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``s1 ``=` `"ddcf"``    ``s2 ``=` `"cedk"``    ` `    ``print``(countManipulations(s1, s2))` `# This code is contributed by ita_c`

## C#

 `// C# Program to find minimum number``// of manipulations required to make``// two strings identical``using` `System;` `public` `class` `GFG {` `    ``// Counts the no of manipulations``    ``// required``    ``static` `int` `countManipulations(``string` `s1,``                                  ``string` `s2)``    ``{``        ``int` `count = 0;` `        ``// store the count of character``        ``int` `[]char_count = ``new` `int``[26];` `        ``// iterate though the first String``        ``// and update count``        ``for` `(``int` `i = 0; i < s1.Length; i++)``            ``char_count[s1[i] - ``'a'``]++;` `        ``// iterate through the second string``        ``// update char_count.``        ``// if character is not found in``        ``// char_count then increase count``        ``for` `(``int` `i = 0; i < s2.Length; i++)``            ``char_count[s2[i] - ``'a'``]--;``      ` `        ``for``(``int` `i = 0; i < 26; ++i)``        ``{``            ``if``(char_count[i] != 0)``            ``{``              ``count+= Math.Abs(char_count[i]);``            ``}``        ``}``        ` `        ``return` `count / 2;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{` `        ``string` `s1 = ``"ddcf"``;``        ``string` `s2 = ``"cedk"``;``        ` `        ``Console.WriteLine(``            ``countManipulations(s1, s2));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`2`

Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(1).

### Approach 2(Sorting and Comparing Characters Approach):

• Sort both strings in lexicographic order.
• Initialize a counter variable to zero.
• Iterate over both strings simultaneously and compare the characters at each position. If they’re not equal, increment the counter.
• Return the counter value divided by 2, since each manipulation affects two characters.

## C++

 `#include ``#include ``#include ` `using` `namespace` `std;` `int` `countManipulations(string s1, string s2)``{``    ``sort(s1.begin(), s1.end());``    ``sort(s2.begin(), s2.end());` `    ``int` `i = 0, j = 0, count = 0;` `    ``while` `(i < s1.size() && j < s2.size())``    ``{``        ``if` `(s1[i] == s2[j])``        ``{``            ``i++;``            ``j++;``        ``}``        ``else` `if` `(s1[i] < s2[j])``        ``{``            ``i++;``            ``count++;``        ``}``        ``else``        ``{``            ``j++;``            ``count++;``        ``}``    ``}` `    ``while` `(i < s1.size())``    ``{``        ``i++;``        ``count++;``    ``}` `    ``while` `(j < s2.size())``    ``{``        ``j++;``        ``count++;``    ``}` `    ``return` `count / 2;``}` `int` `main()``{``    ``string s1 = ``"ddcf"``;``    ``string s2 = ``"cedk"``;``    ``cout << countManipulations(s1, s2) << endl;``    ``return` `0;``}`

Output

```2
```

Time Complexity:

• Sorting the strings takes O(n log n) time, where n is the length of the strings.
• Iterating over the sorted strings takes O(n) time.
• Therefore, the overall time complexity is O(n log n).

Space Complexity: O(n), where n is the length of the longer string. This is because we need to store the sorted versions of both strings in memory, which can take up to n space.

This article is contributed by Sumit Ghosh and improved by Md Istakhar Ansari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up