# 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 : 0Explanation: Both String contains identical charactersInput :       s1 = "ddcf"       s2 = "cedk"Output : 2Explanation : Here, we need to change two charactersin 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. `

## Javascript

 ``

## PHP

 ``

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++

 `// C++ Program to find minimum number ``// of manipulations required to make ``// two strings identical ``#include ``#include ``#include ` `using` `namespace` `std;` `int` `countManipulations(string s1, string s2)``{``    ``// Sort the characters of both strings``    ``sort(s1.begin(), s1.end());``    ``sort(s2.begin(), s2.end());` `    ``int` `i = 0, j = 0, count = 0;` `    ``// Compare characters in sorted strings``    ``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++;``        ``}``    ``}` `    ``// Count the remaining characters in both strings``    ``while` `(i < s1.size())``    ``{``        ``i++;``        ``count++;``    ``}` `    ``while` `(j < s2.size())``    ``{``        ``j++;``        ``count++;``    ``}` `    ``// Return the count divided by 2``    ``return` `count / 2;``}` `int` `main()``{``    ``string s1 = ``"ddcf"``;``    ``string s2 = ``"cedk"``;``    ``cout << countManipulations(s1, s2) << endl;``    ``return` `0;``}`

## Java

 `// Java Program to find minimum number ``// of manipulations required to make ``// two strings identical ``import` `java.util.Arrays;` `public` `class` `StringManipulations {``    ``public` `static` `int` `countManipulations(String s1, String s2) {``        ``// Convert strings to character arrays and sort them``        ``char``[] s1Array = s1.toCharArray();``        ``char``[] s2Array = s2.toCharArray();` `        ``Arrays.sort(s1Array);``        ``Arrays.sort(s2Array);` `        ``int` `i = ``0``, j = ``0``, count = ``0``;` `        ``// Compare characters in sorted arrays``        ``while` `(i < s1Array.length && j < s2Array.length) {``            ``if` `(s1Array[i] == s2Array[j]) {``                ``i++;``                ``j++;``            ``} ``else` `if` `(s1Array[i] < s2Array[j]) {``                ``i++;``                ``count++;``            ``} ``else` `{``                ``j++;``                ``count++;``            ``}``        ``}` `        ``// Count the remaining characters in both arrays``        ``count += (s1Array.length - i) + (s2Array.length - j);``        ``return` `count / ``2``;``    ``}` `    ``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 ``def` `count_manipulations(s1, s2):``    ``# Convert strings to lists and sort them``    ``s1_list ``=` `list``(s1)``    ``s2_list ``=` `list``(s2)` `    ``s1_list.sort()``    ``s2_list.sort()` `    ``i, j, count ``=` `0``, ``0``, ``0` `    ``# Compare characters in sorted lists``    ``while` `i < ``len``(s1_list) ``and` `j < ``len``(s2_list):``        ``if` `s1_list[i] ``=``=` `s2_list[j]:``            ``i ``+``=` `1``            ``j ``+``=` `1``        ``elif` `s1_list[i] < s2_list[j]:``            ``i ``+``=` `1``            ``count ``+``=` `1``        ``else``:``            ``j ``+``=` `1``            ``count ``+``=` `1` `    ``# Count the remaining characters in both lists``    ``count ``+``=` `len``(s1_list) ``-` `i ``+` `len``(s2_list) ``-` `j``    ``return` `count ``/``/` `2` `if` `__name__ ``=``=` `"__main__"``:``    ``s1 ``=` `"ddcf"``    ``s2 ``=` `"cedk"``    ``print``(count_manipulations(s1, s2))`

## C#

 `// C# Program to find minimum number ``// of manipulations required to make ``// two strings identical ``using` `System;` `public` `class` `StringManipulations``{``    ``public` `static` `int` `CountManipulations(``string` `s1, ``string` `s2)``    ``{``        ``// Convert strings to character arrays and sort them``        ``char``[] s1Array = s1.ToCharArray();``        ``char``[] s2Array = s2.ToCharArray();` `        ``Array.Sort(s1Array);``        ``Array.Sort(s2Array);` `        ``int` `i = 0, j = 0, count = 0;` `        ``// Compare characters in sorted arrays``        ``while` `(i < s1Array.Length && j < s2Array.Length)``        ``{``            ``if` `(s1Array[i] == s2Array[j])``            ``{``                ``i++;``                ``j++;``            ``}``            ``else` `if` `(s1Array[i] < s2Array[j])``            ``{``                ``i++;``                ``count++;``            ``}``            ``else``            ``{``                ``j++;``                ``count++;``            ``}``        ``}` `        ``// Count the remaining characters in both arrays``        ``count += (s1Array.Length - i) + (s2Array.Length - j);``        ``return` `count / 2;``    ``}` `    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``string` `s1 = ``"ddcf"``;``        ``string` `s2 = ``"cedk"``;``        ``Console.WriteLine(CountManipulations(s1, s2));``    ``}``}`

## Javascript

 `// Javascript Program to find minimum number ``// of manipulations required to make ``// two strings identical ``function` `countManipulations(s1, s2) {``    ``// Convert strings to arrays and sort them``    ``const s1Array = s1.split(``''``).sort();``    ``const s2Array = s2.split(``''``).sort();` `    ``let i = 0, j = 0, count = 0;` `    ``// Compare characters in sorted arrays``    ``while` `(i < s1Array.length && j < s2Array.length) {``        ``if` `(s1Array[i] === s2Array[j]) {``            ``i++;``            ``j++;``        ``} ``else` `if` `(s1Array[i] < s2Array[j]) {``            ``i++;``            ``count++;``        ``} ``else` `{``            ``j++;``            ``count++;``        ``}``    ``}` `    ``// Count the remaining characters in both arrays``    ``count += (s1Array.length - i) + (s2Array.length - j);``    ``return` `Math.floor(count / 2);``}` `const s1 = ``"ddcf"``;``const s2 = ``"cedk"``;``console.log(countManipulations(s1, s2));`

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.

Previous
Next