# Remove minimum number of characters so that two strings become anagram

Given two strings in lowercase, the task is to make them anagram. The only allowed operation is to remove a character from any string. Find minimum number of characters to be deleted to make both the strings anagram?
If two strings contains same data set in any order then strings are called Anagrams.

Examples :

```Input : str1 = "bcadeh" str2 = "hea"
Output: 3
We need to remove b, c and d from str1.

Input : str1 = "cddgk" str2 = "gcd"
Output: 2

Input : str1 = "bca" str2 = "acb"
Output: 0
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to make character count arrays for both the strings and store frequency of each character. Now iterate the count arrays of both strings and difference in frequency of any character abs(count1[str1[i]-‘a’] – count2[str2[i]-‘a’]) in both the strings is the number of character to be removed in either string.

## C++

 `// C++ program to find minimum number of characters ` `// to be removed to make two strings anagram. ` `#include ` `using` `namespace` `std; ` `const` `int` `CHARS = 26; ` ` `  `// function to calculate minimum numbers of characters ` `// to be removed to make two strings anagram ` `int` `remAnagram(string str1, string str2) ` `{ ` `    ``// make hash array for both string and calculate ` `    ``// frequency of each character ` `    ``int` `count1[CHARS] = {0}, count2[CHARS] = {0}; ` ` `  `    ``// count frequency of each character in first string ` `    ``for` `(``int` `i=0; str1[i]!=``'\0'``; i++) ` `        ``count1[str1[i]-``'a'``]++; ` ` `  `    ``// count frequency of each character in second string ` `    ``for` `(``int` `i=0; str2[i]!=``'\0'``; i++) ` `        ``count2[str2[i]-``'a'``]++; ` ` `  `    ``// traverse count arrays to find number of characters ` `    ``// to be removed ` `    ``int` `result = 0; ` `    ``for` `(``int` `i=0; i<26; i++) ` `        ``result += ``abs``(count1[i] - count2[i]); ` `    ``return` `result; ` `} ` ` `  `// Driver program to run the case ` `int` `main() ` `{ ` `    ``string str1 = ``"bcadeh"``, str2 = ``"hea"``; ` `    ``cout << remAnagram(str1, str2); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find minimum number of ` `// characters to be removed to make two ` `// strings anagram. ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// function to calculate minimum numbers ` `    ``// of characters to be removed to make ` `    ``// two strings anagram ` `    ``static` `int` `remAnagram(String str1, String str2) ` `    ``{ ` `        ``// make hash array for both string  ` `        ``// and calculate frequency of each ` `        ``// character ` `        ``int` `count1[] = ``new` `int``[``26``];  ` `        ``int` `count2[] = ``new` `int``[``26``]; ` ` `  `        ``// count frequency of each character  ` `        ``// in first string ` `        ``for` `(``int` `i = ``0``; i < str1.length() ; i++) ` `            ``count1[str1.charAt(i) -``'a'``]++; ` `     `  `        ``// count frequency of each character  ` `        ``// in second string ` `        ``for` `(``int` `i = ``0``; i < str2.length() ; i++) ` `            ``count2[str2.charAt(i) -``'a'``]++; ` ` `  `        ``// traverse count arrays to find  ` `        ``// number of characters to be removed ` `        ``int` `result = ``0``; ` `        ``for` `(``int` `i = ``0``; i < ``26``; i++) ` `            ``result += Math.abs(count1[i] - ` `                               ``count2[i]); ` `        ``return` `result; ` `    ``} ` ` `  `    ``// Driver program to run the case ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``String str1 = ``"bcadeh"``, str2 = ``"hea"``; ` `        ``System.out.println(remAnagram(str1, str2)); ` `    ``} ` `} ` `// This code is contributed by Prerna Saini `

## Python3

 `# Python 3 program to find minimum  ` `# number of characters ` `# to be removed to make two  ` `# strings anagram. ` ` `  `CHARS ``=` `26` ` `  `# function to calculate minimum  ` `# numbers of characters ` `# to be removed to make two  ` `# strings anagram ` `def` `remAnagram(str1, str2): ` ` `  `    ``# make hash array for both string  ` `    ``# and calculate ` `    ``# frequency of each character ` `    ``count1 ``=` `[``0``]``*``CHARS ` `    ``count2 ``=` `[``0``]``*``CHARS ` ` `  `    ``# count frequency of each character  ` `    ``# in first string ` `    ``i ``=` `0` `    ``while` `i < ``len``(str1): ` `        ``count1[``ord``(str1[i])``-``ord``(``'a'``)] ``+``=` `1` `        ``i ``+``=` `1` ` `  `    ``# count frequency of each character  ` `    ``# in second string ` `    ``i ``=``0` `    ``while` `i < ``len``(str2): ` `        ``count2[``ord``(str2[i])``-``ord``(``'a'``)] ``+``=` `1` `        ``i ``+``=` `1` ` `  `    ``# traverse count arrays to find  ` `    ``# number of characters ` `    ``# to be removed ` `    ``result ``=` `0` `    ``for` `i ``in` `range``(``26``): ` `        ``result ``+``=` `abs``(count1[i] ``-` `count2[i]) ` `    ``return` `result ` ` `  `# Driver program to run the case ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``str1 ``=` `"bcadeh"` `    ``str2 ``=` `"hea"` `    ``print``(remAnagram(str1, str2)) ` `     `  `# This code is contributed by  ` `# ChitraNayal `

## C#

 `// C# program to find minimum  ` `// number of characters to be  ` `// removed to make two strings  ` `// anagram. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// function to calculate  ` `    ``// minimum numbers of  ` `    ``// characters to be removed  ` `    ``// to make two strings anagram ` `    ``static` `int` `remAnagram(``string` `str1,  ` `                          ``string` `str2) ` `    ``{ ` `        ``// make hash array for both  ` `        ``// string and calculate frequency  ` `        ``// of each character ` `        ``int` `[]count1 = ``new` `int``;  ` `        ``int` `[]count2 = ``new` `int``; ` ` `  `        ``// count frequency of each  ` `        ``// character in first string ` `        ``for` `(``int` `i = 0; i < str1.Length ; i++) ` `            ``count1[str1[i] -``'a'``]++; ` `     `  `        ``// count frequency of each   ` `        ``// character in second string ` `        ``for` `(``int` `i = 0; i < str2.Length ; i++) ` `            ``count2[str2[i] -``'a'``]++; ` ` `  `        ``// traverse count arrays to  ` `        ``// find number of characters  ` `        ``// to be removed ` `        ``int` `result = 0; ` `        ``for` `(``int` `i = 0; i < 26; i++) ` `            ``result += Math.Abs(count1[i] - ` `                               ``count2[i]); ` `        ``return` `result; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``string` `str1 = ``"bcadeh"``,  ` `               ``str2 = ``"hea"``; ` `        ``Console.Write(remAnagram(str1, str2)); ` `    ``} ` `} ` ` `  `// This code is contributed ` `// by nitin mittal. `

## PHP

 `

Output :

```3
```

Time Complexity : O(n)
Auxiliary space : O(ALPHABET-SIZE)

The above solution can be optimized to work with single count array.

## C++

 `// C++ implementation to count number of deletions ` `// required from two strings to create an anagram  ` `#include ` `using` `namespace` `std; ` `const` `int` `CHARS = 26; ` ` `  `int` `countDeletions(string str1, string str2) ` `{  ` `    ``int` `arr[CHARS] = {0}; ` `    ``for` `(``int` `i = 0; i < str1.length(); i++)  ` `        ``arr[str1[i] - ``'a'``]++;  ` `     `  `    ``for` `(``int` `i = 0; i < str2.length(); i++)  ` `        ``arr[str2[i] - ``'a'``]--; ` `     `  `    ``long` `long` `int` `ans = 0; ` `    ``for``(``int` `i = 0; i < CHARS; i++) ` `        ``ans +=``abs``(arr[i]); ` `    ``return` `ans; ` `} ` ` `  `int` `main() { ` `    ``string str1 = ``"bcadeh"``, str2 = ``"hea"``; ` `    ``cout << countDeletions(str1, str2); ` `    ``return` `0; ` `}  `

## Java

 `// Java implementation to count number of deletions ` `// required from two strings to create an anagram  ` ` `  `class` `GFG { ` ` `  `    ``final` `static` `int` `CHARS = ``26``; ` ` `  `    ``static` `int` `countDeletions(String str1, String str2) { ` `        ``int` `arr[] = ``new` `int``[CHARS]; ` `        ``for` `(``int` `i = ``0``; i < str1.length(); i++) { ` `            ``arr[str1.charAt(i) - ``'a'``]++; ` `        ``} ` ` `  `        ``for` `(``int` `i = ``0``; i < str2.length(); i++) { ` `            ``arr[str2.charAt(i) - ``'a'``]--; ` `        ``} ` ` `  `        ``int` `ans = ``0``; ` `        ``for` `(``int` `i = ``0``; i < CHARS; i++) { ` `            ``ans += Math.abs(arr[i]); ` `        ``} ` `        ``return` `ans; ` `    ``} ` ` `  `    ``static` `public` `void` `main(String[] args) { ` `        ``String str1 = ``"bcadeh"``, str2 = ``"hea"``; ` `        ``System.out.println(countDeletions(str1, str2)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to find minimum  ` `# number of characters to be  ` `# removed to make two strings  ` `# anagram. ` ` `  `# function to calculate minimum  ` `# numbers of characters to be  ` `# removed to make two strings anagram  ` `def` `makeAnagram(a, b): ` `    ``buffer` `=` `[``0``] ``*` `26` `    ``for` `char ``in` `a: ` `        ``buffer``[``ord``(char) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``for` `char ``in` `b: ` `        ``buffer``[``ord``(char) ``-` `ord``(``'a'``)] ``-``=` `1` `    ``return` `sum``(``map``(``abs``, ``buffer``)) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``str1 ``=` `"bcadeh"` `    ``str2 ``=` `"hea"` `    ``print``(makeAnagram(str1, str2)) ` `     `  `# This code is contributed  ` `# by Raghib Ahsan `

## C#

 `     `  `// C# implementation to count number of deletions ` `// required from two strings to create an anagram  ` `using` `System; ` `public` `class` `GFG { ` `  `  `    ``readonly` `static` `int` `CHARS = 26; ` `  `  `    ``static` `int` `countDeletions(String str1, String str2) { ` `        ``int` `[]arr = ``new` `int``[CHARS]; ` `        ``for` `(``int` `i = 0; i < str1.Length; i++) { ` `            ``arr[str1[i]- ``'a'``]++; ` `        ``} ` `  `  `        ``for` `(``int` `i = 0; i < str2.Length; i++) { ` `            ``arr[str2[i] - ``'a'``]--; ` `        ``} ` `  `  `        ``int` `ans = 0; ` `        ``for` `(``int` `i = 0; i < CHARS; i++) { ` `            ``ans += Math.Abs(arr[i]); ` `        ``} ` `        ``return` `ans; ` `    ``} ` `  `  `    ``static` `public` `void` `Main() { ` `        ``String str1 = ``"bcadeh"``, str2 = ``"hea"``; ` `        ``Console.WriteLine(countDeletions(str1, str2)); ` `    ``} ` `} ` `  ``//This code is contributed by PrinciRaj1992 `

## PHP

 ` `

Output :

```3
```

Thanks to vishal9619 for suggesting this optimized solution.

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