# Count of total anagram substrings

• Difficulty Level : Hard
• Last Updated : 08 Jul, 2022

Given a string of lower alphabet characters, count total substring of this string which are anagram to each other.

Examples:

```Input  : str = “xyyx”
Output : 4
Total substrings of this string which
are anagram to each other are 4 which
can be enumerated as,
{“x”, “x”}, {"y", "y"}, {“xy”, “yx”},
{“xyy”, “yyx”}

Input  : str = "geeg"
Output : 4```

The idea is to create a map. We use character frequencies as keys and corresponding counts as values. We can solve this problem by iterating over all substrings and counting frequencies of characters in every substring. We can update frequencies of characters while looping over substrings i.e. there won’t be an extra loop for counting frequency of characters. In below code, a map of key ‘vector type’ and value ‘int type’ is taken for storing occurrence of ‘frequency array of length 26’ of substring characters.

Once occurrence ‘o’ of each frequency array is stored, total anagrams will be the sum of o*(o-1)/2 for all different frequency arrays because if a particular substring has ‘o’ anagrams in string total o*(o-1)/2 anagram pairs can be formed. Below is the implementation of above idea.

Implementation:

## C++

 `// C++ program to count total anagram``// substring of a string``#include ``using` `namespace` `std;` `// Total number of lowercase characters``#define MAX_CHAR 26` `// Utility method to return integer index``// of character 'c'``int` `toNum(``char` `c)``{``    ``return` `(c - ``'a'``);``}` `// Returns count of total number of anagram``// substrings of string str``int` `countOfAnagramSubstring(string str)``{``    ``int` `N = str.length();` `    ``// To store counts of substrings with given``    ``// set of frequencies.``    ``map, ``int``> mp;` `    ``// loop for starting index of substring``    ``for` `(``int` `i=0; i freq(MAX_CHAR, 0);` `        ``// loop for length of substring``        ``for` `(``int` `j=i; jsecond;``        ``result += ((freq) * (freq-1))/2;``    ``}``    ``return` `result;``}` `//  Driver code to test above methods``int` `main()``{``    ``string str = ``"xyyx"``;``    ``cout << countOfAnagramSubstring(str) << endl;``    ``return` `0;``}`

## Java

 `import` `java.util.Arrays;``import` `java.util.HashMap;` `public` `class` `anagramPairCount {``    ``public` `static` `void` `main(String[] args) {``        ``subString(``"kkkk"``);``    ``}` `    ``static` `void` `subString(String s){``        ``HashMap map= ``new` `HashMap<>();` `        ``for``(``int` `i = ``0``; i < s.length(); i++){``            ``for``(``int` `j = i; j < s.length(); j++){``                ``char``[] valC = s.substring(i, j+``1``).toCharArray();``                ``Arrays.sort(valC);``                ``String val = ``new` `String(valC);``                ``if` `(map.containsKey(val))``                    ``map.put(val, map.get(val)+``1``);``                ``else``                    ``map.put(val, ``1``);``            ``}``        ``}``        ``int` `anagramPairCount = ``0``;``        ``for``(String key: map.keySet()){``            ``int` `n = map.get(key);``            ``anagramPairCount += (n * (n-``1``))/``2``;``        ``}``        ``System.out.println(anagramPairCount);``    ``}``}`

## Python3

 `# Python3 program to count total anagram``# substring of a string``def` `countOfAnagramSubstring(s):``    ` `    ``# Returns total number of anagram``    ``# substrings in s``    ``n ``=` `len``(s)``    ``mp ``=` `dict``()``    ` `    ``# loop for length of substring``    ``for` `i ``in` `range``(n):``        ``sb ``=` `''``        ``for` `j ``in` `range``(i, n):``            ``sb ``=` `''.join(``sorted``(sb ``+` `s[j]))``            ``mp[sb] ``=` `mp.get(sb, ``0``)``            ` `            ``# increase count corresponding``            ``# to this dict array``            ``mp[sb] ``+``=` `1` `    ``anas ``=` `0``    ` `    ``# loop over all different dictionary``    ``# items and aggregate substring count``    ``for` `k, v ``in` `mp.items():``        ``anas ``+``=` `(v``*``(v``-``1``))``/``/``2``    ``return` `anas` `# Driver Code``s ``=` `"xyyx"``print``(countOfAnagramSubstring(s))` `# This code is contributed by fgaim`

## C#

 `using` `System;``using` `System.Collections.Generic;`  `public` `class` `anagramPairCount {``    ``public` `static` `void` `Main() {``        ``subString(``"kkkk"``);``    ``}` `    ``static` `void` `subString(String s){``        ``Dictionary<``string``, ``int``> map= ``new` `Dictionary<``string``, ``int``>();` `        ``for``(``int` `i = 0; i < s.Length; i++){``            ``for``(``int` `j = i; j < s.Length; j++){``                ``char``[] valC = s.Substring(i, j+1-i).ToCharArray();``                ``Array.Sort(valC);``                ``string` `val = ``new` `string``(valC);``                ``if` `(map.ContainsKey(val))``                    ``map[val]=map[val]+1;``                ``else``                    ``map.Add(val, 1);``            ``}``        ``}``        ``int` `anagramPairCount = 0;``        ``foreach``(``string` `key ``in` `map.Keys){``            ``int` `n = map[key];``            ``anagramPairCount += (n * (n-1))/2;``        ``}``        ``Console.Write(anagramPairCount);``    ``}``}` `// This code is contributed by AbhiThakur`

## Javascript

 ``

Output

`4`

Time complexity : O(N2logN)
Auxiliary Space : O(N)

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