# Sherlock and Anagrams HackerRank Solution

Sherlock and Anagrams HackerRank Solution

In this Sherlock and Anagrams HackerRank solution, Two strings are anagrams of each other if the letters of one string can be rearranged to form the other string. Given a string, find the number of pairs of substrings of the string that are anagrams of each other.

Example

The list of all anagrammatic pairs is  at positions  respectively.

Function Description

Complete the function sherlockAndAnagrams in the editor below.

sherlockAndAnagrams has the following parameter(s):

• string s: a string

Returns

• int: the number of unordered anagrammatic pairs of substrings in

Input Format

The first line contains an integer , the number of queries.
Each of the next  lines contains a string  to analyze.

Constraints

contains only lowercase letters in the range ascii[a-z].

## Sherlock and Anagrams HackerRank solution

I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.

### Problem Solution in C#

public static int sherlockAndAnagrams(string s)
{
List<string> substrings = new();
int count = 0;

for (int i = 1; i < s.Length; i++) {
for (int j = 0; j <= s.Length - i; j++) {
string substring = sortString(s.Substring(j, i));
count += substrings.FindAll(str => str == substring).Count;
}
}

return count;
}

public static string sortString(string s) {
List<char> chars = s.ToCharArray().ToList();
chars.Sort();
return new string(chars.ToArray());
}

### Problem Solution in Python

def sherlockAndAnagrams(s : str) -> int:
hashmap = {}
count = 0
for i in range(1,len(s) + 1):
for j in range(len(s) - i + 1):
substring = str(''.join(sorted(s[j : j + i])))
if substring in hashmap:
count += hashmap[substring]
hashmap[substring] += 1
else:
hashmap[substring] = 1
return count

### Problem Solution in JavaScript

function sherlockAndAnagrams(s) {
let count = 0;
let map = new Map();
for(let i = 0; i < s.length; i++){
for(let j = i + 1; j <= s.length; j++){
let substring = s.substring(i,j).split('').sort().join('');
if(map.has(substring)){
count += map.get(substring);
map.set(substring, map.get(substring) + 1);
} else{
map.set(substring, 1);
}
}
}
return count;
}
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