Count of total anagram substrings

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 <bits/stdc++.h>
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<vector<int>, int> mp;

    // loop for starting index of substring
    for (int i=0; i<N; i++)
    {
        vector<int> freq(MAX_CHAR, 0);

        // loop for length of substring
        for (int j=i; j<N; j++)
        {
            // update freq array of current
            // substring
            freq[toNum(str[j])]++;

            // increase count corresponding
            // to this freq array
            mp[freq]++;
        }
    }

    // loop over all different freq array and
    // aggregate substring count
    int result = 0;
    for (auto it=mp.begin(); it!=mp.end(); it++)
    {
        int freq = it->second;
        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<String, Integer> 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

<script>

// JavaScript code to implement the approach

function countOfAnagramSubstring(s){

    // Returns total number of anagram
    // substrings in s
    let n = s.length
    let mp = new Map()

    // loop for length of substring
    for(let i=0;i<n;i++){
        let sb = ''
        for(let j=i;j<n;j++){
            sb = (sb + s[j]).split('').sort().join('')
            if(mp.has(sb))
                mp.set(sb ,mp.get(sb)+1)

            // increase count corresponding
            // to this dict array
            else mp.set(sb, 1)
        }
    }

    let anas = 0

    // loop over all different dictionary
    // items and aggregate substring count
    for(let [k, v] of mp){
        anas += Math.floor((v*(v-1))/2)
    }
    return anas
}

// Driver Code
let s = "xyyx"
document.write(countOfAnagramSubstring(s),"</br>")


// This code is contributed by shinjanpatra

</script>

Output

4

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

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