Count of total anagram substrings Last Updated : 28 Dec, 2023 Comments Improve Suggest changes Like Article Like Report Given a string of lower alphabet characters, count total substring of this string which are anagram to each other. Examples: Input : str = “xyyx”Output : 4Total substrings of this string whichare anagram to each other are 4 which can be enumerated as,{“x”, “x”}, {"y", "y"}, {“xy”, “yx”}, {“xyy”, “yyx”}Input : str = "geeg"Output : 4The 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> Output4 Time complexity : O(N2) Auxiliary Space : O(N) Comment More infoAdvertise with us Next Article Analysis of Algorithms U Utkarsh Trivedi Improve Article Tags : Misc Strings DSA Practice Tags : MiscStrings Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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