Given an array s[] of N strings. The task is to find the number of pairs of indices (i, j) such that s[i] is an anagram of s[j].
Examples:
Input: s[] = {“aaab”, “aaba”, “cde”, “dec”}
Output: 2
(“aaab”, “aaba”) and (“cde”, “dec”) are the only valid pairs.Input: s[] = {“ab”, “bc”, “cd”}
Output: 0
Approach: An efficient approach is to sort each string and increase the count of it in a map. For each string in the map, if k is the count of it then (k * (k – 1)) / 2 is the number of valid pairs.
Below is the implementation of the above approach:
C++
// CPP program to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].#include <bits/stdc++.h>using namespace std;// Function to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].int anagram_pairs(vector<string> s, int n){ // To store count of strings map<string, int> mp; // Traverse all strings and store in the map for (int i = 0; i < n; i++) { // Sort the string sort(s[i].begin(), s[i].end()); // Store in the map mp[s[i]]++; } // To store the number of pairs int ans = 0; // Traverse through the map for (auto i = mp.begin(); i != mp.end(); i++) { int k = i->second; // Count the pairs for each string ans += (k * (k - 1)) / 2; } // Return the required answer return ans;}// Driver codeint main(){ vector<string> s = { "aaab", "aaba", "baaa", "cde", "dec" }; int n = s.size(); // Function call cout << anagram_pairs(s, n); return 0;} |
Java
// Java program to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].import java.util.*;class GFG{ // Function to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].static int anagram_pairs(String []s, int n){ // To store count of strings Map<String, Integer> mp = new HashMap<>(); // Traverse all strings and store in the map for (int i = 0; i < n; i++) { // Sort the string char []chArr = s[i].toCharArray(); Arrays.sort(chArr); s[i] = new String(chArr); // Store in the map if(mp.containsKey(s[i])) { mp.put(s[i], mp.get(s[i]) + 1); } else { mp.put(s[i], 1); } } // To store the number of pairs int ans = 0; // Traverse through the map for (Map.Entry<String, Integer> i : mp.entrySet()) { int k = i.getValue(); // Count the pairs for each string ans += (k * (k - 1)) / 2; } // Return the required answer return ans;}// Driver codepublic static void main(String []args){ String [] s = { "aaab", "aaba", "baaa", "cde", "dec" }; int n = s.length; // Function call System.out.println(anagram_pairs(s, n));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to find # number of pairs of integers i, j # such that s[i] is an anagram of s[j]. # Function to find number of pairs of integers # i, j such that s[i] is an anagram of s[j].def anagram_pairs(s, n): # To store the count of sorted strings mp = dict() # Traverse all strings and store in the map for i in range(n): # Sort the string temp_str = "".join(sorted(s[i])) # If string exists in map, increment count # Else create key value pair with count = 1 if temp_str in mp: mp[temp_str] += 1 else: mp[temp_str] = 1 # To store the number of pairs ans = 0 # Traverse through the map for k in mp.values(): # Count the pairs for each string ans += (k * (k - 1)) // 2 # Return the required answer return ans# Driver codeif __name__ == "__main__": s = ["aaab", "aaba", "baaa", "cde", "dec"] n = len(s) print(anagram_pairs(s, n)) # This code is contributed by AnkitRai01 |
C#
// C# program to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].using System;using System.Collections.Generic;class GFG{ // Function to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].static int anagram_pairs(String []s, int n){ // To store count of strings Dictionary<String, int> mp = new Dictionary<String, int>(); // Traverse all strings and store in the map for (int i = 0; i < n; i++) { // Sort the string char []chArr = s[i].ToCharArray(); Array.Sort(chArr); s[i] = new String(chArr); // Store in the map if(mp.ContainsKey(s[i])) { mp[s[i]] = mp[s[i]] + 1; } else { mp.Add(s[i], 1); } } // To store the number of pairs int ans = 0; // Traverse through the map foreach (KeyValuePair<String, int> i in mp) { int k = i.Value; // Count the pairs for each string ans += (k * (k - 1)) / 2; } // Return the required answer return ans;}// Driver codepublic static void Main(String []args){ String [] s = { "aaab", "aaba", "baaa", "cde", "dec" }; int n = s.Length; // Function call Console.WriteLine(anagram_pairs(s, n));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].// Function to find number of pairs of integers// i, j such that s[i] is an anagram of s[j].function anagram_pairs(s, n){ // To store count of strings let mp = new Map(); // Traverse all strings and store in the map for (let i = 0; i < n; i++) { // Sort the string let chArr = s[i].split(""); chArr.sort(); s[i] = chArr.join(""); // Store in the map if(mp.has(s[i])){ mp.set(s[i], mp.get(s[i]) + 1) }else{ mp.set(s[i], 1) } } // To store the number of pairs let ans = 0; // Traverse through the map for (let i of mp) { let k = i[1]; // Count the pairs for each string ans += Math.floor((k * (k - 1)) / 2); } // Return the required answer return ans;}// Driver code let s = [ "aaab", "aaba", "baaa", "cde", "dec" ]; let n = s.length; // Function call document.write(anagram_pairs(s, n));// This code is contributed by _saurabh_jaiswal</script> |
Output:
4
Time Complexity: O(n2 * logn)
Auxiliary Space: O(n)
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