Count of index pairs with equal elements in an array

Given an array of n elements. The task is to count the total number of indices (i, j) such that arr[i] = arr[j] and i != j

Examples :

Input : arr[] = {1, 1, 2}
Output : 1
As arr[0] = arr[1], the pair of indices is (0, 1)

Input : arr[] = {1, 1, 1}
Output : 3
As arr[0] = arr[1], the pair of indices is (0, 1), 
(0, 2) and (1, 2)

Input : arr[] = {1, 2, 3}
Output : 0

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Brute Force):
For each index i, find element after it with same value as arr[i]. Below is the implementation of this approach:

C++

// C++ program to count of pairs with equal 
// elements in an array. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Return the number of pairs with equal 
// values. 
int countPairs(int arr[], int n) 
{ 
    int ans = 0; 
  
    // for each index i and j 
    for (int i = 0; i < n; i++) 
        for (int j = i+1; j < n; j++) 
  
            // finding the index with same 
            // value but different index. 
            if (arr[i] == arr[j]) 
                ans++; 
    return ans; 
} 
  
// Driven Program 
int main() 
{ 
    int arr[] = { 1, 1, 2 }; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    cout << countPairs(arr, n) << endl; 
    return 0; 
} 

Java

// Java program to count of pairs with equal 
// elements in an array. 
class GFG { 
          
    // Return the number of pairs with equal 
    // values. 
    static int countPairs(int arr[], int n) 
    { 
        int ans = 0; 
      
        // for each index i and j 
        for (int i = 0; i < n; i++) 
            for (int j = i+1; j < n; j++) 
      
                // finding the index with same 
                // value but different index. 
                if (arr[i] == arr[j]) 
                    ans++; 
        return ans; 
    } 
      
    //driver code 
    public static void main (String[] args) 
    { 
        int arr[] = { 1, 1, 2 }; 
        int n = arr.length; 
          
        System.out.println(countPairs(arr, n)); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python3 program to 
# count of pairs with equal 
# elements in an array. 
  
# Return the number of 
# pairs with equal values. 
def countPairs(arr, n): 
  
    ans = 0
  
    # for each index i and j 
    for i in range(0 , n): 
        for j in range(i + 1, n): 
  
            # finding the index  
            # with same value but 
            # different index. 
            if (arr[i] == arr[j]): 
                ans += 1
    return ans 
  
# Driven Code 
arr = [1, 1, 2 ] 
n = len(arr) 
print(countPairs(arr, n)) 
  
# This code is contributed  
# by Smitha 

C#

// C# program to count of pairs with equal 
// elements in an array. 
using System; 
  
class GFG { 
          
    // Return the number of pairs with equal 
    // values. 
    static int countPairs(int []arr, int n) 
    { 
        int ans = 0; 
      
        // for each index i and j 
        for (int i = 0; i < n; i++) 
            for (int j = i+1; j < n; j++) 
      
                // finding the index with same 
                // value but different index. 
                if (arr[i] == arr[j]) 
                    ans++; 
        return ans; 
    } 
      
    // Driver code 
    public static void Main () 
    { 
        int []arr = { 1, 1, 2 }; 
        int n = arr.Length; 
          
        Console.WriteLine(countPairs(arr, n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to count of  
// pairs with equal elements 
// in an array. 
  
// Return the number of pairs 
// with equal values. 
function countPairs( $arr, $n) 
{ 
    $ans = 0; 
  
    // for each index i and j 
    for ( $i = 0; $i < $n; $i++) 
        for ( $j = $i + 1; $j < $n; $j++) 
  
            // finding the index with same 
            // value but different index. 
            if ($arr[$i] == $arr[$j]) 
                $ans++; 
    return $ans; 
} 
  
// Driven Code 
$arr = array( 1, 1, 2 ); 
$n = count($arr); 
echo countPairs($arr, $n) ; 
  
// This code is contributed by anuj_67. 
?> 

Output :

Time Complexity : O(n²)

Method 2 (Efficient approach):
The idea is to count the frequency of each number and then find the number of pairs with equal elements. Suppose, a number x appears k times at index i₁, i₂,….,i_k. Then pick any two indexes i_x and i_y which will be counted as 1 pair. Similarly, i_y and i_x can also be pair. So, choose ⁿC₂ is the number of pairs such that arr[i] = arr[j] = x.

Below is the implementation of this approach:

C++

// C++ program to count of index pairs with 
// equal elements in an array. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Return the number of pairs with equal 
// values. 
int countPairs(int arr[], int n) 
{ 
    unordered_map<int, int> mp; 
  
    // Finding frequency of each number. 
    for (int i = 0; i < n; i++) 
        mp[arr[i]]++; 
  
    // Calculating pairs of each value. 
    int ans = 0; 
    for (auto it=mp.begin(); it!=mp.end(); it++) 
    { 
        int count = it->second; 
        ans += (count * (count - 1))/2; 
    } 
  
    return ans; 
} 
  
// Driven Program 
int main() 
{ 
    int arr[] = {1, 1, 2}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    cout << countPairs(arr, n) << endl; 
    return 0; 
} 

Java

// Java program to count of index pairs with 
// equal elements in an array. 
import java.util.*; 
  
class GFG { 
  
    public static int countPairs(int arr[], int n) 
    {  
        //A method to return number of pairs with 
        // equal values 
          
        HashMap<Integer,Integer> hm = new HashMap<>(); 
          
        // Finding frequency of each number. 
        for(int i = 0; i < n; i++) 
        { 
        if(hm.containsKey(arr[i])) 
            hm.put(arr[i],hm.get(arr[i]) + 1); 
        else
            hm.put(arr[i], 1);  
        } 
        int ans=0;  
          
        // Calculating count of pairs with equal values 
        for(Map.Entry<Integer,Integer> it : hm.entrySet()) 
        {  
            int count = it.getValue(); 
            ans += (count * (count - 1)) / 2; 
        } 
        return ans; 
    } 
      
    // Driver code 
    public static void main(String[] args)  
    { 
        int arr[] = new int[]{1, 2, 3, 1}; 
        System.out.println(countPairs(arr,arr.length)); 
    } 
} 
  
// This Code is Contributed 
// by Adarsh_Verma 

Python3

# Python3 program to count of index pairs  
# with equal elements in an array. 
import math as mt 
  
# Return the number of pairs with  
# equal values. 
def countPairs(arr, n): 
  
    mp = dict() 
  
    # Finding frequency of each number. 
    for i in range(n): 
        if arr[i] in mp.keys(): 
            mp[arr[i]] += 1
        else: 
            mp[arr[i]] = 1
              
    # Calculating pairs of each value. 
    ans = 0
    for it in mp: 
        count = mp[it] 
        ans += (count * (count - 1)) // 2
    return ans 
  
# Driver Code 
arr = [1, 1, 2] 
n = len(arr) 
print(countPairs(arr, n)) 
  
# This code is contributed by mohit kumar 29 

C#

// C# program to count of index pairs with 
// equal elements in an array. 
using System; 
using System.Collections.Generic; 
  
class GFG  
{ 
      
    // Return the number of pairs with  
    // equal values. 
    public static int countPairs(int []arr, int n) 
    {  
        // A method to return number of pairs  
        // with equal values 
        Dictionary<int,  
                   int> hm = new Dictionary<int,  
                                            int>(); 
          
        // Finding frequency of each number. 
        for(int i = 0; i < n; i++) 
        { 
            if(hm.ContainsKey(arr[i])) 
            { 
                int a = hm[arr[i]]; 
                hm.Remove(arr[i]); 
                hm.Add(arr[i], a + 1); 
            } 
            else
                hm.Add(arr[i], 1);  
        } 
        int ans = 0;  
          
        // Calculating count of pairs with  
        // equal values 
        foreach(var it in hm) 
        {  
            int count = it.Value; 
            ans += (count * (count - 1)) / 2; 
        } 
        return ans; 
    } 
      
    // Driver code 
    public static void Main()  
    { 
        int []arr = new int[]{1, 2, 3, 1}; 
        Console.WriteLine(countPairs(arr,arr.Length)); 
    } 
} 
  
// This code is contributed by 29AjayKumar 

Output :

Time Complexity : O(n)

Source:
http://stackoverflow.com/questions/26772364/efficient-algorithm-for-counting-number-of-pairs-of-identical-elements-in-an-arr#comment42124861_26772516

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