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

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

1

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,….,ik. Then pick any two indexes ix and iy which will be counted as 1 pair. Similarly, iy and ix 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output :

1

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

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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