Tutorialspoint.dev

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++

// 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 :



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++

// 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 :

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter