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 = arr, the pair of indices is (0, 1)

Input : arr[] = {1, 1, 1}
Output : 3
As arr = arr, 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++ program to count of pairs with equal // elements in an array. #include 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);     cout << countPairs(arr, n) << endl;     return 0; }

 // 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 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# 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.



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++ program to count of index pairs with // equal elements in an array. #include using namespace std;    // Return the number of pairs with equal // values. int countPairs(int arr[], int n) {     unordered_map 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);     cout << countPairs(arr, n) << endl;     return 0; }

 // 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 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 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 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# 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 hm = new Dictionary();                    // 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)

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