# 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 C++ implementation of this approach:

```// 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;
}
```

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 C++ implementation of this approach:

```// 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;
}

```

Output:

```1
```

Time Complexity : O(n)

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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