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# Count all distinct pairs of repeating elements from the array for every array element

• Last Updated : 10 Nov, 2021

Given an array arr[] of N integers. For each element in the array, the task is to count the possible pairs (i, j), excluding the current element, such that i < j and arr[i] = arr[j].

Examples:

Input: arr[] = {1, 1, 2, 1, 2}
Output: 2 2 3 2 3
Explanation:
For index 1, remaining elements after excluding current element are [1, 2, 1, 2]. So the count is 2.
For index 2, remaining elements after excluding element at index 2 are [1, 2, 1, 2]. So the count is 2.
For index 3, remaining elements after excluding element at index 3 are [1, 1, 1, 2]. So the count is 3.
For index 4, remaining elements after excluding element at index 4 are [1, 1, 2, 2]. So the count is 2.
For index 5, remaining elements after excluding element at index 5 are [1, 1, 2, 1. So the count is 3.

Input: arr[] = {1, 2, 3, 4}
Output: 0 0 0 0
Explanation:
Since all the elements are distinct, so no pair with equal value exists.

Naive Approach: The naive idea is to traverse the given array and for each element exclude the current element from the array and with the remaining array elements find all the possible pairs (i, j) such that arr[i] is equal to arr[j] and i < j. Print the count of pairs for each element.
Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: The idea is to store the frequency of every element and count all the possible pairs(say cnt) with the given conditions. After the above steps for each element remove the count of equal possible pairs from the total count cnt and print that value. Follow the below steps to solve the problem:

1. Store the frequency of each element in Map.
2. Create a variable to store the contribution of each element.
3. Contribution of some number x can be calculated as freq[x]*(freq[x] – 1) divided by 2, where freq[x] is the frequency of x.
4. Traverse the given array and remove the contribution of each element from the total count and store it in ans[].
5. Print all the elements stored in ans[].

Below is the implementation of the above approach:

## C++

 `// C++ program for``// the above approach``#include ``#define int long long int``using` `namespace` `std;` `// Function to print the required``// count of pairs excluding the``// current element``void` `solve(``int` `arr[], ``int` `n)``{``    ``// Store the frequency``    ``unordered_map<``int``, ``int``> mp;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``mp[arr[i]]++;``    ``}` `    ``// Find all the count``    ``int` `cnt = 0;``    ``for` `(``auto` `x : mp) {``        ``cnt += ((x.second)``                ``* (x.second - 1) / 2);``    ``}` `    ``int` `ans[n];` `    ``// Delete the contribution of``    ``// each element for equal pairs``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``ans[i] = cnt - (mp[arr[i]] - 1);``    ``}` `    ``// Print the answer``    ``for` `(``int` `i = 0; i < n; i++) {``        ``cout << ans[i] << ``" "``;``    ``}``}` `// Driver Code``int32_t main()``{``    ``// Given array arr[]``    ``int` `arr[] = { 1, 1, 2, 1, 2 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function Call``    ``solve(arr, N);``    ``return` `0;``}`

## Java

 `// Java program for``// the above approach``import` `java.util.*;``class` `GFG{` `// Function to print the required``// count of pairs excluding the``// current element``static` `void` `solve(``int` `arr[],``                  ``int` `n)``{``  ``// Store the frequency``  ``HashMap mp = ``new` `HashMap();``  ``for` `(``int` `i = ``0``; i < n; i++)``  ``{``    ``if``(mp.containsKey(arr[i]))``    ``{``      ``mp.put(arr[i], mp.get(arr[i]) + ``1``);``    ``}``    ``else``    ``{``      ``mp.put(arr[i], ``1``);``    ``}``  ``}` `  ``// Find all the count``  ``int` `cnt = ``0``;``  ``for` `(Map.Entry x : mp.entrySet())``  ``{``    ``cnt += ((x.getValue()) *``            ``(x.getValue() - ``1``) / ``2``);``  ``}` `  ``int` `[]ans = ``new` `int``[n];` `  ``// Delete the contribution of``  ``// each element for equal pairs``  ``for` `(``int` `i = ``0``; i < n; i++)``  ``{``    ``ans[i] = cnt - (mp.get(arr[i]) - ``1``);``  ``}` `  ``// Print the answer``  ``for` `(``int` `i = ``0``; i < n; i++)``  ``{``    ``System.out.print(ans[i] + ``" "``);``  ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``// Given array arr[]``  ``int` `arr[] = {``1``, ``1``, ``2``, ``1``, ``2``};` `  ``int` `N = arr.length;` `  ``// Function Call``  ``solve(arr, N);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program for``# the above approach` `# Function to print required``# count of pairs excluding the``# current element``def` `solve(arr, n):``    ` `    ``# Store the frequency``    ``mp ``=` `{}``    ``for` `i ``in` `arr:``        ``mp[i] ``=` `mp.get(i, ``0``) ``+` `1` `    ``# Find all the count``    ``cnt ``=` `0``    ``for` `x ``in` `mp:``        ``cnt ``+``=` `((mp[x]) ``*``                ``(mp[x] ``-` `1``) ``/``/` `2``)` `    ``ans ``=` `[``0``] ``*` `n` `    ``# Delete the contribution of``    ``# each element for equal pairs``    ``for` `i ``in` `range``(n):``        ``ans[i] ``=` `cnt ``-` `(mp[arr[i]] ``-` `1``)` `    ``# Print the answer``    ``for` `i ``in` `ans:``        ``print``(i, end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given array arr[]``    ``arr ``=` `[``1``, ``1``, ``2``, ``1``, ``2``]` `    ``N ``=` `len``(arr)` `    ``# Function call``    ``solve(arr, N)``    ` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for``// the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG{` `// Function to print the required``// count of pairs excluding the``// current element``static` `void` `solve(``int` `[]arr,``                  ``int` `n)``{``  ``// Store the frequency``  ``Dictionary<``int``,``             ``int``> mp = ``new` `Dictionary<``int``,``                                      ``int``>();``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``if``(mp.ContainsKey(arr[i]))``    ``{``      ``mp[arr[i]] =  mp[arr[i]] + 1;``    ``}``    ``else``    ``{``      ``mp.Add(arr[i], 1);``    ``}``  ``}` `  ``// Find all the count``  ``int` `cnt = 0;``  ``foreach` `(KeyValuePair<``int``,``                        ``int``> x ``in` `mp)``  ``{``    ``cnt += ((x.Value) *``            ``(x.Value - 1) / 2);``  ``}` `  ``int` `[]ans = ``new` `int``[n];` `  ``// Delete the contribution of``  ``// each element for equal pairs``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``ans[i] = cnt - (mp[arr[i]] - 1);``  ``}` `  ``// Print the answer``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``Console.Write(ans[i] + ``" "``);``  ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``// Given array []arr``  ``int` `[]arr = {1, 1, 2, 1, 2};` `  ``int` `N = arr.Length;` `  ``// Function Call``  ``solve(arr, N);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

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Output:

`2 2 3 2 3`

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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