# Count of all pairs in an Array with minimum absolute difference

• Difficulty Level : Medium
• Last Updated : 01 Apr, 2021

Given an integer array arr[] of size N, the task is to count the total number of distinct pairs having minimum absolute difference.
Examples:

Input: arr[] = {4, 2, 1, 3}
Output:
Explanation:
The minimum absolute difference between the pairs {1, 2}, {2, 3}, {3, 4} is 1.
Input: arr[] = {1, 3, 8, 10, 15}
Output:
Explanation:
The minimum absolute difference between the pairs {1, 3}, {8, 10} is 2.

Approach: The idea is to count the frequency of the minimum absolute difference of the adjacent elements of the sorted elements of the given array. Follow the steps below to solve the problem:

1. Sort the given array arr[].
2. Compare all adjacent pairs in the sorted array and find the minimum absolute difference between all adjacent pairs.
3. Finally, count all the adjacent pairs having difference equal to minimum difference.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to return the count of all``// pairs having minimal absolute difference``int` `numberofpairs(``int` `arr[], ``int` `N)``{``    ``// Stores the count of pairs``    ``int` `answer = 0;` `    ``// Sort the array``    ``sort(arr, arr + N);` `    ``// Stores the minimum difference``    ``// between adjacent pairs``    ``int` `minDiff = INT_MAX;``    ``for` `(``int` `i = 0; i < N - 1; i++)` `        ``// Update the minimum``        ``// difference between pairs``        ``minDiff = min(minDiff,``                      ``arr[i + 1] - arr[i]);` `    ``for` `(``int` `i = 0; i < N - 1; i++) {` `        ``if` `(arr[i + 1] - arr[i] == minDiff)` `            ``// Increase count of``            ``// pairs with difference``            ``// equal to that of``            ``// minimum difference``            ``answer++;``    ``}` `    ``// Return the final count``    ``return` `answer;``}` `// Driver Code``int` `main()``{``    ``// Given array arr[]``    ``int` `arr[] = { 4, 2, 1, 3 };``    ``int` `N = (``sizeof` `arr) / (``sizeof` `arr);` `    ``// Function Call``    ``cout << numberofpairs(arr, N) << ``"\n"``;` `    ``return` `0;``}`

## Java

 `// Java program for the above``import` `java.util.Arrays;``class` `GFG{`` ` `// Function to return the count of all``// pairs having minimal absolute difference``static` `int` `numberofpairs(``int` `[]arr, ``int` `N)``{``    ``// Stores the count of pairs``    ``int` `answer = ``0``;`` ` `    ``// Sort the array``    ``Arrays.sort(arr);`` ` `    ``// Stores the minimum difference``    ``// between adjacent pairs``    ``int` `minDiff = ``10000000``;``    ``for` `(``int` `i = ``0``; i < N - ``1``; i++)`` ` `        ``// Update the minimum``        ``// difference between pairs``        ``minDiff = Math.min(minDiff,``                           ``arr[i + ``1``] - arr[i]);`` ` `    ``for` `(``int` `i = ``0``; i < N - ``1``; i++)``    ``{``        ``if` `(arr[i + ``1``] - arr[i] == minDiff)`` ` `            ``// Increase count of``            ``// pairs with difference``            ``// equal to that of``            ``// minimum difference``            ``answer++;``    ``}`` ` `    ``// Return the final count``    ``return` `answer;``}`` ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// Given array arr[]``    ``int` `arr[] = { ``4``, ``2``, ``1``, ``3` `};``    ``int` `N = arr.length;`` ` `    ``// Function Call``    ``System.out.print(numberofpairs(arr, N));``}``}` `// This code is contributed by rock_cool`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to return the count of all``# pairs having minimal absolute difference``def` `numberofpairs(arr, N):``    ` `    ``# Stores the count of pairs``    ``answer ``=` `0``    ` `    ``# Sort the array``    ``arr.sort()``    ` `    ``# Stores the minimum difference``    ``# between adjacent pairs``    ``minDiff ``=` `10000000``    ``for` `i ``in` `range``(``0``, N ``-` `1``):``        ` `        ``# Update the minimum``        ``# difference between pairs``        ``minDiff ``=` `min``(minDiff,``                      ``arr[i ``+` `1``] ``-` `arr[i])``        ` `    ``for` `i ``in` `range``(``0``, N ``-` `1``):``        ``if` `arr[i ``+` `1``] ``-` `arr[i] ``=``=` `minDiff:``            ` `            ``# Increase count of pairs``            ``# with difference equal to``            ``# that of minimum difference``            ``answer ``+``=` `1``            ` `    ``# Return the final count``    ``return` `answer` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given array arr[]``    ``arr ``=` `[ ``4``, ``2``, ``1``, ``3` `]``    ``N ``=` `len``(arr)``    ` `    ``# Function call``    ``print``(numberofpairs(arr,N))` `# This code is contributed by virusbuddah_`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG{` `// Function to return the count of all``// pairs having minimal absolute difference``static` `int` `numberofpairs(``int` `[]arr, ``int` `N)``{``    ` `    ``// Stores the count of pairs``    ``int` `answer = 0;` `    ``// Sort the array``    ``Array.Sort(arr);` `    ``// Stores the minimum difference``    ``// between adjacent pairs``    ``int` `minDiff = 10000000;``    ``for``(``int` `i = 0; i < N - 1; i++)` `        ``// Update the minimum``        ``// difference between pairs``        ``minDiff = Math.Min(minDiff,``                           ``arr[i + 1] -``                           ``arr[i]);` `    ``for``(``int` `i = 0; i < N - 1; i++)``    ``{``        ``if` `(arr[i + 1] - arr[i] == minDiff)` `            ``// Increase count of``            ``// pairs with difference``            ``// equal to that of``            ``// minimum difference``            ``answer++;``    ``}` `    ``// Return the final count``    ``return` `answer;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ` `    ``// Given array arr[]``    ``int` `[]arr = { 4, 2, 1, 3 };``    ``int` `N = arr.Length;` `    ``// Function Call``    ``Console.Write(numberofpairs(arr, N));``}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up