# Count pairs from an array with absolute difference not less than the minimum element in the pair

Given an array **arr[]** consisting of **N** positive integers, the task is to find the number of pairs **(arr[i], arr[j])** such that absolute difference between the two elements is at least equal to the minimum element in the pair.

**Examples:**

Input:arr[] = {1, 2, 2, 3}Output:3Explanation:

Following are the pairs satisfying the given criteria:

- (arr[0], arr[1]): The absolute difference between the two is abs(1 – 2) = 1, which is at least the minimum of the two i.e., min(1. 2) = 1.
- (arr[0], arr[2]): The absolute difference between the two is abs(1 – 2) = 1, which is at least the minimum of the two i.e., min(1. 2) = 1.
- (arr[0], arr[3]): The absolute difference between the two is abs(1 – 2) = 1, which is at least the minimum of the two i.e., min(1. 2) = 1.
Therefore, the total count of such pairs is 3.

Input:arr[] = {2, 3, 6}Output:2

**Naive Approach:** The simple approach to solve the given problem is to generate all possible pairs from the array and count those pairs that satisfy the given conditions. After checking for all the pairs, print the total count obtained.

**Time Complexity:** O(N^{2})**Auxiliary Space:** O(1)

**Efficient Approach:** The above approach can also be optimized by sorting the given array and then iterate two nested loops such that the first loop, iterate till **N** and the second loop, iterate from **arr[i] – (i%arr[i])** with the increment of **j** as **j += arr[i]** till **N** and count those pairs that satisfy the given conditions. Follow the steps below to solve the problem:

- Initialize the variable, say
**count**as**0**that stores the resultant count of pairs. - Iterate over the range
**[0, N]**using the variable**i**and perform the following steps:- Iterate over the range
**[arr[i] – (i%arr[i]), N]**using the variable**j**with the increment of**j**as**j += arr[i]**and if**i**is less than**j**and**abs(arr[i] – arr[j])**is at least the minimum of**arr[i]**and**arr[j]**then increment the**count**by**1**.

- Iterate over the range
- After performing the above steps, print the value of
**count**as the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the number of pairs` `// (i, j) such that abs(a[i]-a[j]) is` `// at least the minimum of (a[i], a[j])` `int` `getPairsCount(` `int` `arr[], ` `int` `n)` `{` ` ` `// Stores the resultant count of pairs` ` ` `int` `count = 0;` ` ` `// Iterate over the range [0, n]` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// Iterate from arr[i] - (i%arr[i])` ` ` `// till n with an increment` ` ` `// of arr[i]` ` ` `for` `(` `int` `j = arr[i] - (i % arr[i]);` ` ` `j < n; j += arr[i]) {` ` ` `// Count the possible pairs` ` ` `if` `(i < j` ` ` `&& ` `abs` `(arr[i] - arr[j])` ` ` `>= min(arr[i], arr[j])) {` ` ` `count++;` ` ` `}` ` ` `}` ` ` `}` ` ` `// Return the total count` ` ` `return` `count;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = { 1, 2, 2, 3 };` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << getPairsCount(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG` `{` ` ` ` ` `// Function to find the number of pairs` ` ` `// (i, j) such that abs(a[i]-a[j]) is` ` ` `// at least the minimum of (a[i], a[j])` ` ` `static` `int` `getPairsCount(` `int` `arr[], ` `int` `n)` ` ` `{` ` ` ` ` `// Stores the resultant count of pairs` ` ` `int` `count = ` `0` `;` ` ` `// Iterate over the range [0, n]` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `// Iterate from arr[i] - (i%arr[i])` ` ` `// till n with an increment` ` ` `// of arr[i]` ` ` `for` `(` `int` `j = arr[i] - (i % arr[i]); j < n;` ` ` `j += arr[i]) {` ` ` `// Count the possible pairs` ` ` `if` `(i < j` ` ` `&& Math.abs(arr[i] - arr[j])` ` ` `>= Math.min(arr[i], arr[j])) {` ` ` `count++;` ` ` `}` ` ` `}` ` ` `}` ` ` `// Return the total count` ` ` `return` `count;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `1` `, ` `2` `, ` `2` `, ` `3` `};` ` ` `int` `N = arr.length;` ` ` `System.out.println(getPairsCount(arr, N));` ` ` `}` `}` `// This code is contributed by Potta Lokesh` |

## Python3

`# Python 3 program for the above approach` `# Function to find the number of pairs` `# (i, j) such that abs(a[i]-a[j]) is` `# at least the minimum of (a[i], a[j])` `def` `getPairsCount(arr, n):` ` ` ` ` `# Stores the resultant count of pairs` ` ` `count ` `=` `0` ` ` `# Iterate over the range [0, n]` ` ` `for` `i ` `in` `range` `(n):` ` ` ` ` `# Iterate from arr[i] - (i%arr[i])` ` ` `# till n with an increment` ` ` `# of arr[i]` ` ` `for` `j ` `in` `range` `(arr[i] ` `-` `(i ` `%` `arr[i]),n,arr[i]):` ` ` ` ` `# Count the possible pairs` ` ` `if` `(i < j ` `and` `abs` `(arr[i] ` `-` `arr[j]) >` `=` `min` `(arr[i], arr[j])):` ` ` `count ` `+` `=` `1` ` ` `# Return the total count` ` ` `return` `count` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `arr ` `=` `[` `1` `, ` `2` `, ` `2` `, ` `3` `]` ` ` `N ` `=` `len` `(arr)` ` ` `print` `(getPairsCount(arr, N))` ` ` ` ` `# This code is contributed by ipg2016107.` |

## C#

`// C# program for above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the number of pairs` ` ` `// (i, j) such that abs(a[i]-a[j]) is` ` ` `// at least the minimum of (a[i], a[j])` ` ` `static` `int` `getPairsCount(` `int` `[] arr, ` `int` `n)` ` ` `{` ` ` ` ` `// Stores the resultant count of pairs` ` ` `int` `count = 0;` ` ` `// Iterate over the range [0, n]` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// Iterate from arr[i] - (i%arr[i])` ` ` `// till n with an increment` ` ` `// of arr[i]` ` ` `for` `(` `int` `j = arr[i] - (i % arr[i]); j < n;` ` ` `j += arr[i]) {` ` ` `// Count the possible pairs` ` ` `if` `(i < j` ` ` `&& Math.Abs(arr[i] - arr[j])` ` ` `>= Math.Min(arr[i], arr[j])) {` ` ` `count++;` ` ` `}` ` ` `}` ` ` `}` ` ` `// Return the total count` ` ` `return` `count;` ` ` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[] arr = { 1, 2, 2, 3 };` ` ` `int` `N = arr.Length;` ` ` `Console.Write(getPairsCount(arr, N));` ` ` `}` `}` `// This code is contributed by sanjoy_62.` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Function to find the number of pairs` `// (i, j) such that abs(a[i]-a[j]) is` `// at least the minimum of (a[i], a[j])` `function` `getPairsCount(arr, n)` `{` ` ` ` ` `// Stores the resultant count of pairs` ` ` `let count = 0;` ` ` `// Iterate over the range [0, n]` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` ` ` `// Iterate from arr[i] - (i%arr[i])` ` ` `// till n with an increment` ` ` `// of arr[i]` ` ` `for` `(let j = arr[i] - (i % arr[i]);` ` ` `j < n; j += arr[i])` ` ` `{` ` ` ` ` `// Count the possible pairs` ` ` `if` `(i < j && Math.abs(arr[i] - arr[j]) >=` ` ` `Math.min(arr[i], arr[j]))` ` ` `{` ` ` `count++;` ` ` `}` ` ` `}` ` ` `}` ` ` `// Return the total count` ` ` `return` `count;` `}` `// Driver Code` `let arr = [ 1, 2, 2, 3 ];` `let N = arr.length;` `document.write(getPairsCount(arr, N));` `// This code is contributed by sanjoy_62` `</script>` |

**Output:**

3

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

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