Count pairs from an array with absolute difference not less than the minimum element in the pair
Last Updated :
23 Jul, 2025
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: 3
Explanation:
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.
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)
{
int count=0;//TO store the count.
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(abs(arr[i]-arr[j])>=min(arr[i],arr[j]))
count++;// increasing the count when we found the pair
}
}
// 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.util.Arrays;
class Main
{
// 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) {
int count = 0; // To store the count
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (Math.abs(arr[i] - arr[j]) >= Math.min(arr[i], arr[j]))
count++; // increasing the count when we found the pair
}
}
// 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 rishabmalhdijo
Python3
# code
def getPairsCount(arr, n):
count = 0 # To store the count.
for i in range(n):
for j in range(i + 1, n):
if abs(arr[i] - arr[j]) >= min(arr[i], arr[j]):
count += 1 # Increasing the count when we found the pair
# Return the total count
return count
arr = [1, 2, 2, 3]
N = len(arr)
print(getPairsCount(arr, N))
C#
using System;
class MainClass
{
// 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)
{
int count = 0; // To store the count
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
if (Math.Abs(arr[i] - arr[j]) >= Math.Min(arr[i], arr[j]))
{
count++; // increasing the count when we found the pair
}
}
}
// 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.WriteLine(GetPairsCount(arr, N));
}
}
JavaScript
// JavaScript code 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) {
let count = 0; // to store the count.
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
if (Math.abs(arr[i] - arr[j]) >= Math.min(arr[i], arr[j])) {
count++; // increasing the count when we found the pair
}
}
}
// Return the total count
return count;
}
// Driver Code
const arr = [1, 2, 2, 3];
const N = arr.length;
console.log(getPairsCount(arr, N));
// The code is contributed by Arushi Goel.
Time Complexity: O(N2)
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.
- 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>
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
Approach using DP:
The current implementation uses nested loops to iterate over the array elements and count the pairs that satisfy the given condition. However, we can optimize the solution by using a hash table or a frequency array to store the counts of each element in the array.
Here's an updated version of the code that utilizes a hash table to achieve the same result:
C++
#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)
{
// Create a hash table to store the frequency of elements
unordered_map<int, int> freq;
// Iterate over the array and count the frequency of each element
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// Stores the resultant count of pairs
int count = 0;
// Iterate over the array elements
for (int i = 0; i < n; i++) {
// Decrement the frequency of the current element
freq[arr[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
import java.util.HashMap;
import java.util.Map;
public 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) {
// Create a hash map to store the frequency of elements
Map<Integer, Integer> freq = new HashMap<>();
// Iterate over the array and count the frequency of each element
for (int i = 0; i < n; i++) {
freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1);
}
// Stores the resultant count of pairs
int count = 0;
// Iterate over the array elements
for (int i = 0; i < n; i++) {
// Decrement the frequency of the current element
freq.put(arr[i], freq.get(arr[i]) - 1);
// 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));
}
}
Python3
# 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):
# Create a dictionary (hash table) to store the frequency of elements
freq = {}
# Iterate over the array and count the frequency of each element
for i in range(n):
freq[arr[i]] = freq.get(arr[i], 0) + 1
# Stores the resultant count of pairs
count = 0
# Iterate over the array elements
for i in range(n):
# Decrement the frequency of the current element
freq[arr[i]] -= 1
# 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))
C#
using System;
using System.Collections.Generic;
namespace PairsCountExample
{
class Program
{
// 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)
{
// Create a dictionary to store the frequency of elements
Dictionary<int, int> freq = new Dictionary<int, int>();
// Iterate over the array and count the frequency of each element
for (int i = 0; i < n; i++)
{
if (freq.ContainsKey(arr[i]))
freq[arr[i]]++;
else
freq[arr[i]] = 1;
}
// Stores the resultant count of pairs
int count = 0;
// Iterate over the array elements
for (int i = 0; i < n; i++)
{
// Decrement the frequency of the current element
freq[arr[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
static void Main(string[] args)
{
int[] arr = { 1, 2, 2, 3 };
int N = arr.Length;
Console.WriteLine(GetPairsCount(arr, N));
Console.ReadLine();
}
}
}
JavaScript
function getPairsCount(arr, n) {
// Create a Map (equivalent to Dictionary in C#) to store the frequency of elements
const freq = new Map();
// Iterate over the array and count the frequency of each element
for (let i = 0; i < n; i++) {
if (freq.has(arr[i])) {
freq.set(arr[i], freq.get(arr[i]) + 1);
} else {
freq.set(arr[i], 1);
}
}
// Stores the resultant count of pairs
let count = 0;
// Iterate over the array elements
for (let i = 0; i < n; i++) {
// Decrement the frequency of the current element
freq.set(arr[i], freq.get(arr[i]) - 1);
// 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
const arr = [1, 2, 2, 3];
const N = arr.length;
console.log(getPairsCount(arr, N));
Time Complexity: O(n^2)
Auxiliary Space: O(n)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Hereâs the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem