Given an array arr[] consisting of N integers, the task is to find the number of pairs, where i ? j, such that the sum of pairs divides the sum of array elements.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: 3
Explanation:
Below are the pairs that divides the sum of array( = 1 + 2 + 3 + 4 + 5 = 15):
- (1, 2): Sum of pairs = 1 + 2 = 3, that divides sum(= 15).
- (1, 4): Sum of pairs = 1 + 4 = 5, that divides sum(= 15).
- (2, 3): Sum of pairs = 2 + 3 = 5, that divides sum(= 15).
Therefore, the count of pairs is 3.
Input: arr[] = {1, 5, 2}
Output: 0
Approach: The given problem can be solved by generating all possible pairs (arr[i], arr[j]) from the given array such that i ? j and count all those pairs whose sum of elements divides the sum of the array. After checking for all possible pairs, print the total count obtained.
Below is the implementation of the above approach.
C++
// C++ program for the above approach#include <iostream>using namespace std;// Function to find the number of pairs// whose sums divides the sum of arrayvoid countPairs(int arr[], int N){ // Initialize the totalSum and // count as 0 int count = 0, totalSum = 0; // Calculate the total sum of array for (int i = 0; i < N; i++) { totalSum += arr[i]; } // Generate all possible pairs for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // If the sum is a factor // of totalSum or not if (totalSum % (arr[i] + arr[j]) == 0) { // Increment count by 1 count += 1; } } } // Print the total count obtained cout << count;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 5 }; int N = sizeof(arr) / sizeof(arr[0]); countPairs(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to find the number of pairs // whose sums divides the sum of array public static void countPairs(int arr[], int N) { // Initialize the totalSum and // count as 0 int count = 0, totalSum = 0; // Calculate the total sum of array for (int i = 0; i < N; i++) { totalSum += arr[i]; } // Generate all possible pairs for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // If the sum is a factor // of totalSum or not if (totalSum % (arr[i] + arr[j]) == 0) { // Increment count by 1 count += 1; } } } // Print the total count obtained System.out.println(count); } // Driver code public static void main(String[] args) { int arr[] = { 1, 2, 3, 4, 5 }; int N = arr.length; countPairs(arr, N); }} // This code is contributed by Potta Lokesh |
Python3
# Python3 program for the above approach# Function to find the number of pairs# whose sums divides the sum of arraydef countPairs(arr, N): # Initialize the totalSum and # count as 0 count = 0 totalSum = 0 # Calculate the total sum of array for i in range(N): totalSum += arr[i] # Generate all possible pairs for i in range(N): for j in range(i + 1, N, 1): # If the sum is a factor # of totalSum or not if (totalSum % (arr[i] + arr[j]) == 0): # Increment count by 1 count += 1 # Print the total count obtained print(count)# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 3, 4, 5 ] N = len(arr) countPairs(arr, N) # This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the number of pairs // whose sums divides the sum of array public static void countPairs(int[] arr, int N) { // Initialize the totalSum and // count as 0 int count = 0, totalSum = 0; // Calculate the total sum of array for (int i = 0; i < N; i++) { totalSum += arr[i]; } // Generate all possible pairs for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // If the sum is a factor // of totalSum or not if (totalSum % (arr[i] + arr[j]) == 0) { // Increment count by 1 count += 1; } } } // Print the total count obtained Console.WriteLine(count); }// Driver codestatic public void Main(){ int[] arr = { 1, 2, 3, 4, 5 }; int N = arr.Length; countPairs(arr, N);}}// This code is contributed by splevel62. |
Javascript
<script> // JavaScript program for the above approach // Function to find the number of pairs // whose sums divides the sum of array function countPairs(arr, N) { // Initialize the totalSum and // count as 0 let count = 0, totalSum = 0; // Calculate the total sum of array for (let i = 0; i < N; i++) { totalSum += arr[i]; } // Generate all possible pairs for (let i = 0; i < N; i++) { for (let j = i + 1; j < N; j++) { // If the sum is a factor // of totalSum or not if (totalSum % (arr[i] + arr[j]) == 0) { // Increment count by 1 count += 1; } } } // Print the total count obtained document.write(count); } // Driver Code let arr = [1, 2, 3, 4, 5]; let N = arr.length; countPairs(arr, N);// This code is contributed by Potta Lokesh </script> |
Output:
3
Time Complexity: O(N2)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
