# Check if sum of arr[i] / j for all possible pairs (i, j) in an array is 0 or not

Given an array arr[] consisting of N integers, the task is to check if the sum of all possible values of (arr[i] / j) for all pairs (i, j) such that 0 < i ? j < (N – 1) is 0 or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {1, -1, 3, -2, -1}
Output: Yes
Explanation:
For all possible pairs (i, j), such that 0 < i <= j < (N – 1), required sum = 1/1 + -1/2 + 3/3 + -2/4 + -1/5 + -1/2 + 3/3 + -2/4 + -1/5 + 3/3 + -2/4 + -1/5 + -2/ 4 + -1/5 + -1/5 = 0.

Input: arr[] = {1, 2, 3, 4, 5}
Output: No

Approach: The given problem can be solved based on the following observations:

• For every possible value of i over the range [0, N – 1] and for every possible values of j following are the expressions:
• j = 1:
• j = 2:
• j = 3:3rd line and so on…
• Therefore, the sum of all the above expression is given by:

=>

=>

From the above observations, if the sum of the array is 0, then print Yes. Otherwise, print No.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach   #include  using namespace std;   // Function to check if sum of all // values of (arr[i]/j) for all // 0 < i <= j < (N - 1) is 0 or not void check(int arr[], int N) {     // Stores the required sum     int sum = 0;       // Traverse the array     for (int i = 0; i < N; i++)         sum += arr[i];       // If the sum is equal to 0     if (sum == 0)         cout << "Yes";       // Otherwise     else         cout << "No"; }   // Driver Code int main() {     int arr[] = { 1, -1, 3, -2, -1 };     int N = sizeof(arr) / sizeof(arr[0]);     check(arr, N);       return 0; }

## Java

 // Java program for the above approach import java.io.*; import java.lang.*; import java.util.*;   class GFG{   // Function to check if sum of all // values of (arr[i]/j) for all // 0 < i <= j < (N - 1) is 0 or not static void check(int arr[], int N) {           // Stores the required sum     int sum = 0;       // Traverse the array     for(int i = 0; i < N; i++)         sum += arr[i];       // If the sum is equal to 0     if (sum == 0)         System.out.println("Yes");       // Otherwise     else         System.out.println("No"); }   // Driver Code public static void main(String[] args) {     int arr[] = { 1, -1, 3, -2, -1 };     int N = arr.length;           check(arr, N); } }   // This code is contributed by Kingash

## Python3

 # Python3 program for the above approach   # Function to check if sum of all # values of (arr[i]/j) for all # 0 < i <= j < (N - 1) is 0 or not def check(arr, N):           # Stores the required sum     sum = 0       # Traverse the array     for i in range(N):         sum += arr[i]       # If the sum is equal to 0     if (sum == 0):         print("Yes")       # Otherwise     else:         print("No")   # Driver Code if __name__ == '__main__':           arr = [ 1, -1, 3, -2, -1 ]     N = len(arr)           check(arr, N)   # This code is contributed by mohit kumar 29

## C#

 // C# program for the above approach using System;   class GFG {       // Function to check if sum of all     // values of (arr[i]/j) for all     // 0 < i <= j < (N - 1) is 0 or not     static void check(int[] arr, int N)     {           // Stores the required sum         int sum = 0;           // Traverse the array         for (int i = 0; i < N; i++)             sum += arr[i];           // If the sum is equal to 0         if (sum == 0)             Console.WriteLine("Yes");           // Otherwise         else             Console.WriteLine("No");     }       // Driver Code     public static void Main(string[] args)     {         int[] arr = { 1, -1, 3, -2, -1 };         int N = arr.Length;           check(arr, N);     } }   // This code is contributed by ukasp.

## Javascript

 

Output:

Yes

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

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