Skip to content
Related Articles

Related Articles

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

View Discussion
Improve Article
Save Article
Like Article
  • Last Updated : 29 Apr, 2021

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: \frac{a_1}{1}  + \frac{a_2}{2} + \frac{a_3}{3} .... +\frac{a_n}{n}
    • j = 2:\frac{a_2}{2} + \frac{a_3}{3} .... +\frac{a_n}{n}
    • j = 3:\frac{a_3}{3} .... +\frac{a_n}{n} 3rd line and so on…
  • Therefore, the sum of all the above expression is given by:

=> \frac{a_1}{1} + \frac{2*a_2}{2} + \frac{3*a_3}{3} ...+ \frac{n*a_n}{n}

=> a_1 + a_2 + a_3 + ... + a_n

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 <bits/stdc++.h>
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




<script>
// javascript 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
    function check(arr , N) {
 
        // Stores the required sum
        var sum = 0;
 
        // Traverse the array
        for (i = 0; i < N; i++)
            sum += arr[i];
 
        // If the sum is equal to 0
        if (sum == 0)
            document.write("Yes");
 
        // Otherwise
        else
            document.write("No");
    }
 
    // Driver Code
     
        var arr = [ 1, -1, 3, -2, -1 ];
        var N = arr.length;
 
        check(arr, N);
 
// This code contributed by umadevi9616
</script>

Output: 

Yes

 

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


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!