Check if sum of the given array can be reduced to 0 by reducing array elements by K

Given an array arr[] consisting of N integers and an integer K, the task is to check if the sum of the array can be reduced to 0 by subtracting array elements by K any number of times.

Examples:

Input: arr[ ]= {-3, 2, -1, 5, 1}, K=2
Output: “Yes”
Explanation:
Sum of the array is 4.  Therefore, decreasing two elements at any index by K( = 2), makes the sum of the array 0.
Input: arr[ ]= {1, -6, 2, 2}, K=1
Output: “No”

Approach: Follow the steps below to solve the problem:

1. Traverse the array and calculate the sum of the given array.
2. According to the value of the sum, the following cases arise:
   1. If sum = 0: No operation is required. Therefore, the answer is “Yes”.
   2. If sum > 0: Sum can be reduced to 0 only if sum is a multiple of K. If sum is not a multiple of K, print “No”. Otherwise, print “Yes”.
   3. If sum < 0: Simply 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 the
// sum can be made 0 or not
int sumzero(int arr[], int N, int K)
{
    // Stores sum of array elements
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        sum += arr[i];
    }

    if (sum == 0)
        cout << "Yes";

    else if (sum > 0) {

        if (sum % K == 0)
            cout << "Yes";

        else
            cout << "No";
    }

    else
        cout << "No";
    return 0;
}

// Driver Code
int main()
{
    int K, N;

    // Given array arr[]
    int arr1[] = { 1, -6, 2, 2 };
    K = 1;
    N = sizeof(arr1) / sizeof(arr1[0]);

    sumzero(arr1, N, K);

    return 0;
}

Java

// Java program for the above approach

import java.util.*;

class GFG{

// Function to check if the
// sum can be made 0 or not
static int sumzero(int arr[], int N, int K)
{
    // Stores sum of array elements
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        sum += arr[i];
    }

    if (sum == 0)
        System.out.print("Yes");

    else if (sum > 0) {

        if (sum % K == 0)
            System.out.print("Yes");

        else
            System.out.print("No");
    }

    else
        System.out.print("No");
    return 0;
}

// Driver Code
public static void main(String[] args)
{
    int K, N;

    // Given array arr[]
    int arr1[] = { 1, -6, 2, 2 };
    K = 1;
    N = arr1.length;

    sumzero(arr1, N, K);

}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to check if the
# sum can be made 0 or not
def sumzero(arr, N, K) :

    # Stores sum of array elements
    sum = 0;

    # Traverse the array
    for i in range(N) :
        sum += arr[i];
    if (sum == 0) :
        print("Yes");
    elif (sum > 0) :
        if (sum % K == 0) :
            print("Yes");
        else :
            print("No");
    else :
        print("No");

# Driver Code
if __name__ == "__main__" :

    # Given array arr[]
    arr1 = [ 1, -6, 2, 2 ];

    K = 1;
    N = len(arr1);

    sumzero(arr1, N, K);

    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
class GFG{

// Function to check if the
// sum can be made 0 or not
static int sumzero(int []arr, int N, int K)
{

    // Stores sum of array elements
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < N; i++)
    {

        sum += arr[i];
    }

    if (sum == 0)
        Console.Write("Yes");

    else if (sum > 0)
    {

        if (sum % K == 0)
            Console.Write("Yes");

        else
            Console.Write("No");
    }

    else
        Console.Write("No");
    return 0;
}

// Driver Code
public static void Main(String[] args)
{
    int K, N;

    // Given array []arr
    int []arr1 = { 1, -6, 2, 2 };
    K = 1;
    N = arr1.Length;

    sumzero(arr1, N, K);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program for the above approach

// Function to check if the
    // sum can be made 0 or not
    function sumzero(arr , N , K)
    {
        // Stores sum of array elements
        var sum = 0;

        // Traverse the array
        for (i = 0; i < N; i++) {

            sum += arr[i];
        }

        if (sum == 0)
            document.write("Yes");

        else if (sum > 0) {

            if (sum % K == 0)
                document.write("Yes");

            else
                document.write("No");
        }

        else
            document.write("No");
        return 0;
    }

    // Driver Code

        var K, N;

        // Given array arr
        var arr1 = [ 1, -6, 2, 2 ];
        K = 1;
        N = arr1.length;

        sumzero(arr1, N, K);


// This code contributed by gauravrajput1

</script>

Output:

No

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