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

Last Updated : 08 Jan, 2021

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:

Traverse the array and calculate the sum of the given array.
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

Output:

No

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

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