Make all elements zero by decreasing any two elements by one at a time

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Given an array arr[], the task is to check whether it is possible to make all the elements of the array zero by the given operation. In a single operation, any two elements arr[i] and arr[j] can be decremented by one at the same time.
Examples:

Input: arr[] = {1, 2, 1, 2, 2}
Output: Yes
Decrement the 1^st and the 2^nd element, arr[] = {0, 1, 1, 2, 2}
Decrement the 2^nd and the 3^rd element, arr[] = {0, 0, 0, 2, 2}
Decrement the 4^th and the 5^th element, arr[] = {0, 0, 0, 1, 1}
Decrement the 4^th and the 5^th element, arr[] = {0, 0, 0, 0, 0}
Input: arr[] = {1, 2, 3, 4, 5}
Output: No

Approach: The given array can be only be made zero if it holds the following conditions true:

The sum of the elements of the array must be even.
The largest elements must be less than or equal to ?sum / 2? where sum is the sum of the other elements.

Below is the implementation of the above approach:

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if all
// the array elements can be made
// 0 with the given operation
bool checkZeroArray(int* arr, int n)
{
 
    // Find the maximum element
    // and the sum
    int sum = 0, maximum = INT_MIN;
    for (int i = 0; i < n; i++) {
        sum = sum + arr[i];
        maximum = max(maximum, arr[i]);
    }
 
    // Check the required condition
    if (sum % 2 == 0 && maximum <= sum / 2)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 1, 2, 2 };
    int n = sizeof(arr) / sizeof(int);
 
    if (checkZeroArray(arr, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the above approach 
public class GFG 
{
     
    // Function that returns true if all 
    // the array elements can be made 
    // 0 with the given operation 
    static boolean checkZeroArray(int []arr, int n) 
    { 
     
        // Find the maximum element 
        // and the sum 
        int sum = 0, maximum = Integer.MIN_VALUE; 
         
        for (int i = 0; i < n; i++) 
        { 
            sum = sum + arr[i]; 
            maximum = Math.max(maximum, arr[i]); 
        } 
     
        // Check the required condition 
        if (sum % 2 == 0 && maximum <= sum / 2) 
            return true; 
     
        return false; 
    } 
     
    // Driver code 
    public static void main (String[] args)
    { 
        int arr[] = { 1, 2, 1, 2, 2 }; 
        int n = arr.length; 
     
        if (checkZeroArray(arr, n) == true) 
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    } 
}
 
// This code is contributed by AnkitRai01

Python

# Python3 implementation of the approach
 
# Function that returns true if all
# the array elements can be made
# 0 with the given operation
def checkZeroArray(arr,n):
 
    # Find the maximum element
    # and the sum
    sum = 0
    maximum = -10**9
    for i in range(n):
        sum = sum + arr[i]
        maximum = max(maximum, arr[i])
 
    # Check the required condition
    if (sum % 2 == 0 and maximum <= sum // 2):
        return True
 
    return False
 
# Driver code
 
arr = [1, 2, 1, 2, 2]
n = len(arr)
 
if (checkZeroArray(arr, n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of the above approach 
using System;
 
class GFG 
{ 
     
    // Function that returns true if all 
    // the array elements can be made 
    // 0 with the given operation 
    static bool checkZeroArray(int []arr, int n) 
    { 
     
        // Find the maximum element 
        // and the sum 
        int sum = 0, maximum = int.MinValue; 
         
        for (int i = 0; i < n; i++) 
        { 
            sum = sum + arr[i]; 
            maximum = Math.Max(maximum, arr[i]); 
        } 
     
        // Check the required condition 
        if (sum % 2 == 0 && maximum <= sum / 2) 
            return true; 
     
        return false; 
    } 
     
    // Driver code 
    public static void Main () 
    { 
        int []arr = { 1, 2, 1, 2, 2 }; 
        int n = arr.Length; 
     
        if (checkZeroArray(arr, n) == true) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
 
// This code is contributed by AnkitRai01 

Javascript

<script>
 
// JavaScript implementation of the above approach 
     
    // Function that returns true if all 
    // the array elements can be made 
    // 0 with the given operation 
    function checkZeroArray(arr,n)
    {
        // Find the maximum element 
        // and the sum 
        let sum = 0, maximum = Number.MIN_VALUE; 
           
        for (let i = 0; i < n; i++) 
        { 
            sum = sum + arr[i]; 
            maximum = Math.max(maximum, arr[i]); 
        } 
       
        // Check the required condition 
        if (sum % 2 == 0 && maximum <= sum / 2) 
            return true; 
       
        return false; 
    }
     
    // Driver code 
    let arr=[1, 2, 1, 2, 2 ];
    let n = arr.length; 
    if (checkZeroArray(arr, n) == true) 
        document.write("Yes"); 
    else
        document.write("No"); 
       
 
 
// This code is contributed by unknown2108
 
</script>

Output:

Yes

Time Complexity: O(n)

Auxiliary Space: O(1)

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