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

• Last Updated : 25 May, 2021

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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

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

Below is the implementation of the above approach:

## CPP

 // C++ implementation of the approach#include using namespace std; // Function that returns true if all// the array elements can be made// 0 with the given operationbool 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 codeint 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 approachpublic 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 operationdef 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 approachusing 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


Output:
Yes

My Personal Notes arrow_drop_up