Check if all elements of the given array can be made 0 by decrementing value in pairs

Given an array arr[] consisting of positive integers, the task is to check if all elements of the given array can be made 0 by performing the following operation: 
 

• Choose two indices i and j such that i != j and subtract 1 from both arr[i] and arr[j]
• The above operation can be performed any number of times

Examples: 
 

Input: arr[] = {1, 2, 3, 4} 
Output: Yes 
Explanation: 
First, choose values 2 and 4 and perform the above operation 2 times. Then the array becomes 1 0 3 2. 
Now choose 1 and 3 and apply above operation once to get 0 0 2 2. 
Now pick two 2s and perform the above operation twice. 
Finally array becomes 0 0 0 0.
Input: arr[] = {5, 5, 5, 5, 5} 
Output: No 
 

Approach: On observing the problem carefully, it can be observed that if there is only 1 element or the sum of all the elements is odd, then it is not possible to make all elements 0. Since at every iteration, 2 is being subtracted from the sum of all elements, therefore, the array can become 0 only if the sum of all elements of the array is even. And also, it is possible to make the array 0 when the largest number in the array is less than or equal to the sum of remaining elements.
Below is the implementation of the above approach: 
 

// C++ program to make the array zero
// by decrementing value in pairs

#include <bits/stdc++.h>
using namespace std;

// Function to check if all the elements
// can be made 0 in an array
void canMake(int n, int ar[])
{

    // Variable to store
    // sum and maximum element
    // in an array
    int sum = 0, maxx = -1;

    // Loop to calculate the sum and max value
    // of the given array
    for (int i = 0; i < n; i++) {
        sum += ar[i];
        maxx = max(maxx, ar[i]);
    }

    // If n is 1 or sum is odd or
    // sum - max element < max
    // then no solution
    if (n == 1 || sum % 2 == 1
        || sum - maxx < maxx) {
        cout << "No\n";
    }
    else {

        // For the remaining case, print Yes
        cout << "Yes\n";
    }
}

// Driver code
int main()
{

    int n = 6;
    int arr[] = { 1, 1, 2, 3, 6, 11 };

    canMake(n, arr);

    return 0;
}

// Java program to make the array zero
// by decrementing value in pairs
class GFG
{

// Function to check if all the elements
// can be made 0 in an array
static void canMake(int n, int ar[])
{

    // Variable to store
    // sum and maximum element
    // in an array
    int sum = 0, maxx = -1;

    // Loop to calculate the sum and max value
    // of the given array
    for (int i = 0; i < n; i++) 
    {
        sum += ar[i];
        maxx = Math.max(maxx, ar[i]);
    }

    // If n is 1 or sum is odd or
    // sum - max element < max
    // then no solution
    if (n == 1 || sum % 2 == 1
        || sum - maxx < maxx) 
    {
        System.out.print("No\n");
    }
    else 
    {

        // For the remaining case, print Yes
        System.out.print("Yes\n");
    }
}

// Driver code
public static void main(String[] args)
{

    int n = 6;
    int arr[] = { 1, 1, 2, 3, 6, 11 };

    canMake(n, arr);
}
}

// This code is contributed by 29AjayKumar

# Python3 program to make the array zero 
# by decrementing value in pairs 

# Function to check if all the elements 
# can be made 0 in an array 
def canMake(n, ar) :

    # Variable to store 
    # sum and maximum element 
    # in an array 
    sum = 0; maxx = -1; 

    # Loop to calculate the sum and max value 
    # of the given array 
    for i in range(n) :
        sum += ar[i]; 
        maxx = max(maxx, ar[i]); 

    # If n is 1 or sum is odd or 
    # sum - max element < max 
    # then no solution 
    if (n == 1 or sum % 2 == 1
        or sum - maxx < maxx) :
        print("No"); 
    
    else :

        # For the remaining case, print Yes 
        print("Yes"); 

# Driver code 
if __name__ == "__main__" : 

    n = 6; 
    arr = [ 1, 1, 2, 3, 6, 11 ]; 

    canMake(n, arr); 

# This code is contributed by AnkitRai01

// C# program to make the array zero
// by decrementing value in pairs
using System;

class GFG
{

// Function to check if all the elements
// can be made 0 in an array
static void canMake(int n, int []ar)
{

    // Variable to store
    // sum and maximum element
    // in an array
    int sum = 0, maxx = -1;

    // Loop to calculate the sum and max value
    // of the given array
    for (int i = 0; i < n; i++) 
    {
        sum += ar[i];
        maxx = Math.Max(maxx, ar[i]);
    }

    // If n is 1 or sum is odd or
    // sum - max element < max
    // then no solution
    if (n == 1 || sum % 2 == 1
        || sum - maxx < maxx) 
    {
        Console.Write("No\n");
    }
    else
    {

        // For the remaining case, print Yes
        Console.Write("Yes\n");
    }
}

// Driver code
public static void Main(String[] args)
{

    int n = 6;
    int []arr = { 1, 1, 2, 3, 6, 11 };

    canMake(n, arr);
}
}

// This code is contributed by PrinciRaj1992

<script>

// Javascript program to make the array zero
// by decrementing value in pairs

// Function to check if all the elements
// can be made 0 in an array
function canMake(n, ar)
{

    // Variable to store
    // sum and maximum element
    // in an array
    var sum = 0, maxx = -1;

    // Loop to calculate the sum and max value
    // of the given array
    for (var i = 0; i < n; i++) {
        sum += ar[i];
        maxx = Math.max(maxx, ar[i]);
    }

    // If n is 1 or sum is odd or
    // sum - max element < max
    // then no solution
    if (n == 1 || sum % 2 == 1
        || sum - maxx < maxx) {
        document.write( "No");
    }
    else {

        // For the remaining case, print Yes
        document.write( "Yes");
    }
}

// Driver code
var n = 6;
var arr = [1, 1, 2, 3, 6, 11];
canMake(n, arr);

</script> 

Yes

Time Complexity: O(N)

Auxiliary Space: O(1)