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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

Yes

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