Haga que todos los elementos sean cero al disminuir dos elementos cualquiera por uno a la vez

Dada una array arr[] , la tarea es verificar si es posible hacer que todos los elementos de la array sean cero mediante la operación dada. En una sola operación, dos elementos cualesquiera arr[i] y arr[j] pueden reducirse en uno al mismo tiempo. 
Ejemplos: 
 

Entrada: arr[] = {1, 2, 1, 2, 2} 
Salida: Sí 
Decrementa el 1 ^er y el 2 ^do elemento, arr[] = {0, 1, 1, 2, 2} 
Decrementa el 2 ^do y el 3er ^elemento , arr[] = {0, 0, 0, 2, 2} 
Decrementa el 4 ^to y el 5 ^to elemento, arr[] = {0, 0, 0, 1, 1} 
Decrementa el 4 ^to y el 5 ^to elemento, arr[] = {0, 0, 0, 0, 0}
Entrada: arr[] = {1, 2, 3, 4, 5} 
Salida: No 
 

Enfoque: la array dada solo se puede convertir en cero si cumple las siguientes condiciones: 
 

1. La suma de los elementos del arreglo debe ser par.
2. Los elementos más grandes deben ser menores o iguales a ⌊sum / 2⌋ donde sum es la suma de los otros elementos.

A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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

# 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# 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

<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>

Yes

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)