检查1和2组成的数组是否可以分为相等总和的2个部分

给定一个包含N个元素的数组，每个元素为1或2。任务是找出该数组是否可以分为2部分，以使这两部分中的元素总和相等。

例子：

Input : N = 3, arr[] = {1, 1, 2}
Output : YES

Input : N = 4, arr[] = {1, 2, 2, }
Output : NO

这样做的目的是要观察到，只有当数组的总和为偶数(即可被2整除)时，才可以将数组分为相等总和的两个部分。

假设数组的总和由sum表示。

现在，出现两种情况：

• 如果sum / 2是偶数：当sum / 2的值也是偶数时，这意味着两个部分中的每一个的和也是偶数，因此我们不需要考虑任何特殊的问题。因此，在这种情况下返回true。
• 如果sum / 2为奇数：当sum / 2的值为ODD时，表示各部分的和也为奇数。仅当数组的两个部分中的每一个都包含至少一个1时，才有可能。请考虑sum = 2或6或10的情况。因此，当sum / 2为奇数时，请检查数组中是否存在至少一个1。

下面是上述方法的实现：

C++

// C++ implementation of the above
// approach:
  
#include 
using namespace std;
  
// Function to check if it is possible to
// split the array in two parts with
// equal sum
bool isSpiltPossible(int n, int a[])
{
    int sum = 0, c1 = 0;
  
    // Calculate sum of elements
    // and count of 1's
    for (int i = 0; i < n; i++) {
        sum += a[i];
  
        if (a[i] == 1) {
            c1++;
        }
    }
  
    // If total sum is odd, return False
    if (sum % 2)
        return false;
  
    // If sum of each part is even,
    // return True
    if ((sum / 2) % 2 == 0)
        return true;
  
    // If sum of each part is even but
    // there is atleast one 1
    if (c1 > 0)
        return true;
    else
        return false;
}
  
// Driver Code
int main()
{
    int n = 3;
    int a[] = { 1, 1, 2 };
  
    if (isSpiltPossible(n, a))
        cout << "YES";
    else
        cout << "NO";
  
    return 0;
}

Java

// Java implementation of the above
// approach:
class GFG
{
      
// Function to check if it is possible 
// to split the array in two parts with
// equal sum
static boolean isSpiltPossible(int n, 
                               int a[])
{
    int sum = 0, c1 = 0;
  
    // Calculate sum of elements
    // and count of 1's
    for (int i = 0; i < n; i++) 
    {
        sum += a[i];
  
        if (a[i] == 1) 
        {
            c1++;
        }
    }
  
    // If total sum is odd, return False
    if(sum % 2 != 0)
        return false;
  
    // If sum of each part is even,
    // return True
    if ((sum / 2) % 2 == 0)
        return true;
  
    // If sum of each part is even but
    // there is atleast one 1
    if (c1 > 0)
        return true;
    else
        return false;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 3;
    int a[] = { 1, 1, 2 };
  
    if (isSpiltPossible(n, a))
        System.out.println("YES");
    else
        System.out.println("NO");
}
}
  
// This code is contributed by 
// Code Mech

Python3

# Python3 implementation of the above
# approach:
  
# Function to check if it is possible 
# to split the array in two halfs with
# equal Sum
def isSpiltPossible(n, a):
  
    Sum = 0
    c1 = 0
  
    # Calculate Sum of elements
    # and count of 1's
    for i in range(n):
        Sum += a[i]
  
        if (a[i] == 1):
            c1 += 1
  
    # If total Sum is odd, return False
    if (Sum % 2):
        return False
  
    # If Sum of each half is even,
    # return True
    if ((Sum // 2) % 2 == 0):
        return True
  
    # If Sum of each half is even 
    # but there is atleast one 1
    if (c1 > 0):
        return True
    else:
        return False
  
# Driver Code
n = 3
a = [ 1, 1, 2 ]
  
if (isSpiltPossible(n, a)):
    print("YES")
else:
    print("NO")
  
# This code is contributed
# by Mohit Kumar

C#

// C# implementation of the above
// approach:
using System;
  
class GFG
{
      
// Function to check if it is possible 
// to split the array in two parts with
// equal sum
static bool isSpiltPossible(int n, 
                            int[] a)
{
    int sum = 0, c1 = 0;
  
    // Calculate sum of elements
    // and count of 1's
    for (int i = 0; i < n; i++) 
    {
        sum += a[i];
  
        if (a[i] == 1) 
        {
            c1++;
        }
    }
  
    // If total sum is odd, return False
    if(sum % 2 != 0)
        return false;
  
    // If sum of each part is even,
    // return True
    if ((sum / 2) % 2 == 0)
        return true;
  
    // If sum of each part is even but
    // there is atleast one 1
    if (c1 > 0)
        return true;
    else
        return false;
}
  
// Driver Code
public static void Main()
{
    int n = 3;
    int[] a = { 1, 1, 2 };
  
    if (isSpiltPossible(n, a))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
  
// This code is contributed by 
// Code Mech

PHP

 0)
        return true;
    else
        return false;
}
  
// Driver Code
$n = 3;
$a = array( 1, 1, 2 );
  
if (isSpiltPossible($n, $a))
    echo("YES");
else
    echo("NO");
  
// This code is contributed by 
// Code Mech
?>

输出：

YES

时间复杂度： O(N)