3432. Count Partitions with Even Sum Difference

// Approach: Partition sum difference is even iff total sum is even; count valid partitions directly.
// Time: O(n) Space: O(1)

public class Solution
{
    public int CountPartitions(int[] nums)
    {
        // Initialize left sum and right sum
        int leftSum = 0;
        int rightSum = 0;

        // Calculate initial right sum (total sum of all elements)
        foreach (int num in nums)
            rightSum += num;

        // Counter for valid partitions
        int partitionCount = 0;

        // Iterate through possible partition points (excluding the last element)
        // since we need at least one element in the right partition
        for (int i = 0; i < nums.Length - 1; i++)
        {
            // Move current element from right partition to left partition
            leftSum += nums[i];
            rightSum -= nums[i];

            // Check if the difference between left and right sums is even
            if ((leftSum - rightSum) % 2 == 0)
                partitionCount++;
        }

        return partitionCount;
    }
}