3704 Count Partitions With Even Sum Difference

Count Partitions with Even Sum Difference

You are given an integer array nums of length n.

A partition is defined as an index i where 0 <= i < n - 1, splitting the array into two non-empty subarrays such that:

Left subarray contains indices [0, i].
Right subarray contains indices [i + 1, n - 1].

Return the number of partitions where the difference between the sum of the left and right subarrays is even.

 

Example 1:

Input: nums = [10,10,3,7,6]

Output: 4

Explanation:

The 4 partitions are:

[10], [10, 3, 7, 6] with a sum difference of 10 - 26 = -16, which is even.
[10, 10], [3, 7, 6] with a sum difference of 20 - 16 = 4, which is even.
[10, 10, 3], [7, 6] with a sum difference of 23 - 13 = 10, which is even.
[10, 10, 3, 7], [6] with a sum difference of 30 - 6 = 24, which is even.

Example 2:

Input: nums = [1,2,2]

Output: 0

Explanation:

No partition results in an even sum difference.

Example 3:

Input: nums = [2,4,6,8]

Output: 3

Explanation:

All partitions result in an even sum difference.

 

Constraints:

2 <= n == nums.length <= 100
1 <= nums[i] <= 100

class Solution:
    def countPartitions(self, nums: List[int]) -> int:
        left = []
        right = []
        temp = 0
        for k in nums:
            temp += k
            left.append(temp)

        temp = 0
        for k in nums[::-1]:
            temp += k
            right.append(temp)
        right.reverse()
        res = 0
        #print(left, right)
        for i , k in enumerate(left[:-1]):
            if abs(k - right[i+1]) % 2 == 0:
                res += 1

        return res