java.lang.Object
g3401_3500.s3432_count_partitions_with_even_sum_difference.Solution

public class Solution extends Object
3432 - Count Partitions with Even Sum Difference.

Easy

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
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • countPartitions

      public int countPartitions(int[] nums)