3432. Count Partitions with Even Sum Difference

Description

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

 

Solutions

Solution: Math

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const countPartitions = function (nums) {
  const n = nums.length;
  const sum = nums.reduce((result, num) => result + num);

  return sum % 2 ? 0 : n - 1;
};