# Solving Leetcode Interviews in Seconds with AI: Count Partitions with Even Sum Difference


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3432" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Calculate Total Sum:** First, compute the sum of the entire array. This allows us to efficiently calculate the right subarray sum by subtracting the left subarray sum from the total sum.
*   **Iterate and Check:** Iterate through all possible partition indices. For each index `i`, calculate the sum of the left subarray (from index 0 to `i`).  Then calculate the right subarray sum (total sum - left subarray sum) and check if the difference between the left and right sums is even.
*   **Count Even Differences:** Maintain a counter to keep track of the number of partitions that result in an even difference.

*   **Runtime Complexity:** O(n), where n is the length of the array.
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	def count_even_partition(nums):
    """
    Calculates the number of partitions where the difference between the sum of the left and right subarrays is even.

    Args:
        nums: An integer array of length n.

    Returns:
        The number of partitions with an even sum difference.
    """

    n = len(nums)
    total_sum = sum(nums)
    count = 0
    left_sum = 0

    for i in range(n - 1):
        left_sum += nums[i]
        right_sum = total_sum - left_sum
        if (left_sum - right_sum) % 2 == 0:
            count += 1

    return count
	```