# Solving Leetcode Interviews in Seconds with AI: Partition Array into Disjoint Intervals


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "915" 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
	> Given an integer array nums, partition it into two (contiguous) subarrays left and right so that:  Every element in left is less than or equal to every element in right. left and right are non-empty. left has the smallest possible size.  Return the length of left after such a partitioning. Test cases are generated such that partitioning exists.   Example 1:  Input: nums = [5,0,3,8,6] Output: 3 Explanation: left = [5,0,3], right = [8,6]  Example 2:  Input: nums = [1,1,1,0,6,12] Output: 4 Explanation: left = [1,1,1,0], right = [6,12]    Constraints:  2 <= nums.length <= 105 0 <= nums[i] <= 106 There is at least one valid answer for the given input.  

	# Explanation
	Here's the solution to find the optimal partitioning of an integer array into two subarrays, as described:

*   **High-Level Approach:**
    *   Iterate through possible lengths of the left subarray.
    *   For each length, check if the condition (max(left) <= min(right)) is satisfied.
    *   Return the smallest length that satisfies the condition.

*   **Complexity:**
    *   Runtime Complexity: O(n), where n is the length of the input array `nums`.
    *   Storage Complexity: O(1) - constant extra space.

	
	# Code
	```python
	def partitionDisjoint(nums: list[int]) -> int:
    """
    Partitions an integer array into two subarrays, left and right, such that:
    - Every element in left is less than or equal to every element in right.
    - left and right are non-empty.
    - left has the smallest possible size.

    Returns the length of left after such a partitioning.
    """

    n = len(nums)
    max_left = nums[0]
    overall_max = nums[0]
    partition_index = 1

    for i in range(1, n):
        if nums[i] < max_left:
            partition_index = i + 1
            max_left = overall_max
        else:
            overall_max = max(overall_max, nums[i])

    return partition_index
	```