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, 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

    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