Introduction

In this blog post, we will explore how to solve the LeetCode problem "456" 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 array of n integers nums, a 132 pattern is a subsequence of three integers nums[i], nums[j] and nums[k] such that i < j < k and nums[i] < nums[k] < nums[j]. Return true if there is a 132 pattern in nums, otherwise, return false. Example 1: Input: nums = [1,2,3,4] Output: false Explanation: There is no 132 pattern in the sequence. Example 2: Input: nums = [3,1,4,2] Output: true Explanation: There is a 132 pattern in the sequence: [1, 4, 2]. Example 3: Input: nums = [-1,3,2,0] Output: true Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0]. Constraints: n == nums.length 1 <= n <= 2 * 105 -109 <= nums[i] <= 109

Explanation

• Monotonic Stack for '2': Utilize a stack to maintain decreasing elements from right to left. These represent potential '2' values in the 132 pattern.
  • Finding '3': Iterate from right to left. For each number encountered, check if it can be the '3' by comparing it with the elements in the stack. If it's greater than an element in the stack, that element becomes a candidate for '2', and we compare the current number (potential '3') with the largest such '2' encountered so far.
  • Finding '1': If the current number (potential '3') is less than the largest '2' we found, a 132 pattern exists.

• Time Complexity: O(n), Space Complexity: O(n)

Code

    def find132pattern(nums):
    """
    Finds if a 132 pattern exists in the given array.

    Args:
        nums: A list of integers.

    Returns:
        True if a 132 pattern exists, False otherwise.
    """
    stack = []
    second = float('-inf')  # Initialize '2' to negative infinity

    for i in range(len(nums) - 1, -1, -1):
        if nums[i] < second:
            return True

        while stack and nums[i] > stack[-1]:
            second = stack.pop()

        stack.append(nums[i])

    return False