1. Brute Force Approach
2. Using Min Array and Binary Search
3. Optimal Approach using Stack

The brute force approach involves a straightforward examination of all possible triplets in the array to determine whether they form the 132 pattern.

The idea is to iterate over all possible combinations of three indices such that i < j < k and check if the values at these indices satisfy the condition nums[i] < nums[k] < nums[j].

def find132pattern(nums):
    n = len(nums)

    # Traverse each triplet combination
    for i in range(n - 2):
        for j in range(i + 1, n - 1):
            for k in range(j + 1, n):
                # Check if it's 132 pattern
                if nums[i] < nums[k] < nums[j]:
                    return True
    return False

• Time Complexity: (O(n^3)) - Due to three nested loops iterating over the list.
• Space Complexity: (O(1)) - No extra data structures are used.

This approach reduces the time complexity by maintaining an array of minimum values up to each index and using binary search to find a potential nums[k].

1. Maintain a min_array where min_array[i] is the minimum value from the start to the current index i.
2. For each index j, we need to find an index k such that nums[k] is greater than min_array[j] and less than nums[j].

from bisect import bisect_left

def find132pattern(nums):
    n = len(nums)
    if n < 3:
        return False
    
    min_array = [nums[0]] * n
    for i in range(1, n):
        min_array[i] = min(min_array[i-1], nums[i])
    
    sorted_list = []

    for j in range(n-1, -1, -1):
        if nums[j] > min_array[j]:
            pos = bisect_left(sorted_list, min_array[j] + 1)
            if pos < len(sorted_list) and sorted_list[pos] < nums[j]:
                return True
            bisect.insort(sorted_list, nums[j])
    
    return False

• Time Complexity: (O(n \log n)) - Due to the use of binary search for potential candidates.
• Space Complexity: (O(n)) - For the min_array and the sorted list.

The most efficient way utilizes a stack to keep track of potential nums[k] from the end to the start.

We move backwards from the end of the array and use a stack to track all possible values of nums[k]. As we iterate each number as a potential nums[j], we check the condition against a maintained "second" number, which acts as our nums[k].

1. Use a stack to maintain potential nums[k].
2. Traverse the list from the right to find feasible nums[j] and nums[k].

def find132pattern(nums):
    stack = []
    second = float('-inf')

    for num in reversed(nums):
        # Check if the current number can be nums[i]
        if num < second:
            return True
        # Update the stack with potential nums[k]
        while stack and num > stack[-1]:
            second = stack.pop()
        stack.append(num)

    return False

• Time Complexity: (O(n)) - Each element is pushed and popped from the stack at most once.
• Space Complexity: (O(n)) - The stack can hold up to n elements.