Solution.py

"""

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.
Follow up: The O(n^2) is trivial, could you come up with the O(n logn) or the O(n) solution?
 
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 <= 104
-109 <= nums[i] <= 109


"""


class Solution:
    def find132pattern(self, nums):
        min_list = list(accumulate(nums, min))
        stack, n = [], len(nums)
        
        for j in range(n-1, -1, -1):
            if nums[j] > min_list[j]:
                while stack and stack[-1] <= min_list[j]:
                    stack.pop()
                if stack and stack[-1] < nums[j]:
                    return True
                stack.append(nums[j])           
        return False