31-Days-October-Leetcode-master/23.132_Pattern/23.132_Pattern.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: List[int]) -> bool:       
        if len(set(nums)) < 3:
            return False
        min_nums = [nums[0]]
        for i in range(1, len(nums)):
            min_nums.append(min(nums[i], min_nums[-1]))
        stack = []  
        i = len(nums) - 1
        for i in range(len(nums)-1, -1, -1):
            if( nums[i] > min_nums[i] ):
                while( stack and stack[-1] <= min_nums[i] ):
                    stack.pop()
                if(stack and min_nums[i] < stack[-1] < nums[i] ):
                    return True
                stack.append(nums[i])
        return False