# 456. 132 Pattern

https://leetcode.com/problems/132-pattern/

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
``````

• code
``````class Solution:
def find132pattern(self, nums: List[int]) -> bool:
minlist = [nums] * len(nums)
minsofar = nums
for i in range(1, len(nums)):
minlist[i] = min(minlist[i-1], nums[i])

stack = []
for i in range(len(nums)-1, -1, -1):
while stack and stack[-1] < nums[i] # stack[-1] is the middle one
if stack[-1] > minlist[i]:
return True
stack.pop()
stack.append(nums[i])
return False
``````