Leetcode #456: 132 Pattern
In this guide, we solve Leetcode #456 132 Pattern in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Array, Binary Search, Ordered Set, Monotonic Stack
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: nums = [1,2,3,4]
Output: false
Explanation: There is no 132 pattern in the sequence.
Python Solution
class Solution:
def find132pattern(self, nums: List[int]) -> bool:
vk = -inf
stk = []
for x in nums[::-1]:
if x < vk:
return True
while stk and stk[-1] < x:
vk = stk.pop()
stk.append(x)
return False
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.