leetcode - 456. 132 Pattern

Description

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.
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Example 2:

Input: nums = [3,1,4,2]
Output: true
Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
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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].
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Constraints:

n == nums.length
1 <= n <= 2 * 10^5
-10^9 <= nums[i] <= 10^9
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Solution

Didn’t even come up with $n^2$ solution, I feel so bad.

$o(n^2)$ solution

Scan from left to right, keep track of the minimum value, and at each point scan the right of the list, to find the middle element.

Time complexity: $o(n^2)$
Space complexity: $o(1)$

Monotonic Stack

Solved after help.

ref: https://leetcode.com/problems/132-pattern/solutions/94071/single-pass-c-o-n-space-and-time-solution-8-lines-with-detailed-explanation/

We are going to search for a subsequence s1, s2, s3, that s1 < s3 < s2. s3 should be as large as possible if we could find a s2 that satisfies s2 > s3. So we could use a decreasing stack to store all the elements from right to left, if the current number is larger than the top element, then the element we pop from the stack is a candidate for s3.

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

$o(n^2)$ solution

class Solution:
    def find132pattern(self, nums: List[int]) -> bool:
        min_val = nums[0]
        for j in range(1, len(nums)):
            if nums[j] <= min_val:
                min_val = nums[j]
                continue
            else:
                for k in range(j + 1, len(nums)):
                    if min_val < nums[k] < nums[j]:
                        return True
        return False
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Monotonic Stack

class Solution:
    def find132pattern(self, nums: List[int]) -> bool:
        stack = []
        s3 = None
        for each_num in nums[::-1]:
            while stack and stack[-1] < each_num:
                if s3:
                    s3 = max(s3, stack.pop())
                else:
                    s3 = stack.pop()
            stack.append(each_num)
            if s3 and each_num < s3:
                return True
        return False
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