132_Pattern.swift

/*
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
*/

/*
Solution 1:
Stack

from end to begin,
use stack to keep recording current largest and smallest element
then see if current element is smallest one or not

Time Complexity: O(n)
Space Complexity: O(n)
*/
class Solution {
    func find132pattern(_ nums: [Int]) -> Bool {
        let n = nums.count
        if n < 3 { return false }

        var k = Int.min

        // track j
        var stack = [Int]()
        for i in stride(from: n-1, through: 0, by: -1) {
            if nums[i] < k { return true }
            while !stack.isEmpty, stack.last! < nums[i] {
                k = stack.removeLast()
            }

            stack.append(nums[i])
        }
        return false
    }
}