README_EN.md

456. 132 Pattern

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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.

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

Solutions

Python3

class Solution:
    def find132pattern(self, nums: List[int]) -> bool:
        ak = float('-inf')
        stack = []
        for num in nums[::-1]:
            if num < ak:
                return True
            while stack and num > stack[-1]:
                ak = stack.pop()
            stack.append(num)
        return False

Java

class Solution {
    public boolean find132pattern(int[] nums) {
        int ak = Integer.MIN_VALUE;
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = nums.length - 1; i >= 0; --i) {
            if (nums[i] < ak) {
                return true;
            }
            while (!stack.isEmpty() && nums[i] > stack.peek()) {
                ak = stack.pop();
            }
            stack.push(nums[i]);
        }
        return false;
    }
}

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