456. 132 Pattern

Problem

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

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var find132pattern = function(nums) {
    if (nums.length < 3) return false;
    var stack = [];
    var min = Array(nums.length);
    min[0] = nums[0];
    for (var i = 1; i < nums.length; i++) {
        min[i] = Math.min(min[i - 1], nums[i]);
    }
    for (var j = nums.length - 1; j >= 0; j--) {
        if (nums[j] > min[j]) {
            while (stack.length && stack[stack.length - 1] <= min[j]) stack.pop();
            if (stack.length && stack[stack.length - 1] < nums[j]) return true;
            stack.push(nums[j]);
        }
    }
    return false;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).