2454. Next Greater Element IV

Problem

You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer.

The second greater integer of nums[i] is nums[j] such that:

• j > i

• nums[j] > nums[i]

• There exists exactly one index k such that nums[k] > nums[i] and i < k < j.

If there is no such nums[j], the second greater integer is considered to be -1.

• For example, in the array [1, 2, 4, 3], the second greater integer of 1 is 4, 2 is 3, and that of 3 and 4 is -1.

Return** an integer array answer, where answer[i] is the second greater integer of nums[i].**

  Example 1:

Input: nums = [2,4,0,9,6]
Output: [9,6,6,-1,-1]
Explanation:
0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
Thus, we return [9,6,6,-1,-1].

Example 2:

Input: nums = [3,3]
Output: [-1,-1]
Explanation:
We return [-1,-1] since neither integer has any integer greater than it.

  Constraints:

• 1 <= nums.length <= 105

• 0 <= nums[i] <= 109

Solution

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var secondGreaterElement = function(nums) {
    var res = Array(nums.length).fill(-1);
    var stack1 = []; // first greater integer
    var stack2 = []; // second greater integer
    for (var i = 0; i < nums.length; i++) {
        while (stack2.length && nums[i] > nums[stack2[stack2.length - 1]]) {
            res[stack2.pop()] = nums[i];
        }
        var tempArr = [];
        while (stack1.length && nums[i] > nums[stack1[stack1.length - 1]]) {
            tempArr.unshift(stack1.pop());
        }
        stack2.push(...tempArr);
        stack1.push(i);
    }
    return res;
};

Explain:

nope.

Complexity:

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