README.md

2454. Next Greater Element IV

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

Solutions (Rust)

1. Solution

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    pub fn second_greater_element(nums: Vec<i32>) -> Vec<i32> {
        let mut heap0 = BinaryHeap::new();
        let mut heap1 = BinaryHeap::new();
        let mut answer = vec![-1; nums.len()];

        for i in 0..nums.len() {
            while let Some(&Reverse((x, j))) = heap1.peek() {
                if x >= nums[i] {
                    break;
                }
                answer[j] = nums[i];
                heap1.pop();
            }
            while let Some(&Reverse((x, j))) = heap0.peek() {
                if x >= nums[i] {
                    break;
                }
                heap1.push(heap0.pop().unwrap());
            }
            heap0.push(Reverse((nums[i], i)));
        }

        answer
    }
}