LeetCode in Kotlin

503. Next Greater Element II

Medium

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn’t exist, return -1 for this number.

Example 1:

Input: nums = [1,2,1]

Output: [2,-1,2] Explanation: The first 1’s next greater number is 2; The number 2 can’t find next greater number. The second 1’s next greater number needs to search circularly, which is also 2.

Example 2:

Input: nums = [1,2,3,4,3]

Output: [2,3,4,-1,4]

Constraints:

Solution

import java.util.ArrayDeque
import java.util.Deque

class Solution {
    fun nextGreaterElements(nums: IntArray): IntArray {
        val result = IntArray(nums.size)
        val stack: Deque<Int> = ArrayDeque()
        for (i in nums.size * 2 - 1 downTo 0) {
            while (stack.isNotEmpty() && nums[stack.peek()] <= nums[i % nums.size]) {
                stack.pop()
            }
            result[i % nums.size] = if (stack.isEmpty()) -1 else nums[stack.peek()]
            stack.push(i % nums.size)
        }
        return result
    }
}
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