503. Next Greater Element II #

Problem #

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. 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, output -1 for this number.

Example 1:

Input: [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.

Note: The length of given array won’t exceed 10000.

Summary #

The problem gives an array A. For each element in array A, find a number greater than that element in array A. A is a circular array. If found, output this value; if not found, output -1.

Solution Approach #

This problem is an enhanced version of Problem 496, adding the condition of a circular array on top of Problem 496. This problem can still be simulated using the same approach as Problem 496. A better approach is to use a monotonic stack, where the stack records indices in monotonically increasing order.

Code #

package leetcode

// Solution 1 Monotonic stack
func nextGreaterElements(nums []int) []int {
	res := make([]int, 0)
	indexes := make([]int, 0)
	for i := 0; i < len(nums); i++ {
		res = append(res, -1)
	}
	for i := 0; i < len(nums)*2; i++ {
		num := nums[i%len(nums)]
		for len(indexes) > 0 && nums[indexes[len(indexes)-1]] < num {
			index := indexes[len(indexes)-1]
			res[index] = num
			indexes = indexes[:len(indexes)-1]
		}
		indexes = append(indexes, i%len(nums))
	}
	return res
}

// Solution 2
func nextGreaterElements1(nums []int) []int {
	if len(nums) == 0 {
		return []int{}
	}
	res := []int{}
	for i := 0; i < len(nums); i++ {
		j, find := (i+1)%len(nums), false
		for j != i {
			if nums[j] > nums[i] {
				find = true
				res = append(res, nums[j])
				break
			}
			j = (j + 1) % len(nums)
		}
		if !find {
			res = append(res, -1)
		}
	}
	return res
}