Next Greater Element 2

Problem

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.

Examples

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]

Solution

Method 1 - Monotonic Decreasing Stack on Indices with Twice the Loop

This is similar to [NGR - Nearest Greater Element to Right of every element](ngr-nearest-greater-element-to-right-of-every-element) OR [Next Greater Element 1](next-greater-element-1). But we need to use indices as input to the stack and also we have to loop twice:

Code

Java

public int[] nextGreaterElements(int[] nums) {
	int n = nums.length, ans[] = new int[n];

	Arrays.fill(ans, -1);

	Stack<Integer> stack = new Stack<>();
	
	for (int i = 0; i <n * 2; i++) {
		while (!stack.isEmpty() && nums[stack.peek()]<nums[i % n]) {
			ans[stack.pop()] = nums[i % n];
		}
		stack.push(i % n);
	}
	return ans;
}

Complexity

• ⏰ Time complexity: O(n)
• 🧺 Space complexity: O(n)