Next_Greater_Element_II.java

/*

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:

1 <= nums.length <= 104
-109 <= nums[i] <= 109

*/
import java.util.*;
class Next_Greater_Element_II {
    public int[] nextGreaterElements(int[] nums) {
        Stack<Integer> stack = new Stack<>();
        int n = nums.length;
        int[] res = new int[n];

        for (int i = 2 * n - 1; i >= 0; i--) {
            int idx = i % n;

           
            while (!stack.isEmpty() && stack.peek() <= nums[idx]) {
                stack.pop();
            }

            
            if (i < n) {
                if (stack.isEmpty()) {
                    res[idx] = -1;
                } else {
                    res[idx] = stack.peek();
                }
            }

            
            stack.push(nums[idx]);
        }

        return res;
    }
}