503. Next Greater Element II

Description

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.

Constraints

• 1 <= nums.length <= 104

• -109 <= nums[i] <= 109

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

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]

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        if(nums == null || nums.length == 0) {
            return new int[0];
        }
        
        int n = nums.length;
        int[] result = new int[n];
        Stack<Integer> stack = new Stack<>();
        Arrays.fill(result, -1);
        
        for(int i = 0; i < 2*n; i++) {
            while(!stack.isEmpty() && nums[stack.peek()] < nums[i%n]) {
                result[stack.pop()] = nums[i%n];
            }
            if(i < n) {
                stack.push(i);
            }
        }
        
        return result;
    }
}

Follow up