Problem Statement

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

Problem Link

Next Greater Element II - LeetCode

Code

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        
        int n = nums.size();
        vector<int> res(n,-1);
        
        stack<int> s;
        
        for(int i=0;i<2*n;i++)
        {
            while(s.size()&&nums[s.top()]<nums[i%n])
            {
                res[s.top()] = nums[i%n];
                s.pop();
            }
            
            s.push(i%n);
        }
      
        return res;
        
        
    }
};