503 - Next Greater Element II

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]

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        vector<int> result( nums.size(), -1 );
        stack<pair<int,int>> st;
        for ( int i = 0; i < 2 * nums.size(); i++ ) {
            int idx = i % nums.size();
            while ( !st.empty() && st.top().first < nums[idx] ) {
                pair<int,int> cur = st.top();
                st.pop();
                // Because we go through nums twice
                // we might update some ans twice (which we do not desire)
                // so we only update if ans[cur.second] == -1
                if ( result[cur.second] == -1 ) result[cur.second] = nums[idx];
            }

            if ( result[idx] == -1 ) st.push( make_pair( nums[idx], idx ) );
        }


        return result;
    }
};