7.Next Greater Element II

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. 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, output -1 for this number.

Example 1:
Input: [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.

Solution I: (Brute Force)

Creating an array of size 2 * n and searching

class Solution
{
public:
    vector<int> nextGreaterElements(vector<int> &nums)
    {

        vector<int> res;
        int n = nums.size();
        vector<int> v(2 * n);

        for (int i = 0; i < 2 * n; i++)
        {
            v[i] = nums[i % n];
        }

        for (int i = 0; i < n; i++)
        {
            int max = -1;
            for (int j = i+1; j < v.size(); j++)
            {
                if (v[j] > nums[i])
                {
                    max = v[j];
                    break;
                }
            }
            res.push_back(max);
        }
        return res;
    }
};

Time Complexity: O(N^2)

Solution II: (Using stack)

To get in a circular array we iterate through the array twice

class Solution
{
public:
    vector<int> nextGreaterElements(vector<int> &nums)
    {

        stack<int> s;
        int n = nums.size();
        vector<int> res(n, -1);

        for (int i = 0; i < 2 * n; i++)
        {

            while (!s.empty() && nums[s.top()] < nums[i % n])
            {
                int idx = s.top();
                res[idx] = nums[i % n];
                s.pop();
            }
            s.push(i % n);
        }

        return res;
    }
};

Time Complexity: O(N)