18.Queue Reconstruction by Height

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.

Note: The number of people is less than 1,100.

Example

Input:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

Output:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

Solution : (Using Sorting)

class Solution
{
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>> &people)
    {
        
        int n = people.size();
        
        vector<vector<int>> v(n, vector<int>(2, -1));
     
        sort(people.begin(), people.end());

        for (int i = 0; i < people.size(); i++)
        {
            int count = people[i][1];
            int j = 0;
            while (count > 0)
            {
                if (v[j][0] == -1 || v[j][0] >= people[i][0])
                {
                    count--;
                }
                j++;
            }

            while (v[j][0] != -1)
            {
                j++;
            }
            v[j][0] = people[i][0];
            v[j][1] = people[i][1];
        }

        return v;
    }
};

Time Complexity: O(n^2)