Queue Reconstruction by Height -- LeetCode

[Question]

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]]

[Analysis]
Be Greedy. Person (h,k) at position 0 should have k=0. So group all (h,0), the one with smallest h' is at position 0. Then look at all other persons (h,k) whose h is greater or equal to h', decrease their k by 1 (because h' contribute 1 to their k values). Repeat the previous steps and find person at position 1, .., n. The time complexity is O(N^2).

Another way to reconstruct the queue is by sorting. Suppose a person (h',k'), all persons (h,k) with greater or equal height have been in a sorted queue Q, the k' is the right position to insert (h',k') into the Q and create a Q' while maintaining all existing (h,k). Repeat the same process on Q' and remaining persons. The time complexity is O(N^2).

[Solution]
//-- using sorting --
class Solution {
public:
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        auto comp=[](pair<int,int> &a, pair<int,int> &b)
            { return a.first<b.first || a.first==b.first && a.second > b.second;};
        sort (people.begin(), people.end(), comp);
        vector<int,int> res;
        for (int i=people.size(); i>=0; i++)
            res.(res.begin()+people[i].second, people[i]);
    }
};

//--greedy --
class Solution {
public:
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        vector<pair<int, int>> rslt;
     
        vector<pair<int, int>> bak (people);
        auto comp= [](pair<int, int> a, pair<int,int> b)
            { return a.second < b.second || a.second ==b.second && a.first<b.first;};
         
        while (rslt.size()!= people.size() ) {
            auto it = min_element(bak.begin(), bak.end(), comp);
            rslt.push_back(people[it-bak.begin()] );
            it->second = INT_MAX;
            for (auto &p: bak) {
                if (p.second!=INT_MAX && p.first<=it->first) {
                    p.second --;
                }
            }
        }
        return rslt;
    }
};