870. Advantage Shuffle

Given two arrays A and B of equal size, the advantage of A with respect to B is the number of indices i for which A[i] > B[i].

Return any permutation of A that maximizes its advantage with respect to B.

Example 1:

Input: A = [2,7,11,15], B = [1,10,4,11]

Output: [2,11,7,15]

Example 2:

Input: A = [12,24,8,32], B = [13,25,32,11]

Output: [24,32,8,12]

Note:

1. 1 <= A.length = B.length <= 10000
2. 0 <= A[i] <= 10^9
3. 0 <= B[i] <= 10^9

Approach #1: C++.

class Solution {

public:

    vector<int> advantageCount(vector<int>& A, vector<int>& B) {

        map<int, int> m;

        for (int i : A) m[i]++;

        map<int, int>::iterator it;

        vector<int> res;

        for (int i : B) {

            it = m.upper_bound(i);

            int x = it != m.end() ? it->first : m.begin()->first;

            if (--m[x] == 0) m.erase(x);

            res.push_back(x);

        }

        return res;

    }

};