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;
    }
};