870. Advantage Shuffle

870. Advantage Shuffle (Medium)

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

Solution 1.

// OJ: https://leetcode.com/problems/advantage-shuffle/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> advantageCount(vector<int>& A, vector<int>& B) {
        sort(A.begin(), A.end());
        vector<int> copy(B.begin(), B.end());
        sort(B.begin(), B.end());
        unordered_map<int, queue<int>> m;
        queue<int> leftover;
        int i = 0;
        for (int a : A) {
            if (a > B[i]) m[B[i++]].push(a);
            else leftover.push(a);
        }
        vector<int> ans;
        for (int b : copy) {
            if (m[b].size()) {
                ans.push_back(m[b].front());
                m[b].pop();
            } else {
                ans.push_back(leftover.front());
                leftover.pop();
            }
        }
        return ans;
    }
};