870

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

Related Topics:
Array, Greedy

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) {
        int N = A.size(), i = 0;
        vector<int> ia(N), ib(N), ans(N, -1);
        iota(begin(ia), end(ia), 0);
        iota(begin(ib), end(ib), 0);
        sort(begin(ia), end(ia), [&](int a, int b) { return A[a] > A[b]; });
        sort(begin(ib), end(ib), [&](int a, int b) { return B[a] > B[b]; });
        for (int j = 0; i < N && j < N; ++i) {
            while (j < N && B[ib[j]] >= A[ia[i]]) ++j;
            if (j < N) {
                ans[ib[j++]] = A[ia[i]];
            } else break;
        }
        for (int j = 0; i < N && j < N; ++i) {
            while (j < N && ans[j] != -1) ++j;
            if (j < N) {
                ans[j++] = A[ia[i]];
            }
        }
        return ans;
    }
};