-
Notifications
You must be signed in to change notification settings - Fork 523
/
Copy path24. Advantage Shuffle.cpp
54 lines (45 loc) · 1.03 KB
/
24. Advantage Shuffle.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
/*
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 <= A.length = B.length <= 10000
0 <= A[i] <= 10^9
0 <= B[i] <= 10^9
*/
class Solution
{
public:
vector<int> advantageCount(vector<int> &A, vector<int> &B)
{
vector<vector<int>> b;
for (int i = 0; i < B.size(); ++i)
b.push_back({B[i], i});
sort(b.begin(), b.end(), greater<vector<int>>());
sort(A.begin(), A.end());
int i = 0, j = A.size() - 1;
vector<int> ans(A.size(), -1);
for (int k = 0; k < b.size(); ++k)
{
int pos = b[k][1];
if (A[j] > b[k][0])
{
ans[pos] = A[j];
j--;
}
else
{
ans[pos] = A[i];
i++;
}
}
return ans;
}
};