# Advantage Shuffle

Updated: Mar 24, 2021

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

**Solution**:

```
class Solution {
public int[] advantageCount(int[] A, int[] B) {
int[] result = new int[A.length];
Arrays.sort(A);
PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> b[1] - a[1]);
for (int i = 0; i < B.length; i++) {
maxHeap.offer(new int[] {i, B[i]});
}
int slow = 0, fast = A.length - 1;
while (!maxHeap.isEmpty()) {
int[] b = maxHeap.poll();
result[b[0]] = b[1] >= A[fast] ? A[slow++] : A[fast--];
}
return result;
}
}
```