870. Advantage Shuffle
Description
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
.
Constraints
1 <= A.length = B.length <= 10000
0 <= A[i] <= 10^9
0 <= B[i] <= 10^9
Approach
Links
GeeksforGeeks
ProgramCreek
YouTube
Examples
Input: A = [2, 7, 11, 15], B = [1, 10, 4, 11]
Output: [2, 11, 7, 15]
Solutions
/**
* Time complexity :
* Space complexity :
*/
class Solution {
public int[] advantageCount(int[] A, int[] B) {
int[] result = new int[A.length];
Arrays.sort(A);
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[1] - a[1]);
for(int i = 0; i < B.length; i++) {
pq.offer(new int[]{i, B[i]});
}
int slow = 0, fast = A.length-1;
while(!pq.isEmpty()) {
int[] b = pq.poll();
result[b[0]] = b[1] >= A[fast]? A[slow++]: A[fast--];
}
return result;
}
}
Follow up
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