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.
题目链接：https://leetcode.com/problems/advantage-shuffle/
Difficulty: medium
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
分析
- updating
参考代码
class Solution:
def advantageCount(self, A, B):
s_a=sorted(A)
s_b=sorted(B)
assigned={b:[] for b in B}
li=[]
j=0
for a in s_a:
if(a > s_b[j]):
assigned[s_b[j]].append(a)
j += 1
else:
li.append(a)
return [assigned[b].pop() if assigned[b] else li.pop() for b in B]