870. 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. 1 <= A.length = B.length <= 10000
2. 0 <= A[i] <= 10^9
3. 0 <= B[i] <= 10^9

分别排序，优先用a的小的处理b的小的，反正最后要的是个数

注意b排序前先求出排序后的index变化

class Solution:
    def advantageCount(self, a, b):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: List[int]
        """
        b_idx = sorted(range(len(b)), key=lambda k:b[k])
        a.sort()
        b.sort()
        res=[]
        not_used=[]
        
        j=0
        for i in range(len(b)):
            while j<len(a) and a[j]<=b[i]: 
                not_used.append(a[j])
                j+=1
            if j==len(a): break
            res.append(a[j])
            j+=1
        a=res+not_used
        
        a2=[-1]*len(a)
        for v, i in zip(a, b_idx):
            a2[i]=v
        return a2