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Tested Python solution for LeetCode 870 with 20 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 870 # by problem number
lcpy gen -s advantage_shuffle # by problem name
nums1 and nums2 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i].
Return any permutation of nums1 that maximizes its advantage with respect to nums2.
Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
Output: [2,11,7,15]
Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]
Output: [24,32,8,12]
1 <= nums1.length <= 10^5nums2.length == nums1.length0 <= nums1[i], nums2[i] <= 10^9nums1 that reaches the maximum possible advantage count.
class Solution:
# Time: O(n log n)
# Space: O(n)
def advantage_count(self, nums1: list[int], nums2: list[int]) -> list[int]:
sorted1 = sorted(nums1)
order = sorted(range(len(nums2)), key=lambda i: nums2[i])
result = [0] * len(nums1)
low = 0
high = len(nums1) - 1
for idx in reversed(order):
if sorted1[high] > nums2[idx]:
result[idx] = sorted1[high]
high -= 1
else:
result[idx] = sorted1[low]
low += 1
return result
| Time | Space |
|---|---|
| O(n log n) | O(n) |