Given an integer array nums, return an integer array counts where counts[i] is the number of smaller elements to the right of nums[i].
Example 1:
Input: nums = [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
Example 2:
Input: nums = [-1]
Output: [0]
Example 3:
Input: nums = [-1,-1]
Output: [0,0]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 104
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
|
class Solution:
def countSmaller(self, nums: List[int]) -> List[int]:
def sort(enums):
length = len(enums)
half = length // 2
if not half:
return enums
left, right = sort(enums[:half]), sort(enums[half:])
# merge in-place
for i in range(length-1, -1, -1):
if not right or (left and left[-1][1] > right[-1][1]):
# nums remained in right need to shift
# to the left of left[-1]
res[left[-1][0]] += len(right)
enums[i] = left.pop()
else:
enums[i] = right.pop()
return enums
res = [0] * len(nums)
sort(list(enumerate(nums)))
return res
|