forked from YuriSpiridonov/LeetCode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
1365.HowManyNumbersAreSmallerThantheCurrentNumber.py
34 lines (30 loc) · 1.39 KB
/
1365.HowManyNumbersAreSmallerThantheCurrentNumber.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
"""
Given the array nums, for each nums[i] find out how many numbers in the
array are smaller than it. That is, for each nums[i] you have to count the
number of valid j's such that j != i and nums[j] < nums[i].
Return the answer in an array.
Example:
Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation:
- For nums[0]=8 there exist four smaller numbers than it
(1, 2, 2 and 3).
- For nums[1]=1 does not exist any smaller number than it.
- For nums[2]=2 there exist one smaller number than it (1).
- For nums[3]=2 there exist one smaller number than it (1).
- For nums[4]=3 there exist three smaller numbers than it
(1, 2 and 2).
Constraints:
- 2 <= nums.length <= 500
- 0 <= nums[i] <= 100
"""
#Difficulty: Easy
#103 / 103 test cases passed.
#Runtime: 92 ms
#Memory Usage: 13.7 MB
#Runtime: 92 ms, faster than 56.51% of Python3 online submissions for How Many Numbers Are Smaller Than the Current Number.
#Memory Usage: 13.7 MB, less than 93.98% of Python3 online submissions for How Many Numbers Are Smaller Than the Current Number.
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
sorted_nums = sorted(nums)
return [len(sorted_nums[:sorted_nums.index(n)]) for n in nums]