LeetCodeCount of Smaller Numbers After Self
Sorting
Asked at 6 companies
2 recent
- Amazon
- Infosys
- Bloomberg
- Microsoft
- Meta
Community-reported · source · updated 2026-10-08. ● asked in the last 6 months.
Go step by step: try it yourself before the video and the visual. Progress saves on this device.
0 / 160 pointsSolved (Hard) 0/70Reviews 0/50Mastered 0/25Practice steps 0/15
Step 1
Know these before you start
Practise each one here, or tick what you already know
Concept videoStep 2
Try it on paper
5–10 minutes, no code yet
In the wild · Every “Sort by” button: Sorting a spreadsheet by price, then by name, needs a stable sort, which is why Python and JavaScript both use merge-sort-based TimSort.
About this problem
Write merge sort and partitioning yourself, then use sorted order (and in-place tricks) to make problems easy.
Read the full statement and examples on LeetCode, try it for 20–30 minutes, then come back for the walkthrough.
Examples
Full statement on LeetCode ↗Example 1
- Input
- nums = [5, 2, 6, 1]
- Output
- [2, 1, 1, 0]
Example 2
Just one itemNegative numbers- Input
- nums = [-1]
- Output
- [0]
Example 3
Negative numbersAll the same- Input
- nums = [-1, -1]
- Output
- [0, 0]
Code it here
Tab indents · ⌘/Ctrl+Enter runs · Esc leaves the editor
Step 6
Get it accepted
All test cases pass
Step 7
Review it so it sticks
Re-solve from memory after
The step that matters most: without reviews most of a solved problem fades within weeks. Three short re-solves, each just as you start to forget, are what make it yours.
Your notes
Three lines in your own words. They come back with this problem's reviews, so you can check yourself after recalling the solution.
Solve it yourself before you look
Reading the solution first feels like progress, but it makes the next similar problem — and the interview version — much harder, because you skipped the part where you figure it out. Give it an honest 20–30 minutes. Stuck? Re-read the pattern, watch the concept video, or try the brute force first.
Hidden: optimal solution
Solution code
Optimal solutionthe indices by value. While merging, a number from the left half is bigger than every right-half number already taken, so add that count to it. Each number collects its count across all the merges.
class Solution:
def countSmaller(self, nums: List[int]) -> List[int]:
counts = [0] * len(nums)
def sort(idx: List[int]) -> List[int]:
if len(idx) <= 1:
return idx
mid = len(idx) // 2
left, right = sort(idx[:mid]), sort(idx[mid:])
out, j = [], 0
for i in left:
# right-half numbers smaller than this one go first
while j < len(right) and nums[right[j]] < nums[i]:
out.append(right[j])
j += 1
counts[i] += j
out.append(i)
return out + right[j:]
sort(list(range(len(nums))))
return countsReference solution, tested in Python, Java and C++.
Other ways to solve it
- Thought of another approach (slower, or cleverer)? Code it here: the same tests check any approach.
- Community solutions on LeetCode ↗ for many more versions, in many languages. Read them after you've solved it.