LeetCodeFind All Numbers Disappeared in an Array
Sorting
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Community-reported · source · updated 2026-10-08. ● asked in the last 6 months.
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0 / 110 pointsSolved (Easy) 0/20Reviews 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 and the step-by-step visual.
Examples
Full statement on LeetCode ↗Example 1
- Input
- nums = [4, 3, 2, 7, 8, 2, 3, 1]
- Output
- [5, 6]
Walk through it step by step(13 steps · shows the approach, so try first)
- Every number is between 1 and 8, so each one has a home: number v belongs at index v − 1. Pass 1 marks a home as visited by making its value negative; pass 2 reads off the homes nobody visited. No extra memory needed.
- nums[0] is 4, whose home is index 4 − 1 = 3. Negate nums[3] (7 becomes -7): 4 is here.
- nums[1] is 3, whose home is index 3 − 1 = 2. Negate nums[2] (2 becomes -2): 3 is here.
- nums[2] is -2 (use |-2| = 2), whose home is index 2 − 1 = 1. Negate nums[1] (3 becomes -3): 2 is here.
- nums[3] is -7 (use |-7| = 7), whose home is index 7 − 1 = 6. Negate nums[6] (3 becomes -3): 7 is here.
- nums[4] is 8, whose home is index 8 − 1 = 7. Negate nums[7] (1 becomes -1): 8 is here.
- nums[5] is 2, whose home is index 1. nums[1] is already negative (we've seen 2 before), so leave it.
- nums[6] is -3 (use |-3| = 3), whose home is index 2. nums[2] is already negative (we've seen 3 before), so leave it.
- nums[7] is -1 (use |-1| = 1), whose home is index 1 − 1 = 0. Negate nums[0] (4 becomes -4): 1 is here.
- Pass 2: any value still positive sits at a home nobody visited, so the number that belongs there (index + 1) never appeared.
- nums[4] = 8 is positive, so 4 + 1 = 5 is missing.
- nums[5] = 2 is positive, so 5 + 1 = 6 is missing.
- Every other index is negative, so the missing numbers are [5, 6].
Example 2
All the same- Input
- nums = [1, 1]
- Output
- [2]
Example 3
No valid answer- Input
- nums = [1, 2, 3]
- Output
- []
- The answer can be in any order.
Code it here
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Step 6
Get it accepted
All test cases pass
Step 7
Review it so it sticks
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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 solutionEvery value v belongs at index v − 1. Mark that home as seen by making its value negative; afterwards, any index that's still positive is a number that never appeared.
class Solution:
def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
for x in nums:
home = abs(x) - 1
nums[home] = -abs(nums[home]) # seen
return [i + 1 for i, x in enumerate(nums) if x > 0]Reference 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.