LeetCode (448)Find All Numbers Disappeared in an Array
In-Place Marking
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Community-reported · source · updated 2026-10-08. ● asked in the last 6 months.
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Step 1
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Concept videoStep 2
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5–10 minutes, no code yet
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ProblemProblem statementGiven an array nums of n integers where each value is in the range [1, n], return an array of all the integers in [1, n] that do not appear in nums.ReadHide
Problem Statement
Given an array nums of n integers where each value is in the range [1, n], return an array of all the integers in [1, n] that do not appear in nums.
Added 2026-11-05 from Srujan's LeetCode Recent AC list (photo) — not previously in the repo.
Input / Output
Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]
Input: nums = [1,1]
Output: [2]
Input: nums = [1,2,3]
Output: []
Code it here
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Hidden: approach · solution code · optimal solution
Approach
Pattern Recognition
In-Place Marking — the values are already valid indices (1..n), so the array can be its own checklist. For each value v, flip the sign of the element at index v - 1. After one pass, the indices that are still positive were never visited — their numbers are the missing ones. No extra set needed.
Approach
Brute Force — Check Every Value, O(n²)
for x from 1 to n:
if x never appears in nums (linear scan): add x to the answer
Optimized — In-Place Sign Marking, O(n)
for v in nums:
i = abs(v) - 1
if nums[i] > 0: nums[i] = -nums[i] # mark value (i + 1) as seen
answer = [i + 1 for i where nums[i] > 0] # never marked = missing
Complexity
| Approach | Time | Space | Notes |
|---|---|---|---|
| Check every value | O(n²) | O(1) extra | Re-scans the array per value |
| In-place marking | O(n) | O(1) extra | The array is its own checklist |
Solution code
from typing import List
class SolutionBruteForce:
"""O(n^2) time | O(1) extra space — check every value 1..n by scanning."""
def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
answer = []
for x in range(1, len(nums) + 1):
found = False
for v in nums:
if v == x:
found = True
break
if not found:
answer.append(x)
return answer
class Solution:
"""O(n) time | O(1) extra space — mark seen values by flipping signs."""
def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
for v in nums:
i = abs(v) - 1
if nums[i] > 0:
nums[i] = -nums[i] # value (i + 1) is present
return [i + 1 for i in range(len(nums)) if nums[i] > 0]Other ways to solve it
- Thought of another approach (slower, or cleverer)? Code it here: the same tests check any approach.
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