0418EasyArrayHash Table

Bibs the Mat Never Saw

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Trains the technique from

LeetCode 448Find All Numbers Disappeared in an Array

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A fun run gives out bibs numbered 1 through n, one per runner, and a timing mat at the finish writes one line into its log for every runner who crosses. So nums holds exactly n readings, and every reading is a bib number between 1 and n.

The mat is not perfect. It sometimes fires twice on the same bib and misses another one entirely, so a bib number may appear in nums several times while some bib numbers appear nowhere at all.

Return every bib number from 1 to n that does not appear in nums, listed in increasing order. Return an empty list when the mat caught them all.

The log may be reorganised in place; aim to answer in a number of steps proportional to n and without any working storage beyond the list you return.

Examples

Example 1

Input: nums = [3, 6, 1, 6, 3, 5]
Output: [2, 4]

Six bibs were given out. The log names 1, 3, 5 and 6, with 3 and 6 written twice, and never names 2 or 4.

Example 2

Input: nums = [1, 2, 3]
Output: []

Each of the three bibs appears once, so nothing is missing.

Example 3

Input: nums = [2, 2]
Output: [1]

Two bibs were given out and the mat wrote bib 2 on both lines, so bib 1 is absent.

Example 4

Input: nums = [5, 5, 5, 5, 5]
Output: [1, 2, 3, 4]

Five bibs were given out and every line of the log reads 5, so bibs 1 through 4 never appear.

Constraints

n == nums.length
1 <= n <= 10^5
1 <= nums[i] <= n

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def find_disappeared_numbers(nums: list[int]) -> list[int]:

Java

public List<Integer> findDisappearedNumbers(int[] nums)

The intended approach

Time O(n): two passes over the log.Space O(1) beyond the returned list; the sign flips reuse the log's own slots.

Every reading is between 1 and n, which is exactly the range of slot numbers in the log itself, so the log can carry its own bookkeeping. For each reading, make the slot it names hold a negative value, taking the reading's magnitude so a slot that was already flipped is still read correctly and guarding the flip so a repeated bib does not undo it. A second pass reports the slot numbers still holding a positive value, which come out in increasing order for free.

Pythonentry point: find_disappeared_numbers

def find_disappeared_numbers(nums):
    # Every reading names a slot of the log itself, so the log doubles as the
    # seen-or-not table: flip the sign of the slot a reading points at, using
    # the magnitude so an already flipped slot still reads correctly. Slots left
    # positive are the bibs nothing pointed at.
    size = len(nums)
    for pos in range(size):
        slot = abs(nums[pos]) - 1
        if nums[slot] > 0:
            nums[slot] = -nums[slot]
    missing = []
    for pos in range(size):
        if nums[pos] > 0:
            missing.append(pos + 1)
    return missing

The obvious version

Take each bib number from 1 to n in turn and scan the whole log looking for it, collecting the ones the scan never finds. O(n^2).

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_find_disappeared_numbers(nums):
    size = len(nums)
    answer = []
    for bib in range(1, size + 1):
        found = False
        for reading in nums:
            if reading == bib:
                found = True
                break
        if not found:
            answer.append(bib)
    return answer

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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