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LeetCode 2996Smallest Missing Integer Greater Than Sequential Prefix Sum

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 gym's handover log lists the locker codes handed out today, in the order they were handed out, as nums.

Look at the front of the log. A run of codes starting at the very first entry is called a ladder when every code in it is exactly one above the code before it; the first entry on its own always counts as a ladder. Let total be the sum of the codes in the longest ladder the log begins with.

Return the smallest integer that is at least total and does not appear anywhere in nums.

Examples

Example 1

Input: nums = [4,5,6,2,9]
Output: 15

The log begins with the ladder 4, 5, 6, because 2 does not follow 6. Its codes add up to 15, and 15 never appears in the log.

Example 2

Input: nums = [7,8,3,15,16,17]
Output: 18

The longest ladder at the front is 7, 8, which totals 15. The log already shows 15, 16 and 17, so the first value at or above 15 that is free is 18.

Example 3

Input: nums = [9,20]
Output: 10

20 does not follow 9, so the ladder holds only the code 9 and the total is 9. The log shows 9 itself, and 10 is free.

Constraints

1 <= nums.length <= 50
1 <= nums[i] <= 50

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 missing_integer(nums: list[int]) -> int:

Java

public int missingInteger(int[] nums)

The intended approach

Time O(n + r), where r is the number of values the upward step triesSpace O(n) for the membership set

Extend the ladder from the first entry while each code is one above its predecessor, summing the codes it covers. Put every logged code in a set, then step upward from that sum until a value is not in the set.

Pythonentry point: missing_integer

def missing_integer(nums):
    # Walk the ladder from the front while each code is one above the last, adding
    # as we go, then climb from that sum looking for a value the log never shows.
    # Time O(n + r), where r is how far the climb goes (bounded by the totals the
    # constraints allow).  Space O(n) for the membership set.
    total = nums[0]
    i = 1
    while i < len(nums) and nums[i] == nums[i - 1] + 1:
        total += nums[i]
        i += 1
    present = set(nums)
    candidate = total
    while candidate in present:
        candidate += 1
    return candidate

The obvious version

Test every prefix length for the ladder rule by re-scanning it, keep the longest that passes, add up its codes, then step upward checking each candidate against the log with a linear search.

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_first_free_code(nums):
    n = len(nums)
    best_len = 1
    for length in range(1, n + 1):
        ok = True
        for j in range(1, length):
            if nums[j] != nums[j - 1] + 1:
                ok = False
        if ok and length > best_len:
            best_len = length
    total = 0
    for j in range(best_len):
        total += nums[j]
    candidate = total
    while True:
        hit = False
        for v in nums:
            if v == candidate:
                hit = True
        if not hit:
            return candidate
        candidate += 1

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
3 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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