0747MediumArrayBinary SearchStackMonotonic Stack

Reliable Lookups On An Unordered Shelf

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Trains the technique from

LeetCode 1966Binary Searchable Numbers in an Unsorted 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 stockroom shelf holds parts left to right, and codes[i] is the part code in position i. All part codes on the shelf are different. The shelf is not in code order, but the stockroom's lookup routine assumes it is.

Looking up a code v works on a range of positions lo through hi, starting with the whole shelf, lo = 0 and hi = n - 1:

If lo > hi the range is empty and the lookup fails.
Otherwise the routine picks a split position mid with lo <= mid <= hi. Which position it picks is up to the routine and you may not assume anything about the choice.
If codes[mid] == v the lookup succeeds.
If codes[mid] < v the routine repeats from step 1 on the range mid + 1 through hi.
If codes[mid] > v the routine repeats from step 1 on the range lo through mid - 1.

Call position i reliable when looking up codes[i] succeeds for every possible sequence of split positions the routine could pick. A single unlucky sequence that fails is enough to make the position unreliable.

Return how many positions on the shelf are reliable.

Examples

Example 1

Input: codes = [12, 30, 20, 45, 70]
Output: 3

Looking up the 30 at position 1 can start by splitting at position 2, whose code 20 is below 30, which sends the search to positions 3 through 4 and leaves position 1 out of range for good. Looking up the 20 at position 2 can start by splitting at position 1, whose code 30 is above 20, which sends the search to positions 0 through 0. Positions 0, 3 and 4 come out successful however the splits are chosen, so the count is 3.

Example 2

Input: codes = [3, 2, 1, 4]
Output: 1

Looking up the 3 at position 0 can split at position 2, whose code 1 is below 3, moving the search to positions 3 through 3. Looking up the 2 at position 1 can split at position 0, whose code 3 is above 2, moving the search to an empty range. Looking up the 1 at position 2 fails after the same first split. Position 3 succeeds whatever the routine picks, so the count is 1.

Example 3

Input: codes = [8, 6, 4, 2]
Output: 0

Looking up 8 can split at position 3, whose code 2 is below 8, moving the search past the end of the shelf. Looking up 6, 4 or 2 can each split at position 0, whose code 8 is above all of them, moving the search before the start of the shelf. No position is reliable.

Constraints

1 <= codes.length <= 10^5
-10^5 <= codes[i] <= 10^5
All part codes in codes are different from one another.

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 reliable_lookups(codes: list[int]) -> int:

Java

public int reliableLookups(int[] codes)

The intended approach

Time O(n) for n positions: two linear passes.Space O(n) for the suffix table.

A position survives every split sequence only when no split can ever push it out of the live range, and the split that would do so is one landing on a code that points the search the wrong way. That turns into two facts about each position, one about everything to its left and one about everything to its right, so precompute both with two linear passes: a right-to-left pass storing the smallest code from each position onward, and a left-to-right pass carrying the largest code seen so far. A position counts when the largest code before it is below its own code and the smallest code after it is above its own code. Both passes are flat loops, so a shelf of 10^5 positions is safe.

Pythonentry point: reliable_lookups

def reliable_lookups(codes):
    # one right-to-left pass records the smallest code from each position on,
    # then one left-to-right pass carries the largest code seen so far.
    # O(n) time, O(n) space.
    n = len(codes)
    smallest_after = [0] * n
    running = codes[n - 1]
    for i in range(n - 1, -1, -1):
        if codes[i] < running:
            running = codes[i]
        smallest_after[i] = running

    total = 0
    largest_before = None
    for i in range(n):
        left_clear = largest_before is None or largest_before < codes[i]
        right_clear = i == n - 1 or smallest_after[i + 1] > codes[i]
        if left_clear and right_clear:
            total += 1
        if largest_before is None or codes[i] > largest_before:
            largest_before = codes[i]
    return total

The obvious version

Encode the lookup rules directly: for each position, recurse over every split the routine could pick at every stage, memoising ranges, and accept the position only when no sequence of splits fails. Exponentially many sequences collapse into a table of ranges, but it still reruns the whole search per position.

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_reliable_lookups(codes):
    """Run the lookup rules literally: for each code, ask whether every possible
    choice of split index, at every stage, still ends in success."""
    n = len(codes)
    total = 0
    for idx in range(n):
        target = codes[idx]
        memo = {}

        def always(lo, hi):
            if lo > hi:
                return False
            key = (lo, hi)
            if key in memo:
                return memo[key]
            memo[key] = True
            answer = True
            for mid in range(lo, hi + 1):
                if codes[mid] == target:
                    continue
                if codes[mid] < target:
                    if not always(mid + 1, hi):
                        answer = False
                        break
                else:
                    if not always(lo, mid - 1):
                        answer = False
                        break
            memo[key] = answer
            return answer

        if always(0, n - 1):
            total += 1
    return total

How this problem was checked

Reference solution executed against 14 test cases — 14 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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