0738MediumBacktrackingBreadth-First Search

Badge Codes With A Fixed Digit Step

Tracked in this browser only

Trains the technique from

LeetCode 967Numbers With Same Consecutive Differences

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 site office issues badge codes. A code is a whole number written with exactly size digits, and the badge printer refuses a code whose first digit is 0, so no code may carry a leading zero.

The office also insists that neighbouring digits step by a fixed amount: for every pair of digits standing next to each other in the code, the absolute difference between them must equal gap exactly. A difference smaller or larger than gap is rejected, and the step may go up or down at each position.

Return every badge code the office would accept, as integers, in any order. When gap is 0 each digit has to match the one before it, so exactly one code exists per leading digit.

Examples

Example 1

Input: size = 2, gap = 7
Output: [18, 29, 70, 81, 92]

Every listed code has two digits whose absolute difference is 7: 1 and 8, 2 and 9, 7 and 0, 8 and 1, 9 and 2. The pair 0 and 7 also differs by 7 but `07` would start with a zero, so it is not a code.

Example 2

Input: size = 4, gap = 0
Output: [1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999]

A gap of 0 forces every digit to equal the digit before it, so each code is one digit written four times. `0000` is left out because a code may not begin with 0.

Example 3

Input: size = 3, gap = 9
Output: [909]

The only two digits nine apart are 0 and 9, so the digits have to alternate between them, and the code cannot begin with 0.

Constraints

2 <= size <= 9
0 <= gap <= 9

The values you return may be in any order.

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 badge_codes(size: int, gap: int) -> list[int]:

Java

public int[] badgeCodes(int size, int gap)

The intended approach

Time O(2^size): each extension round at most doubles the working list, and each code is extended in constant time.Space O(2^size) for the working list and the returned codes.

Keep a working list of the partial codes that are legal so far, seeded with the nine allowed leading digits. Extend the list one digit position at a time: from a partial code whose last digit is `d`, the only continuations are `d - gap` and `d + gap`, and each is kept only when it lands inside 0 through 9. When `gap` is `0` those two candidates coincide, so a single candidate is offered to avoid listing the same code twice. After `size - 1` extension rounds the working list holds exactly the accepted codes.

Pythonentry point: badge_codes

def badge_codes(size, gap):
    # grow the codes one digit at a time, keeping only partial codes that are
    # still legal. O(2^size) codes are produced, each extended in O(1).
    cur = list(range(1, 10))
    for _ in range(size - 1):
        nxt = []
        for val in cur:
            last = val % 10
            steps = [last] if gap == 0 else [last - gap, last + gap]
            for nd in steps:
                if 0 <= nd <= 9:
                    nxt.append(val * 10 + nd)
        cur = nxt
    return cur

The obvious version

Enumerate the full cross product of a leading digit and one up-or-down choice per remaining position, discard any choice sequence that leaves the 0 through 9 range, and collect the survivors in a set so repeated codes collapse. No pruning while building.

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_badge_codes(size, gap):
    """Enumerate the full cross product of a leading digit and one up-or-down
    choice per following digit, keep whatever lands in range, and de-duplicate.
    No pruning during construction."""
    from itertools import product

    found = set()
    for start in range(1, 10):
        for signs in product((-1, 1), repeat=size - 1):
            digits = [start]
            ok = True
            for s in signs:
                nd = digits[-1] + s * gap
                if not 0 <= nd <= 9:
                    ok = False
                    break
                digits.append(nd)
            if ok:
                val = 0
                for d in digits:
                    val = val * 10 + d
                found.add(val)
    return sorted(found)

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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in