0222MediumHash TableStringStackGreedy

Typesetter's Rack

Tracked in this browser only

Trains the technique from

LeetCode 2434Using a Robot to Print the Lexicographically Smallest String

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 hand typesetter is composing a single line of type. Letter slugs reach the bench on a belt in a fixed order, given as the string slugs. Next to the bench stands a narrow upright rack that holds slugs one above another, and the composed line starts out empty as well.

Each step is one of exactly two moves:

take the slug at the head of the belt and set it on top of the rack;
take the slug currently on top of the rack and add it to the end of the line.

The typesetter keeps making moves until the belt and the rack are both empty, so every slug ends up somewhere in the line. Many orders of moves are possible and they compose different lines.

Return the composed line that comes first in dictionary order out of all lines the typesetter can produce.

Examples

Example 1

Input: slugs = "tqrq"
Output: "qqrt"

Set `t` and then `q` on the rack, add the top `q` to the line, set `r` and the last `q` on the rack, then add `q`, `r` and `t` to the line in that order. The line reads `qqrt`.

Example 2

Input: slugs = "dgcae"
Output: "acegd"

All five slugs travel from the belt through the rack and into the line, which ends up reading `acegd`.

Example 3

Input: slugs = "mfemf"
Output: "effmm"

Each of the five slugs is added to the line exactly once and the finished line reads `effmm`.

Constraints

1 <= slugs.length <= 10^5
slugs consists of lowercase English letters only

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 smallest_printed_line(slugs: str) -> str:

Java

public String smallestPrintedLine(String slugs)

The intended approach

Time O(n + 26)Space O(n) for the rack and the composed line

Keep a count per letter of how many slugs are still on the belt. After setting a slug on the rack, move a pointer forward to the smallest letter that still has copies waiting on the belt; the pointer never moves back because those counts only fall. While the slug on top of the rack is no later than that letter, add it to the line, and once the belt is empty drain the rack from the top down.

Pythonentry point: smallest_printed_line

def smallest_printed_line(slugs):
    letters = "abcdefghijklmnopqrstuvwxyz"
    waiting = [0] * 26
    for ch in slugs:
        waiting[ord(ch) - 97] += 1
    rack = []
    line = []
    lowest = 0
    for ch in slugs:
        waiting[ord(ch) - 97] -= 1
        rack.append(ch)
        while lowest < 26 and waiting[lowest] == 0:
            lowest += 1
        while rack and (lowest == 26 or rack[-1] <= letters[lowest]):
            line.append(rack.pop())
    while rack:
        line.append(rack.pop())
    return "".join(line)

The obvious version

Search every reachable state, keyed by how many slugs have left the belt plus the exact rack contents, and return the smaller of the two continuations at each state. Every completion from a state has the same length, so comparing whole strings is enough. Exponential without the memo, fine on tiny inputs.

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_smallest_printed_line(slugs):
    total = len(slugs)
    memo = {}

    def best_from(taken, rack):
        if taken == total and not rack:
            return ""
        key = (taken, rack)
        if key in memo:
            return memo[key]
        best = None
        if taken < total:
            best = best_from(taken + 1, rack + (slugs[taken],))
        if rack:
            lifted = rack[-1] + best_from(taken, rack[:-1])
            if best is None or lifted < best:
                best = lifted
        memo[key] = best
        return best

    return best_from(0, ())

How this problem was checked

Reference solution executed against 11 test cases — 11 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