0035EasyHash TableStringQueueCounting

First Solo Stop

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Trains the technique from

LeetCode 387First Unique Character in a 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 courier's shift is written down as the string route. Each position holds one lowercase letter naming the loading bay that was visited at that point in the shift, and a bay can be visited more than once.

A bay is solo when it shows up exactly once in the whole shift. Return the position of the earliest solo bay in route, counting positions from 0. If every bay in the shift was visited two or more times, return -1.

Examples

Example 1

Input: route = "swiss"
Output: 1

Bay s is visited three times, so the earliest bay visited once is w at position 1.

Example 2

Input: route = "kayak"
Output: 2

Bays k and a are each visited twice, leaving y at position 2 as the only solo bay.

Example 3

Input: route = "gg"
Output: -1

The single bay in this shift is visited twice, so no solo bay exists.

Constraints

1 <= route.length <= 10^5
route contains 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 first_solo_stop(route: str) -> int:

Java

public int firstSoloStop(String route)

The intended approach

Time O(n)Space O(1)

Tally how many times each of the 26 letters occurs in one pass. Walk the shift a second time and hand back the first position whose letter has a tally of one; a single pass cannot decide, because a bay revisited later still looks solo when you first meet it.

Pythonentry point: first_solo_stop

def first_solo_stop(route):
    tally = {}
    for bay in route:
        tally[bay] = tally.get(bay, 0) + 1
    for i, bay in enumerate(route):
        if tally[bay] == 1:
            return i
    return -1

The obvious version

For each position, rescan the whole shift and count how often that letter appears, returning the first position with a count of one. 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_first_solo_stop(route):
    for i in range(len(route)):
        hits = 0
        for bay in route:
            if bay == route[i]:
                hits += 1
        if hits == 1:
            return i
    return -1

How this problem was checked

Reference solution executed against 10 test cases — 10 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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