How hard is LeetCode 2687. Bikes Last Time Used?

LeetCode 2687. Bikes Last Time Used is rated Easy on LeetCode.

What topics does LeetCode 2687. Bikes Last Time Used cover?

LeetCode 2687. Bikes Last Time Used is tagged Database on LeetCode.

Is LeetCode 2687. Bikes Last Time Used a premium problem?

Yes. LeetCode 2687. Bikes Last Time Used is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

EasyNo. 2687LeetCode Premium

Bikes Last Time Used — LeetCode 2687 Python Solution

Topics: Database
Reading time: 3 min
Source: leetcode.com

Table schema

SQLTable schema

Table: Bikes +-------------+----------+ | Column Name | Type | +-------------+----------+ | ride_id | int | | bike_number | int | | start_time | datetime | | end_time | datetime | +-------------+----------+ ride_id column contains unique values. Each row contains a ride information that includes ride_id, bike number, start and end time of the ride.

Example

SQLExample

+-------------+----------+ 
| Column Name | Type     | 
+-------------+----------+ 
| ride_id     | int      | 
| bike_number | int      | 
| start_time  | datetime |
| end_time    | datetime |
+-------------+----------+
ride_id column contains unique values.
Each row contains a ride information that includes ride_id, bike number, start and end time of the ride.
It is guaranteed that start_time and end_time are valid datetime values.

Python solution

PythonSolution

import duckdb
import pandas as pd

# Pass input tables as keyword arguments matching the SQL table names.
def solution(**tables) -> pd.DataFrame:
    con = duckdb.connect()
    for name, df in tables.items():
        con.register(name, df)
    return con.execute("""SELECT
    bike_number,
    MAX(end_time) AS end_time
FROM Bikes
GROUP BY bike_number
ORDER BY end_time DESC;""").df()

Complexity

Time: O(n log n)(typical)
Space · auxiliary: O(n)

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Frequently asked questions

How hard is LeetCode 2687. Bikes Last Time Used?: LeetCode 2687. Bikes Last Time Used is rated Easy on LeetCode.
What topics does LeetCode 2687. Bikes Last Time Used cover?: LeetCode 2687. Bikes Last Time Used is tagged Database on LeetCode.
Is LeetCode 2687. Bikes Last Time Used a premium problem?: Yes. LeetCode 2687. Bikes Last Time Used is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

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Transcribing...

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

Table schema

Example

Python solution

Complexity

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Frequently asked questions

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Approach

Approach

Approach