How hard is LeetCode 2238. Number of Times a Driver Was a Passenger?

LeetCode 2238. Number of Times a Driver Was a Passenger is rated Medium on LeetCode.

What topics does LeetCode 2238. Number of Times a Driver Was a Passenger cover?

LeetCode 2238. Number of Times a Driver Was a Passenger is tagged Database on LeetCode.

Is LeetCode 2238. Number of Times a Driver Was a Passenger a premium problem?

Yes. LeetCode 2238. Number of Times a Driver Was a Passenger is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 2238LeetCode Premium

Number of Times a Driver Was a Passenger — LeetCode 2238 Python Solution

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

Table schema

SQLTable schema

Table: Rides +--------------+------+ | Column Name | Type | +--------------+------+ | ride_id | int | | driver_id | int | | passenger_id | int | +--------------+------+ ride_id is the column with unique values for this table. Each row of this table contains the ID of the driver and the ID of the passenger that rode in ride_id.

Example

SQLExample

+--------------+------+
| Column Name  | Type |
+--------------+------+
| ride_id      | int  |
| driver_id    | int  |
| passenger_id | int  |
+--------------+------+
ride_id is the column with unique values for this table.
Each row of this table contains the ID of the driver and the ID of the passenger that rode in ride_id.
Note that driver_id != passenger_id.

Python solution

PythonSolution

import duckdb
import pandas as pd

def solution(rides: pd.DataFrame) -> pd.DataFrame:
    con = duckdb.connect()
    con.register("Rides", rides)
    return con.execute("""WITH T AS (SELECT DISTINCT driver_id FROM Rides)
SELECT t.driver_id, COUNT(passenger_id) AS cnt
FROM
    T AS t
    LEFT JOIN Rides AS r ON t.driver_id = r.passenger_id
GROUP BY 1;""").df()

Complexity

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

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Frequently asked questions

How hard is LeetCode 2238. Number of Times a Driver Was a Passenger?: LeetCode 2238. Number of Times a Driver Was a Passenger is rated Medium on LeetCode.
What topics does LeetCode 2238. Number of Times a Driver Was a Passenger cover?: LeetCode 2238. Number of Times a Driver Was a Passenger is tagged Database on LeetCode.
Is LeetCode 2238. Number of Times a Driver Was a Passenger a premium problem?: Yes. LeetCode 2238. Number of Times a Driver Was a Passenger 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