How hard is LeetCode 2783. Flight Occupancy and Waitlist Analysis?

LeetCode 2783. Flight Occupancy and Waitlist Analysis is rated Medium on LeetCode.

What topics does LeetCode 2783. Flight Occupancy and Waitlist Analysis cover?

LeetCode 2783. Flight Occupancy and Waitlist Analysis is tagged Database on LeetCode.

Is LeetCode 2783. Flight Occupancy and Waitlist Analysis a premium problem?

Yes. LeetCode 2783. Flight Occupancy and Waitlist Analysis is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 2783LeetCode Premium

Flight Occupancy and Waitlist Analysis — LeetCode 2783 Python Solution

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

Table schema

SQLTable schema

Table: Flights +-------------+------+ | Column Name | Type | +-------------+------+ | flight_id | int | | capacity | int | +-------------+------+ flight_id is the column with unique values for this table. Each row of this table contains flight id and its capacity.

Example

SQLExample

+-------------+------+
| Column Name | Type |
+-------------+------+
| flight_id   | int  |
| capacity    | int  |
+-------------+------+
flight_id is the column with unique values for this table.
Each row of this table contains flight id and its capacity.

Python solution

PythonSolution

import duckdb
import pandas as pd

def solution(passengers: pd.DataFrame) -> pd.DataFrame:
    con = duckdb.connect()
    con.register("Passengers", passengers)
    return con.execute("""SELECT
    flight_id,
    LEAST(COUNT(passenger_id), capacity) AS booked_cnt,
    GREATEST(COUNT(passenger_id) - capacity, 0) AS waitlist_cnt
FROM
    Flights
    LEFT JOIN Passengers USING (flight_id)
GROUP BY 1
ORDER 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 2783. Flight Occupancy and Waitlist Analysis?: LeetCode 2783. Flight Occupancy and Waitlist Analysis is rated Medium on LeetCode.
What topics does LeetCode 2783. Flight Occupancy and Waitlist Analysis cover?: LeetCode 2783. Flight Occupancy and Waitlist Analysis is tagged Database on LeetCode.
Is LeetCode 2783. Flight Occupancy and Waitlist Analysis a premium problem?: Yes. LeetCode 2783. Flight Occupancy and Waitlist Analysis 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.

Type your message...

GPT-6.1 Terra

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