How hard is LeetCode 1571. Warehouse Manager?

LeetCode 1571. Warehouse Manager is rated Easy on LeetCode.

What topics does LeetCode 1571. Warehouse Manager cover?

LeetCode 1571. Warehouse Manager is tagged Database on LeetCode.

Is LeetCode 1571. Warehouse Manager a premium problem?

Yes. LeetCode 1571. Warehouse Manager is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

EasyNo. 1571LeetCode Premium

Warehouse Manager — LeetCode 1571 Python Solution

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

Table schema

SQLTable schema

Table: Warehouse +--------------+---------+ | Column Name | Type | +--------------+---------+ | name | varchar | | product_id | int | | units | int | +--------------+---------+ (name, product_id) is the primary key (combination of columns with unique values) for this table. Each row of this table contains the information of the products in each warehouse.

Example

SQLExample

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| name         | varchar |
| product_id   | int     |
| units        | int     |
+--------------+---------+
(name, product_id) is the primary key (combination of columns with unique values) for this table.
Each row of this table contains the information of the products in each warehouse.

Python solution

PythonSolution

import duckdb
import pandas as pd

def solution(warehouse: pd.DataFrame, products: pd.DataFrame) -> pd.DataFrame:
    con = duckdb.connect()
    con.register("Warehouse", warehouse)
    con.register("Products", products)
    return con.execute("""SELECT
    name AS warehouse_name,
    SUM(width * length * height * units) AS volume
FROM
    Warehouse
    JOIN Products USING (product_id)
GROUP BY 1;""").df()

Complexity

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

Related problems
No.	Problem	Difficulty
0175	Combine Two Tables	Easy
0176	Second Highest Salary	Medium
0177	Nth Highest Salary	Medium
0178	Rank Scores	Medium
0180	Consecutive Numbers	Medium
0181	Employees Earning More Than Their Managers	Easy

Frequently asked questions

How hard is LeetCode 1571. Warehouse Manager?: LeetCode 1571. Warehouse Manager is rated Easy on LeetCode.
What topics does LeetCode 1571. Warehouse Manager cover?: LeetCode 1571. Warehouse Manager is tagged Database on LeetCode.
Is LeetCode 1571. Warehouse Manager a premium problem?: Yes. LeetCode 1571. Warehouse Manager is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

Stuck on problems like this in a live interview?

Stealth Interview is a desktop app for macOS and Windows. It reads the problem off your screen and returns a working solution with a step-by-step explanation and its time and space complexity, invisible to screen sharing.

Download for Mac Download for Windows

Free to download. Check it runs on your machine, then pick a plan.

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."

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...

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...

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

Related problems

Frequently asked questions

Stuck on problems like this in a live interview?

Approach

Approach

Approach