How hard is LeetCode 1821. Find Customers With Positive Revenue this Year?

LeetCode 1821. Find Customers With Positive Revenue this Year is rated Easy on LeetCode.

What topics does LeetCode 1821. Find Customers With Positive Revenue this Year cover?

LeetCode 1821. Find Customers With Positive Revenue this Year is tagged Database on LeetCode.

EasyNo. 1821LeetCode Premium

Find Customers With Positive Revenue this Year — LeetCode 1821 Python Solution

Topics: Database
Reading time: 2 min
Source: leetcode.com

Table schema

SQLTable schema

Table: Customers +--------------+------+ | Column Name | Type | +--------------+------+ | customer_id | int | | year | int | | revenue | int | +--------------+------+ (customer_id, year) is the primary key (combination of columns with unique values) for this table. This table contains the customer ID and the revenue of customers in different years.

Example

SQLExample

+--------------+------+
| Column Name  | Type |
+--------------+------+
| customer_id  | int  |
| year         | int  |
| revenue      | int  |
+--------------+------+
(customer_id, year) is the primary key (combination of columns with unique values) for this table.
This table contains the customer ID and the revenue of customers in different years.
Note that this revenue can be negative.

Python solution

PythonSolution

import duckdb
import pandas as pd

def solution(customers: pd.DataFrame) -> pd.DataFrame:
    con = duckdb.connect()
    con.register("Customers", customers)
    return con.execute("""SELECT
    customer_id
FROM Customers
WHERE year = '2021' AND revenue > 0;""").df()

Complexity

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

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Frequently asked questions

How hard is LeetCode 1821. Find Customers With Positive Revenue this Year?: LeetCode 1821. Find Customers With Positive Revenue this Year is rated Easy on LeetCode.
What topics does LeetCode 1821. Find Customers With Positive Revenue this Year cover?: LeetCode 1821. Find Customers With Positive Revenue this Year is tagged Database on LeetCode.
Is LeetCode 1821. Find Customers With Positive Revenue this Year a premium problem?: Yes. LeetCode 1821. Find Customers With Positive Revenue this Year 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

Related problems

Frequently asked questions

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Approach

Approach

Approach