How hard is LeetCode 1069. Product Sales Analysis II?

LeetCode 1069. Product Sales Analysis II is rated Easy on LeetCode.

What topics does LeetCode 1069. Product Sales Analysis II cover?

LeetCode 1069. Product Sales Analysis II is tagged Database on LeetCode.

Is LeetCode 1069. Product Sales Analysis II a premium problem?

Yes. LeetCode 1069. Product Sales Analysis II is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

EasyNo. 1069LeetCode Premium

Product Sales Analysis II — LeetCode 1069 Python Solution

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

The problem

For every product, report its id and the total quantity sold across all years, in any order. The Sales table has one row per sale: sale_id (int), product_id (int), year (int), quantity (int) and price (int, the price per unit), with the pair (sale_id, year) as the primary key. The Product table maps product_id (int, the primary key) to product_name (varchar).

Example

Sales table:

| sale_id | product_id | year | quantity | price |
| ------- | ---------- | ---- | -------- | ----- |
| 11      | 300        | 2019 | 25       | 1200  |
| 12      | 300        | 2020 | 30       | 1150  |
| 13      | 400        | 2020 | 8        | 4300  |

Product table:

| product_id | product_name |
| ---------- | ------------ |
| 300        | Router       |
| 400        | Laptop       |

Result:

| product_id | total_quantity |
| ---------- | -------------- |
| 300        | 55             |
| 400        | 8              |

Product 300 sold 25 units in 2019 and 30 in 2020 for a total of 55, and product 400 sold 8 units in a single year.

Python solution

PythonSolution

import pandas as pd

def sales_analysis(sales: pd.DataFrame, product: pd.DataFrame) -> pd.DataFrame:
    res = sales.groupby('product_id', as_index=False)['quantity'].sum()
    res.columns = ['product_id', 'total_quantity']
    return res

Complexity

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

Related problems
No.	Problem	Difficulty
1068	Product Sales Analysis I	Easy
0175	Combine Two Tables	Easy
0176	Second Highest Salary	Medium
0177	Nth Highest Salary	Medium
0178	Rank Scores	Medium
0180	Consecutive Numbers	Medium

Frequently asked questions

How hard is LeetCode 1069. Product Sales Analysis II?: LeetCode 1069. Product Sales Analysis II is rated Easy on LeetCode.
What topics does LeetCode 1069. Product Sales Analysis II cover?: LeetCode 1069. Product Sales Analysis II is tagged Database on LeetCode.
Is LeetCode 1069. Product Sales Analysis II a premium problem?: Yes. LeetCode 1069. Product Sales Analysis II 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+

The problem

Example

Python solution

Complexity

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