How hard is LeetCode 2880. Select Data?

LeetCode 2880. Select Data is rated Easy on LeetCode.

What topics does LeetCode 2880. Select Data cover?

LeetCode 2880. Select Data is tagged Pandas on LeetCode.

EasyNo. 2880

Select Data — LeetCode 2880 Python Solution

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

Table schema

SQLTable schema

DataFrame students +-------------+--------+ | Column Name | Type | +-------------+--------+ | student_id | int | | name | object | | age | int | +-------------+--------+ Write a solution to select the name and age of the student with student_id = 101. The result format is in the following example.

Example

SQLExample

DataFrame students
+-------------+--------+
| Column Name | Type   |
+-------------+--------+
| student_id  | int    |
| name        | object |
| age         | int    |
+-------------+--------+

Python solution

PythonSolution

import pandas as pd


def selectData(students: pd.DataFrame) -> pd.DataFrame:
    return students[students['student_id'] == 101][['name', 'age']]

Complexity

Time: O(n)
Space · auxiliary: O(1) to O(n)

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Frequently asked questions

How hard is LeetCode 2880. Select Data?: LeetCode 2880. Select Data is rated Easy on LeetCode.
What topics does LeetCode 2880. Select Data cover?: LeetCode 2880. Select Data is tagged Pandas on LeetCode.

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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...

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

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Approach

Approach

Approach