How hard is LeetCode 614. Second Degree Follower?

LeetCode 614. Second Degree Follower is rated Medium on LeetCode.

What topics does LeetCode 614. Second Degree Follower cover?

LeetCode 614. Second Degree Follower is tagged Database on LeetCode.

Is LeetCode 614. Second Degree Follower a premium problem?

Yes. LeetCode 614. Second Degree Follower is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 0614LeetCode Premium

Second Degree Follower — LeetCode 614 Python Solution

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

The problem

A second-degree follower is a user who follows somebody and is also followed by somebody. Report every second-degree follower with the number of followers they have, ordered by name. The Follow table has one row per relationship: followee (varchar) and follower (varchar), meaning that follower follows followee, with the pair as the primary key. Nobody follows themself.

Example

Follow table:

| followee | follower |
| -------- | -------- |
| Maya     | Noah     |
| Noah     | Omar     |
| Noah     | Priya    |
| Priya    | Quinn    |

Result:

| follower | num |
| -------- | --- |
| Noah     | 2   |
| Priya    | 1   |

Noah follows Maya and is followed by Omar and Priya, and Priya follows Noah and is followed by Quinn; Omar and Quinn follow somebody but have no followers, and Maya has a follower but follows nobody.

Python solution

PythonSolution

import pandas as pd

def second_degree_follower(follow: pd.DataFrame) -> pd.DataFrame:
    active = set(follow['follower'])
    res = (
        follow[follow['followee'].isin(active)]
        .groupby('followee')['follower']
        .nunique()
        .reset_index()
    )
    res.columns = ['follower', 'num']
    return res.sort_values('follower')

Complexity

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

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Frequently asked questions

How hard is LeetCode 614. Second Degree Follower?: LeetCode 614. Second Degree Follower is rated Medium on LeetCode.
What topics does LeetCode 614. Second Degree Follower cover?: LeetCode 614. Second Degree Follower is tagged Database on LeetCode.
Is LeetCode 614. Second Degree Follower a premium problem?: Yes. LeetCode 614. Second Degree Follower 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...

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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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