How hard is LeetCode 2924. Find Champion II?

LeetCode 2924. Find Champion II is rated Medium on LeetCode.

What is the time complexity of LeetCode 2924. Find Champion II?

The Python solution on this page runs in O(n).

What is the space complexity of LeetCode 2924. Find Champion II?

The Python solution on this page uses O(n) auxiliary space.

What topics does LeetCode 2924. Find Champion II cover?

LeetCode 2924. Find Champion II is tagged Graph on LeetCode.

MediumNo. 2924

Find Champion II — LeetCode 2924 Python Solution

Pattern: Depth-First Search
Topics: Graph
Reading time: 2 min
Source: leetcode.com

The problem

There are n teams numbered from 0 to n - 1 in a tournament; each team is also a node in a DAG. You are given the integer n and a 0-indexed 2D integer array edges of length m representing the DAG, where edges[i] = [ui, vi] indicates that there is a directed edge from team ui to team vi in the graph.

Example

Input: n = 3, edges = [[0,1],[1,2]]
Output: 0
Explanation: Team 1 is weaker than team 0. Team 2 is weaker than team 1. So the champion is team 0.

Python solution

PythonSolution

class Solution:
    def findChampion(self, n: int, edges: List[List[int]]) -> int:
        indeg = [0] * n
        for _, v in edges:
            indeg[v] += 1
        return -1 if indeg.count(0) != 1 else indeg.index(0)

Complexity

Time: O(n)
Space · auxiliary: O(n)

Pattern: Depth-First Search

Follow one path to its end before trying the next — the default way to explore a graph. LeetCode 2924. Find Champion II is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Graph.

The depth-first search guide has the Python template for the pattern and the 366 LeetCode problems that use it.

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Frequently asked questions

How hard is LeetCode 2924. Find Champion II?: LeetCode 2924. Find Champion II is rated Medium on LeetCode.
What is the time complexity of LeetCode 2924. Find Champion II?: The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 2924. Find Champion II?: The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 2924. Find Champion II cover?: LeetCode 2924. Find Champion II is tagged Graph 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.

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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

Pattern: Depth-First Search

Related problems

Frequently asked questions

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Approach

Approach

Approach