How hard is LeetCode 2923. Find Champion I?

LeetCode 2923. Find Champion I is rated Easy on LeetCode.

What is the time complexity of LeetCode 2923. Find Champion I?

The Python solution on this page runs in O(n^2), where n is the number of teams.

What is the space complexity of LeetCode 2923. Find Champion I?

The Python solution on this page uses O(1) auxiliary space.

What topics does LeetCode 2923. Find Champion I cover?

LeetCode 2923. Find Champion I is tagged Array and Matrix on LeetCode.

EasyNo. 2923

Find Champion I — LeetCode 2923 Python Solution

Pattern: Matrix and Grid
Topics: Array, Matrix
Reading time: 2 min
Source: leetcode.com

The problem

There are n teams numbered from 0 to n - 1 in a tournament. Given a 0-indexed 2D boolean matrix grid of size n * n.

Example

Input: grid = [[0,1],[0,0]]
Output: 0
Explanation: There are two teams in this tournament.

Python solution

PythonSolution

class Solution:
    def findChampion(self, grid: List[List[int]]) -> int:
        for i, row in enumerate(grid):
            if all(x == 1 for j, x in enumerate(row) if i != j):
                return i

Complexity

Time: O(n2)where n is the number of teams
Space · auxiliary: O(1)

Pattern: Matrix and Grid

Treat a 2-D grid as a graph whose neighbours are the four adjacent cells. LeetCode 2923. Find Champion I is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Matrix.

The matrix and grid guide has the Python template for the pattern and the 216 LeetCode problems that use it.

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0566	Reshape the Matrix	Easy
0661	Image Smoother	Easy

Frequently asked questions

How hard is LeetCode 2923. Find Champion I?: LeetCode 2923. Find Champion I is rated Easy on LeetCode.
What is the time complexity of LeetCode 2923. Find Champion I?: The Python solution on this page runs in O(n²), where n is the number of teams.
What is the space complexity of LeetCode 2923. Find Champion I?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2923. Find Champion I cover?: LeetCode 2923. Find Champion I is tagged Array and Matrix 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: Matrix and Grid

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Frequently asked questions

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Approach

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