How hard is LeetCode 1688. Count of Matches in Tournament?

LeetCode 1688. Count of Matches in Tournament is rated Easy on LeetCode.

What is the time complexity of LeetCode 1688. Count of Matches in Tournament?

The Python solution on this page runs in O(1).

What is the space complexity of LeetCode 1688. Count of Matches in Tournament?

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

What topics does LeetCode 1688. Count of Matches in Tournament cover?

LeetCode 1688. Count of Matches in Tournament is tagged Math and Simulation on LeetCode.

EasyNo. 1688

Count of Matches in Tournament — LeetCode 1688 Python Solution

Pattern: Math and Number Theory
Topics: Math, Simulation
Reading time: 2 min
Source: leetcode.com

The problem

You are given an integer n, the number of teams in a tournament that has strange rules: If the current number of teams is even, each team gets paired with another team. A total of n / 2 matches are played, and n / 2 teams advance to the next round.

Example

Input: n = 7
Output: 6
Explanation: Details of the tournament:

Python solution

PythonSolution

class Solution:
    def numberOfMatches(self, n: int) -> int:
        return n - 1

Complexity

Time: O(1)
Space · auxiliary: O(1)

Pattern: Math and Number Theory

Find the closed form, the invariant, or the modular identity — and skip the loop entirely. LeetCode 1688. Count of Matches in Tournament is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Math.

The math and number theory guide has the Python template for the pattern and the 485 LeetCode problems that use it.

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Frequently asked questions

How hard is LeetCode 1688. Count of Matches in Tournament?: LeetCode 1688. Count of Matches in Tournament is rated Easy on LeetCode.
What is the time complexity of LeetCode 1688. Count of Matches in Tournament?: The Python solution on this page runs in O(1).
What is the space complexity of LeetCode 1688. Count of Matches in Tournament?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 1688. Count of Matches in Tournament cover?: LeetCode 1688. Count of Matches in Tournament is tagged Math and Simulation 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...

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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: Math and Number Theory

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