How hard is LeetCode 2660. Determine the Winner of a Bowling Game?

LeetCode 2660. Determine the Winner of a Bowling Game is rated Easy on LeetCode.

What is the time complexity of LeetCode 2660. Determine the Winner of a Bowling Game?

The Python solution on this page runs in O(n), where n is the length of the array.

What is the space complexity of LeetCode 2660. Determine the Winner of a Bowling Game?

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

What topics does LeetCode 2660. Determine the Winner of a Bowling Game cover?

LeetCode 2660. Determine the Winner of a Bowling Game is tagged Array and Simulation on LeetCode.

EasyNo. 2660

Determine the Winner of a Bowling Game — LeetCode 2660 Python Solution

Topics: Array, Simulation
Reading time: 2 min
Source: leetcode.com

The problem

You are given two 0-indexed integer arrays player1 and player2, representing the number of pins that player 1 and player 2 hit in a bowling game, respectively. The bowling game consists of n turns, and the number of pins in each turn is exactly 10.

Python solution

PythonSolution

class Solution:
    def isWinner(self, player1: List[int], player2: List[int]) -> int:
        def f(arr: List[int]) -> int:
            s = 0
            for i, x in enumerate(arr):
                k = 2 if (i and arr[i - 1] == 10) or (i > 1 and arr[i - 2] == 10) else 1
                s += k * x
            return s

        a, b = f(player1), f(player2)
        return 1 if a > b else (2 if b > a else 0)

Complexity

Time: O(n)where n is the length of the array
Space · auxiliary: O(1)

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Frequently asked questions

How hard is LeetCode 2660. Determine the Winner of a Bowling Game?: LeetCode 2660. Determine the Winner of a Bowling Game is rated Easy on LeetCode.
What is the time complexity of LeetCode 2660. Determine the Winner of a Bowling Game?: The Python solution on this page runs in O(n), where n is the length of the array.
What is the space complexity of LeetCode 2660. Determine the Winner of a Bowling Game?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2660. Determine the Winner of a Bowling Game cover?: LeetCode 2660. Determine the Winner of a Bowling Game is tagged Array 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.

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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

Python solution

Complexity

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