How hard is LeetCode 2069. Walking Robot Simulation II?

LeetCode 2069. Walking Robot Simulation II is rated Medium on LeetCode.

What is the time complexity of LeetCode 2069. Walking Robot Simulation II?

The Python solution on this page runs in Varies by operation.

What is the space complexity of LeetCode 2069. Walking Robot Simulation II?

The Python solution on this page uses Varies by operation auxiliary space.

What topics does LeetCode 2069. Walking Robot Simulation II cover?

LeetCode 2069. Walking Robot Simulation II is tagged Design and Simulation on LeetCode.

MediumNo. 2069

Walking Robot Simulation II — LeetCode 2069 Python Solution

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

The problem

A width x height grid is on an XY-plane with the bottom-left cell at (0, 0) and the top-right cell at (width - 1, height - 1). The grid is aligned with the four cardinal directions ("North", "East", "South", and "West").

Example

Input: ["Robot", "step", "step", "getPos", "getDir", "step", "step", "step", "getPos", "getDir"]
Output: [null, null, null, [4, 0], "East", null, null, null, [1, 2], "West"]
Explanation: Robot robot = new Robot(6, 3); // Initialize the grid and the robot at (0, 0) facing East.

Complexity

Time: Varies by operation
Space · auxiliary: Varies by operation

Related problems
No.	Problem	Difficulty
0874	Walking Robot Simulation	Medium
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0295	Find Median from Data Stream	Hard
0307	Range Sum Query - Mutable	Medium
0715	Range Module	Hard
2276	Count Integers in Intervals	Hard

Frequently asked questions

How hard is LeetCode 2069. Walking Robot Simulation II?: LeetCode 2069. Walking Robot Simulation II is rated Medium on LeetCode.
What is the time complexity of LeetCode 2069. Walking Robot Simulation II?: The Python solution on this page runs in Varies by operation.
What is the space complexity of LeetCode 2069. Walking Robot Simulation II?: The Python solution on this page uses Varies by operation auxiliary space.
What topics does LeetCode 2069. Walking Robot Simulation II cover?: LeetCode 2069. Walking Robot Simulation II is tagged Design 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

Complexity

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Approach

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