How hard is LeetCode 2705. Compact Object?

LeetCode 2705. Compact Object is rated Medium on LeetCode.

What is the time complexity of LeetCode 2705. Compact Object?

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

What is the space complexity of LeetCode 2705. Compact Object?

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

What topics does LeetCode 2705. Compact Object cover?

LeetCode 2705. Compact Object is tagged JavaScript on LeetCode.

MediumNo. 2705

Compact Object — LeetCode 2705 Python Solution

Topics: JavaScript
Reading time: 2 min
Source: leetcode.com

The problem

Given an object or array obj, return a compact object. A compact object is the same as the original object, except with keys containing falsy values removed.

Example

Input: obj = [null, 0, false, 1]
Output: [1]
Explanation: All falsy values have been removed from the array.

Complexity

Time: O(n)
Space · auxiliary: O(n)

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Frequently asked questions

How hard is LeetCode 2705. Compact Object?: LeetCode 2705. Compact Object is rated Medium on LeetCode.
What is the time complexity of LeetCode 2705. Compact Object?: The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 2705. Compact Object?: The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 2705. Compact Object cover?: LeetCode 2705. Compact Object is tagged JavaScript 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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