How hard is LeetCode 2150. Find All Lonely Numbers in the Array?

LeetCode 2150. Find All Lonely Numbers in the Array is rated Medium on LeetCode.

What is the time complexity of LeetCode 2150. Find All Lonely Numbers in the Array?

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

What is the space complexity of LeetCode 2150. Find All Lonely Numbers in the Array?

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

What topics does LeetCode 2150. Find All Lonely Numbers in the Array cover?

LeetCode 2150. Find All Lonely Numbers in the Array is tagged Array, Hash Table and Counting on LeetCode.

MediumNo. 2150

Find All Lonely Numbers in the Array — LeetCode 2150 Python Solution

Pattern: Hash Map
Topics: Array, Hash Table, Counting
Reading time: 2 min
Source: leetcode.com

The problem

You are given an integer array nums. A number x is lonely when it appears only once, and no adjacent numbers (i.e.

Example

Input: nums = [10,6,5,8]
Output: [10,8]
Explanation: - 10 is a lonely number since it appears exactly once and 9 and 11 does not appear in nums.

Python solution

PythonSolution

class Solution:
    def findLonely(self, nums: List[int]) -> List[int]:
        cnt = Counter(nums)
        return [
            x for x, v in cnt.items() if v == 1 and cnt[x - 1] == 0 and cnt[x + 1] == 0
        ]

Complexity

Time: O(n)
Space · auxiliary: O(n)

Pattern: Hash Map

Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 2150. Find All Lonely Numbers in the Array is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Hash Table and Counting.

The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.

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Frequently asked questions

How hard is LeetCode 2150. Find All Lonely Numbers in the Array?: LeetCode 2150. Find All Lonely Numbers in the Array is rated Medium on LeetCode.
What is the time complexity of LeetCode 2150. Find All Lonely Numbers in the Array?: The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 2150. Find All Lonely Numbers in the Array?: The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 2150. Find All Lonely Numbers in the Array cover?: LeetCode 2150. Find All Lonely Numbers in the Array is tagged Array, Hash Table and Counting 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: Hash Map

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Frequently asked questions

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Approach

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