How hard is LeetCode 1929. Concatenation of Array?

LeetCode 1929. Concatenation of Array is rated Easy on LeetCode.

What is the time complexity of LeetCode 1929. Concatenation of Array?

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

What is the space complexity of LeetCode 1929. Concatenation of Array?

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

What topics does LeetCode 1929. Concatenation of Array cover?

LeetCode 1929. Concatenation of Array is tagged Array and Simulation on LeetCode.

EasyNo. 1929

Concatenation of Array — LeetCode 1929 Python Solution

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

The problem

Given an integer array nums of length n, you want to create an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed). Specifically, ans is the concatenation of two nums arrays.

Example

Input: nums = [1,2,1]
Output: [1,2,1,1,2,1]
Explanation: The array ans is formed as follows:

Python solution

PythonSolution

class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        return nums + nums

Complexity

Time: O(n)
Space · auxiliary: O(n)

Related problems
No.	Problem	Difficulty
1920	Build Array from Permutation	Easy
0495	Teemo Attacking	Easy
0985	Sum of Even Numbers After Queries	Medium
1389	Create Target Array in the Given Order	Easy
1409	Queries on a Permutation With Key	Medium
1503	Last Moment Before All Ants Fall Out of a Plank	Medium

Frequently asked questions

How hard is LeetCode 1929. Concatenation of Array?: LeetCode 1929. Concatenation of Array is rated Easy on LeetCode.
What is the time complexity of LeetCode 1929. Concatenation of Array?: The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 1929. Concatenation of Array?: The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 1929. Concatenation of Array cover?: LeetCode 1929. Concatenation of Array is tagged Array and Simulation on LeetCode.

Stuck on problems like this in a live interview?

Stealth Interview is a desktop app for macOS and Windows. It reads the problem off your screen and returns a working solution with a step-by-step explanation and its time and space complexity, invisible to screen sharing.

Download for Mac Download for Windows

Free to download. Check it runs on your machine, then pick a plan.

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."

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...

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...

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

Related problems

Frequently asked questions

Stuck on problems like this in a live interview?

Approach

Approach

Approach