How hard is LeetCode 1920. Build Array from Permutation?

LeetCode 1920. Build Array from Permutation is rated Easy on LeetCode.

What is the time complexity of LeetCode 1920. Build Array from Permutation?

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

What is the space complexity of LeetCode 1920. Build Array from Permutation?

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

What topics does LeetCode 1920. Build Array from Permutation cover?

LeetCode 1920. Build Array from Permutation is tagged Array and Simulation on LeetCode.

EasyNo. 1920

Build Array from Permutation — LeetCode 1920 Python Solution

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

The problem

Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it. A zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).

Example

Input: nums = [0,2,1,5,3,4]
Output: [0,1,2,4,5,3]
Explanation: The array ans is built as follows:

Python solution

PythonSolution

class Solution:
    def buildArray(self, nums: List[int]) -> List[int]:
        return [nums[num] for num in nums]

Complexity

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

Related problems
No.	Problem	Difficulty
1929	Concatenation of Array	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 1920. Build Array from Permutation?: LeetCode 1920. Build Array from Permutation is rated Easy on LeetCode.
What is the time complexity of LeetCode 1920. Build Array from Permutation?: The Python solution on this page runs in O(n), where n is the length of the array \textit{nums}.
What is the space complexity of LeetCode 1920. Build Array from Permutation?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 1920. Build Array from Permutation cover?: LeetCode 1920. Build Array from Permutation 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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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

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Approach

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