How hard is LeetCode 2113. Elements in Array After Removing and Replacing Elements?

LeetCode 2113. Elements in Array After Removing and Replacing Elements is rated Medium on LeetCode.

What is the time complexity of LeetCode 2113. Elements in Array After Removing and Replacing Elements?

The Python solution on this page runs in O(m), where m is the length of the array queries.

What is the space complexity of LeetCode 2113. Elements in Array After Removing and Replacing Elements?

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

What topics does LeetCode 2113. Elements in Array After Removing and Replacing Elements cover?

LeetCode 2113. Elements in Array After Removing and Replacing Elements is tagged Array on LeetCode.

MediumNo. 2113LeetCode Premium

Elements in Array After Removing and Replacing Elements — LeetCode 2113 Python Solution

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

The problem

You are given a 0-indexed integer array nums. Initially on minute 0, the array is unchanged.

Example

Input: nums = [0,1,2], queries = [[0,2],[2,0],[3,2],[5,0]]
Output: [2,2,-1,0]
Explanation: Minute 0: [0,1,2] - All elements are in the nums.

Python solution

PythonSolution

class Solution:
    def elementInNums(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        n, m = len(nums), len(queries)
        ans = [-1] * m
        for j, (t, i) in enumerate(queries):
            t %= 2 * n
            if t < n and i < n - t:
                ans[j] = nums[i + t]
            elif t > n and i < t - n:
                ans[j] = nums[i]
        return ans

Complexity

Time: O(m)where m is the length of the array queries
Space · auxiliary: O(1)

Related problems
No.	Problem	Difficulty
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0228	Summary Ranges	Easy
0485	Max Consecutive Ones	Easy
0665	Non-decreasing Array	Medium
0674	Longest Continuous Increasing Subsequence	Easy
0699	Falling Squares	Hard

Frequently asked questions

How hard is LeetCode 2113. Elements in Array After Removing and Replacing Elements?: LeetCode 2113. Elements in Array After Removing and Replacing Elements is rated Medium on LeetCode.
What is the time complexity of LeetCode 2113. Elements in Array After Removing and Replacing Elements?: The Python solution on this page runs in O(m), where m is the length of the array queries.
What is the space complexity of LeetCode 2113. Elements in Array After Removing and Replacing Elements?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2113. Elements in Array After Removing and Replacing Elements cover?: LeetCode 2113. Elements in Array After Removing and Replacing Elements is tagged Array on LeetCode.
Is LeetCode 2113. Elements in Array After Removing and Replacing Elements a premium problem?: Yes. LeetCode 2113. Elements in Array After Removing and Replacing Elements is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

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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...

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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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