How hard is LeetCode 2917. Find the K-or of an Array?

LeetCode 2917. Find the K-or of an Array is rated Easy on LeetCode.

What is the time complexity of LeetCode 2917. Find the K-or of an Array?

The Python solution on this page runs in O(n \times \log M), where n and M are the length of the array nums and the maximum value in nums, respectively.

What is the space complexity of LeetCode 2917. Find the K-or of an Array?

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

What topics does LeetCode 2917. Find the K-or of an Array cover?

LeetCode 2917. Find the K-or of an Array is tagged Bit Manipulation and Array on LeetCode.

EasyNo. 2917

Find the K-or of an Array — LeetCode 2917 Python Solution

Pattern: Bit Manipulation
Topics: Bit Manipulation, Array
Reading time: 2 min
Source: leetcode.com

The problem

You are given an integer array nums, and an integer k. Let's introduce K-or operation by extending the standard bitwise OR.

Python solution

PythonSolution

class Solution:
    def findKOr(self, nums: List[int], k: int) -> int:
        ans = 0
        for i in range(32):
            cnt = sum(x >> i & 1 for x in nums)
            if cnt >= k:
                ans |= 1 << i
        return ans

Complexity

Time: O(n × log M)where n and M are the length of the array nums and the maximum value in nums, respectively
Space · auxiliary: O(1)

Pattern: Bit Manipulation

Use XOR, masks and the low-bit trick to replace whole data structures with an integer. LeetCode 2917. Find the K-or of an Array is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Bit Manipulation.

The bit manipulation guide has the Python template for the pattern and the 194 LeetCode problems that use it.

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0393	UTF-8 Validation	Medium

Frequently asked questions

How hard is LeetCode 2917. Find the K-or of an Array?: LeetCode 2917. Find the K-or of an Array is rated Easy on LeetCode.
What is the time complexity of LeetCode 2917. Find the K-or of an Array?: The Python solution on this page runs in O(n × log M), where n and M are the length of the array nums and the maximum value in nums, respectively.
What is the space complexity of LeetCode 2917. Find the K-or of an Array?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2917. Find the K-or of an Array cover?: LeetCode 2917. Find the K-or of an Array is tagged Bit Manipulation and Array 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

Python solution

Complexity

Pattern: Bit Manipulation

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Frequently asked questions

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Approach

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