How hard is LeetCode 1550. Three Consecutive Odds?

LeetCode 1550. Three Consecutive Odds is rated Easy on LeetCode.

What is the time complexity of LeetCode 1550. Three Consecutive Odds?

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

What is the space complexity of LeetCode 1550. Three Consecutive Odds?

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

What topics does LeetCode 1550. Three Consecutive Odds cover?

LeetCode 1550. Three Consecutive Odds is tagged Array on LeetCode.

EasyNo. 1550

Three Consecutive Odds — LeetCode 1550 Python Solution

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

The problem

Given an integer array arr, return true if there are three consecutive odd numbers in the array. Otherwise, return false.

Example

Input: arr = [2,6,4,1]
Output: false
Explanation: There are no three consecutive odds.

Python solution

PythonSolution

class Solution:
    def threeConsecutiveOdds(self, arr: List[int]) -> bool:
        cnt = 0
        for x in arr:
            if x & 1:
                cnt += 1
                if cnt == 3:
                    return True
            else:
                cnt = 0
        return False

Complexity

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

Related problems
No.	Problem	Difficulty
0057	Insert Interval	Medium
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 1550. Three Consecutive Odds?: LeetCode 1550. Three Consecutive Odds is rated Easy on LeetCode.
What is the time complexity of LeetCode 1550. Three Consecutive Odds?: The Python solution on this page runs in O(n), where n is the length of the array \textit{arr}.
What is the space complexity of LeetCode 1550. Three Consecutive Odds?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 1550. Three Consecutive Odds cover?: LeetCode 1550. Three Consecutive Odds is tagged 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

Example

Python solution

Complexity

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