LeetCode (485)Max Consecutive Ones
Single Pass / Running Count
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Community-reported · source · updated 2026-10-08. ● asked in the last 6 months.
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0 / 110 pointsSolved (Easy) 0/20Reviews 0/50Mastered 0/25Practice steps 0/15
Step 1
Know these before you start
Practise each one here, or tick what you already know
Concept videoWatch the concept video if the pattern is new to you.
Step 2
Try it on paper
5–10 minutes, no code yet
ProblemProblem statementGiven a binary array nums, return the maximum number of consecutive 1s in the array.ReadHide
Problem Statement
Given a binary array nums, return the maximum number of consecutive 1s in the array.
Added 2026-11-01 from Srujan's LeetCode Recent AC list (photo) — not previously in the repo.
Input / Output
Input: nums = [1,1,0,1,1,1]
Output: 3
Input: nums = [1,0,1,1,0,1]
Output: 2
Input: nums = [0]
Output: 0
Step 6
Get it accepted
All test cases pass
Step 7
Review it so it sticks
Re-solve from memory after
The step that matters most: without reviews most of a solved problem fades within weeks. Three short re-solves, each just as you start to forget, are what make it yours.
Your notes
Three lines in your own words. They come back with this problem's reviews, so you can check yourself after recalling the solution.
Solve it yourself before you look
Reading the solution first feels like progress, but it makes the next similar problem — and the interview version — much harder, because you skipped the part where you figure it out. Give it an honest 20–30 minutes. Stuck? Re-read the pattern, watch the concept video, or try the brute force first.
Hidden: approach · solution code
Stuck? One hint at a time.
Approach
Pattern Recognition
Single Pass / Running Count — you only need two numbers while walking the array once: the length of the run you are in right now, and the best run you have seen so far. A 0 resets the current run; a 1 grows it.
Approach
Brute Force — Count From Every Start, O(n²)
best = 0
for i in range(len(nums)):
count = 0
for j in range(i, len(nums)):
if nums[j] == 0: break
count += 1
best = max(best, count)
return best
Optimized — One Pass, O(n)
count = 0
best = 0
for v in nums:
count = count + 1 if v == 1 else 0
best = max(best, count)
return best
Complexity
| Approach | Time | Space | Notes |
|---|---|---|---|
| Count every start | O(n²) | O(1) | Re-counts runs from each index |
| One pass | O(n) | O(1) | Reset on 0, grow on 1 |
Solution code
from typing import List
class SolutionBruteForce:
"""O(n^2) time | O(1) space — count the run starting at every index."""
def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
best = 0
for i in range(len(nums)):
count = 0
for j in range(i, len(nums)):
if nums[j] == 0:
break
count += 1
best = max(best, count)
return best
class Solution:
"""O(n) time | O(1) space — one pass, reset the run on 0."""
def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
count = 0
best = 0
for v in nums:
count = count + 1 if v == 1 else 0
best = max(best, count)
return bestOther ways to solve it
- Community solutions on LeetCode ↗ for many more versions, in many languages. Read them after you've solved it.
Revising this one?
- →Solve it from memory before peeking
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