How do you solve Max Consecutive Ones on LeetCode?

Max Consecutive Ones (LeetCode #485) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem can be solved by counting streaks of 1's.

What is the brute force solution for Max Consecutive Ones?

The brute force approach for Max Consecutive Ones is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible subarray to count consecutive 1's. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Max Consecutive Ones?

Max Consecutive Ones uses the following concepts: Array. The recommended patterns to study are: Sliding Window, Array.

What are the interview tips for Max Consecutive Ones?

Always consider edge cases, like arrays with all 0's or all 1's. Explain your thought process clearly during the interview. Practice dry runs of your code with sample inputs.

What are common mistakes when solving Max Consecutive Ones?

Not resetting the count when encountering a 0. Overcomplicating the logic with unnecessary variables.

#485

Max Consecutive Ones

Easy

ArraySliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the array to count consecutive 1's. This is efficient and avoids unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Initialize two variables: max_count to track the maximum consecutive 1's and current_count to count the current streak of 1's.
2Step 2: Iterate through each element in the array.
3Step 3: If the current element is 1, increment current_count. If it's 0, reset current_count to 0.
4Step 4: After each increment, update max_count if current_count exceeds it.

solution.py12 lines

1# Full working Python code
2
3def findMaxConsecutiveOnes(nums):
4    max_count = 0
5    current_count = 0
6    for num in nums:
7        if num == 1:
8            current_count += 1
9            max_count = max(max_count, current_count)
10        else:
11            current_count = 0
12    return max_count

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array. The space complexity is O(1) as we only use a few variables for counting.

1The problem can be solved by counting streaks of 1's.
2Using a single pass is more efficient than nested loops.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.