LeetCode Solutions Blog

Max Consecutive Ones

Number: 485

Difficulty: Easy

Paid? No

Problem Description

Given a binary array, return the maximum number of consecutive 1's in the array.

Key Insights

Iterate over the array while keeping track of the current sequence length of 1's.
Reset the counter when a 0 is encountered.
Update the maximum value during the iteration.
This direct scanning results in a simple O(n) solution with constant space.

Space and Time Complexity

Time Complexity: O(n) - We traverse the array once. Space Complexity: O(1) - Only constant extra variables are used.

Solution

We iterate through the binary array and maintain two counters: one for the current sequence of consecutive 1's (currentCount) and one for the maximum sequence found so far (maxConsecutive). As we scan the array, we increment currentCount when we see a 1 and update maxConsecutive. When a 0 is found, we reset currentCount to 0. This approach leverages a simple linear pass through the array and uses a constant amount of extra space.

Code Solutions

# Function to return maximum consecutive 1s in an array
def findMaxConsecutiveOnes(nums):
    # Initialize counters for consecutive ones
    max_consecutive = 0
    current_count = 0
    # Iterate through each number in the list
    for num in nums:
        if num == 1:
            # Increase the count when the current number is 1
            current_count += 1
            # Update maximum if the current count is higher
            max_consecutive = max(max_consecutive, current_count)
        else:
            # Reset current count when a 0 is encountered
            current_count = 0
    return max_consecutive

# Example usage:
if __name__ == "__main__":
    example_nums = [1, 1, 0, 1, 1, 1]
    print(findMaxConsecutiveOnes(example_nums))  # Expected output: 3