The Problem

The problem asks us to find the maximum number of consecutive 1s in a binary array. You know, like if you have something like this:

[1, 1, 0, 1, 1, 1]

The maximum consecutive ones here is 3, right? Simple enough, that’s how we roll.

So, here’s the actual problem statement from LeetCode:

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

My Solution

Here’s a straightforward solution I came up with while deep in the zone (you know what I mean). I’ve written it in Java, but if you’re a Python aficionado or something else, you’ll get the gist!

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int fc = 0;  // This is for the final count
        int lc = 0;  // This is for the current count
        
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) {
                lc = 0; // If it's a zero, reset the current count
            } else {
                lc++; // Otherwise, just increment the current count
            }
            fc = Math.max(lc, fc); // Update max count
        }
        return fc; // Return the max consecutive ones
    }
}

Complexity Analysis

Let’s break down the complexity a bit because that’s always a fun part, right?

Time Complexity: O(n), where n is the length of the array. We only loop through the array once, so it’s pretty efficient.
Space Complexity: O(1), since we’re using just a couple of extra variables and not any additional data structures.

Conclusion

And there you have it, folks! My very own solution to LeetCode problem 485. It’s clean, it’s neat, and hell, it works! Each time I solve one of these LeetCode problems, I feel like I’ve leveled up my coding skills, even if just a smidge. It’ll certainly make my day job a bit easier, I hope!

If you want to dive deeper or see examples of more coding problems, check out the following references: