The O(n) Club: Max Consecutive Ones III: Where Sliding Windows Beat Your WiFi and Your Bugs

The O(n) Club: Max Consecutive Ones III—Where Sliding Windows Beat Your WiFi and Your Bugs

⚡ TL;DR

Want to binge 1s in your binary array by flipping up to k zeros? Don’t cry—slide! Two pointers. Window never gets stuck in traffic. Java magic here:

int longestOnes(int[] nums, int k) {
    int left = 0, zeros = 0, max = 0;
    for (int right = 0; right < nums.length; right++) {
        if (nums[right] == 0) zeros++;
        while (zeros > k) {
            if (nums[left++] == 0) zeros--;
        }
        max = Math.max(max, right - left + 1);
    }
    return max;
}

🧠 How Devs Usually Mess This Up

Ah, the classics. You treat “flip” as “delete”, so you skip zeros and create an algorithm so slow it’s basically a screensaver. Or, you forget to “refund” your precious flips when sliding off a 0—suddenly your window is shrinking faster than your caffeine supply. Sometimes folks try to count 1s, just to make sure the universe is as confusing as possible. Spoiler: it never works.

🔍 What’s Actually Going On

Think streaming video: you get k chances to patch up packet drops (zeroes), and you want max binge before buffering. The array is like episodes, the window is your attention span, and those zeros? Neighbors microwaving soup. Slide the window right, use flips on 0s, and every time you run out, push the left side forward until you’re good again. At each step, brag about your streak. Algorithms: fixing WiFi before the router can.

🛠️ PseudoCode

• Initialize window stuff:

  left = 0; zeros = 0; maxLen = 0;

  Track where the window starts, how many zeros in it, and the best streak.
• Expand right end:

  for (right = 0; right < nums.length; right++)

  Slide in new elements at right.
• Count zeros you flip:

  if nums[right] == 0: zeros++;

  Every 0 is a “WiFi fail” you can patch—until you run out.
• Shrink left if broke:

  while zeros > k:

  Sliding too many fails? Move the left side:

  if nums[left] == 0: zeros--;
  left++;

• Track the peak binge:

  maxLen = max(maxLen, right - left + 1);

  Check if this is your longest window yet.

💻 The Code

public int longestOnes(int[] nums, int k) {
    int left = 0, zeros = 0, max = 0;
    for (int right = 0; right < nums.length; right++) {
        if (nums[right] == 0) zeros++;
        while (zeros > k) {
            if (nums[left++] == 0) zeros--;
        }
        max = Math.max(max, right - left + 1);
    }
    return max;
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Test Case Roulette: All zeros? All ones? What if k is huge? Relax, window still slides.
• Painful common bug: Forgetting to decrement zeros when skipping a 0 at left. This turns your O(n) dream into a night terror.
• Time Cost: Both pointers each stroll across the array just once—O(n). Space: A handful of ints. Fits on a cocktail napkin.

🧠 Interviewer Brain-Tattoo

“If your sliding window gets stuck, it’s not the algorithm—it’s you. Don’t count 1s. Slide the window and let it do the flexing.”