The O(n) Club: Maximum Sum Circular Subarray Madness

The O(n) Club: Maximum Sum Circular Subarray Madness

⚡ TL;DR

If you’ve ever dreamed of wrapping your arrays like burritos, here’s your shot. To find the maximum sum of a circular array segment, run Kadane’s for max and for min, compare, and don’t forget: if the whole thing is negative, return the largest number and move on with your life.
Here’s how you cheat like a pro:

// Java: Donut-style Kadane’s
int donutKadane(int[] nums) {
    int max = nums[0], min = nums[0], total = nums[0], curMax = nums[0], curMin = nums[0];
    for (int i = 1; i < nums.length; i++) {
        curMax = Math.max(nums[i], curMax + nums[i]);
        max = Math.max(max, curMax);
        curMin = Math.min(nums[i], curMin + nums[i]);
        min = Math.min(min, curMin);
        total += nums[i];
    }
    return (max < 0) ? max : Math.max(max, total - min);
}

🧠 How Devs Usually Mess This Up

Let the facepalms commence:

• Trying every single wrap by brute force (O(n²)). If your CPU had feelings it would file a complaint.
• Getting creative with prefix/suffix combos—then waterboarding themselves with edge cases.
• Thinking subarrays mean you can pick-n-mix elements. Sorry, only unbroken segments—no hopscotch allowed.
• Assuming negative-only arrays will somehow reward you with a wrap-around miracle. You just end up with zero—aka, disappointment in numeric form.

🔍 What’s Actually Going On

Think of your circular array as a 24-hour diner counter: you can start eating anywhere, keep munching until you’re full, and even loop back to breakfast if you want. Kadane’s helps you find the best feast (max segment). For the circular twist, imagine just skipping the worst greasy patch (the minimum subarray) and devouring the rest. That’s total sum – min subarray sum. It’s not magic. It’s not even microwaved. Just classic algorithmic leftovers – but with the best bits kept.

But beware: if the entire menu is rotten (all negatives), you’re forced to pick the least disgusting dish. There’s no wrap that saves you from food poisoning.

🛠️ PseudoCode

• Let maxSum, curMax, minSum, curMin, and total all start at nums[0].
• Loop from 1 to n-1:
  • Max subarray: curMax = max(nums[i], curMax + nums[i]); maxSum = max(maxSum, curMax)
  • Min subarray: curMin = min(nums[i], curMin + nums[i]); minSum = min(minSum, curMin)
  • Add to total: total += nums[i]
• If maxSum < 0 (everything is negative), return maxSum.
• Otherwise, answer is max(maxSum, total - minSum).

💻 The Code

public int maxSubarraySumCircular(int[] nums) {
    int total = nums[0];
    int curMax = nums[0], maxSum = nums[0];
    int curMin = nums[0], minSum = nums[0];
    for (int i = 1; i < nums.length; i++) {
        curMax = Math.max(nums[i], curMax + nums[i]);
        maxSum = Math.max(maxSum, curMax);
        curMin = Math.min(nums[i], curMin + nums[i]);
        minSum = Math.min(minSum, curMin);
        total += nums[i];
    }
    if (maxSum < 0) return maxSum;
    return Math.max(maxSum, total - minSum);
}

⚠️ Pitfalls, Traps & Runtime Smacks

• All negatives? Don’t use the circular trick; it’ll hand you a big fat zero. Take the least-awful number.
• Prefix/Suffix overthinkers: It’s O(n). Don’t build towers of index arithmetic unless you want a hard debug session.
• Always contiguous. Subarrays are whatever you’d get from a single uninterrupted stomachache.
• Watch for off-by-one in min/max updates. Otherwise, it’s endless “Why is my test case failing?” therapy sessions.

🧠 Interviewer Brain-Tattoo

“Want to find the best part of a circular array? Subtract out the worst, and see what’s left. Even arrays know teamwork means firing the weakest link.”