The O(n) Club: Why Splitting Arrays Feels Like Filing Taxes (LeetCode 410 Explained)

The O(n) Club: Why Splitting Arrays Feels Like Filing Taxes (LeetCode 410 Explained)

⚡ TL;DR

You’ve got a rowdy array, and you want to split it into k continuous, non-empty groups. The catch: make the fattest group as skinny as possible. Brute-force? Only if your hobby is watching paint dry. Use binary search with a pinch of greedy.

// Painfully slow approach – for masochists only:
int bruteSplit(int[] nums, int k) {
    // Try every split. Weep. Repeat.
    return -1;
}

🧠 How Devs Usually Mess This Up

First-timers often mess up by:

• Trying to force every group to have the same sum. Newsflash: Arrays aren’t pizza, you can’t slice them evenly every time.
• Building an epic, memory-eating DP table that runs slower than Windows Update.
• Mistaking the point: the real goal is minimizing the largest group sum, not perfect balance or using all your splits just for sport.

🔍 What’s Actually Going On

Think about assigning workloads. Imagine splitting a giant spreadsheet among k interns. You want to make sure nobody gets stuck with that one tab that makes them hate Mondays. That’s the algorithm: keep the biggest chunk as small as possible.

Here’s the Jedi trick: Binary search a possible “max group sum”, then check if you can split into k or fewer groups without going over. Don’t like a guess? Nudge it up or down. It’s like haggling at a garage sale, but with numbers.

🛠️ PseudoCode

• Set your hunting grounds:
  

  int low = max(nums);
  int high = sum(nums);

  Start the search at the biggest single value (every group must have at least one number), cap it at the full sum (one group, no splits).
• Guess and check using binary search:
  

  while (low < high) {
      int mid = (low + high) / 2;
      if (canSplit(nums, k, mid)) {
          high = mid;
      } else {
          low = mid + 1;
      }
  }

  Move the search window according to whether your guess is too optimistic (need more groups than allowed) or too fat (fits easily, try smaller).
• The greedy checker:
  

  int groups = 1, sum = 0;
  for (int n : nums) {
      if (sum + n > maxAllowed) {
          groups++;
          sum = n;
      } else {
          sum += n;
      }
  }
  return groups <= k;

  Keep a running sum. Go over the limit? Start a new group. Count your groups. If you run out of k before the end, your guess was too low.

💻 The Code

public int splitArray(int[] nums, int k) {
    int low = 0, high = 0;
    for (int n : nums) {
        low = Math.max(low, n);
        high += n;
    }
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (canSplit(nums, k, mid)) {
            high = mid;
        } else {
            low = mid + 1;
        }
    }
    return low;
}
 private boolean canSplit(int[] nums, int k, int maxAllowed) {
    int groups = 1, sum = 0;
    for (int n : nums) {
        if (sum + n > maxAllowed) {
            groups++;
            sum = n;
        } else {
            sum += n;
        }
    }
    return groups <= k;
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Edge-caser alert: If k == 1, no splits, you get the sum. If k >= nums.length, your biggest single element is your answer.
• Off-by-one syndrome: Mixing up < and <= in the checker or binary search can cause nastier bugs than a week-old doughnut.
• Complexity check: O(n log(sum(nums))): The binary search window depends on the spread between biggest element and total sum; the greedy walk takes O(n) per check. Space? No awkward piles, it’s a breezy O(1).

🧠 Interviewer Brain-Tattoo

“Array splitting isn’t about perfection—it’s about who gets stuck with the least-bad piece of the mess. Code like a chef dividing up the last donut.”