The O(n) Club: Partition to K Equal Sum Subsets Without Losing Sanity

The O(n) Club: Partition to K Equal Sum Subsets Without Losing Sanity

⚡ TL;DR

Divide an array into k non-empty groups with equal sums. If sum(nums) % k != 0 or any number is bigger than the target, save yourself the headache and bail immediately.
When in doubt, backtrack—Java style:

public boolean canPartitionKSubsets(int[] nums, int k) {
    int sum = 0;
    for (int n : nums) sum += n;
    if (sum % k != 0) return false;
    // Now for the real circus: backtracking...
}

🧠 How Devs Usually Mess This Up

If you like chasing bugs, ignore these classics:

• Backtracking bravely, without checking whether sum(nums) divides evenly into k. Nothing says ‘Sunday wasted’ like an impossible state.
• Missing that any number greater than the target means failure. No, Java can’t shrink a 10 into a 5 by “creative” casting.
• Fixating on the actual partitions, not just whether it’s possible. Just tell your PM if it can be done, not which intern is on which team.

🔍 What’s Actually Going On

Pretend you’re splitting an overstuffed piñata’s loot among k impatient kids. Every kid wants the same amount or they riot. You get one shot: if the loot divides clean, and none of it’s too chunky, you’re golden. Otherwise, have fun with chaos.

This is the classic Partition problem, but with a multipack. Before you spiral into recursive misery:

🛠️ PseudoCode

• Pre-check: Save future-you hours
  

  int sum = total(nums);
  if (sum % k != 0) return false;

  If it doesn’t divide, you’re done.

• Size matters
  

  int target = sum / k;
  for (int n : nums) if (n > target) return false;

  No oversized candies in the bucket, please.

• Sort for easier fails
  

  Arrays.sort(nums); reverse(nums);

  Tackle big numbers first—fail fast, nap sooner.

• Backtracking with buckets
  

  boolean backtrack(nums, used, k, sum, idx, target) {
    // If all but one bucket is full, relax
    // Try placing nums[idx] into any bucket
    // Undo if it doesn't fit, try next
  }

  Classic fill, undo, repeat cycle (also known as debugging in production).

💻 The Code

public boolean canPartitionKSubsets(int[] nums, int k) {
    int sum = 0;
    for (int n : nums) sum += n;
    if (sum % k != 0) return false;
    int target = sum / k;
    Arrays.sort(nums);
    int n = nums.length;
    // Reverse for descending order
    for (int i = 0, j = n - 1; i < j; i++, j--) {
        int t = nums[i]; nums[i] = nums[j]; nums[j] = t;
    }
    if (nums[0] > target) return false;
    boolean[] used = new boolean[n];
    return backtrack(nums, used, k, 0, 0, target);
}
 private boolean backtrack(int[] nums, boolean[] used, int k, int bucket, int start, int target) {
    if (k == 1) return true;
    if (bucket == target) return backtrack(nums, used, k - 1, 0, 0, target);
    for (int i = start; i < nums.length; i++) {
        if (used[i] || bucket + nums[i] > target) continue;
        used[i] = true;
        if (backtrack(nums, used, k, bucket + nums[i], i + 1, target)) return true;
        used[i] = false;
        if (bucket == 0) break;
    }
    return false;
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Skipping divisibility check: Leads to existential debugging crises.
• Letting big numbers in: Spoiler alert: they never fit.
• Not sorting: You’ll backtrack way more than your laptop’s fan can handle. Always sort descending.
• Overcomplicating with fancy DP: If n ≤ 16, standard backtracking’s fine. Your interviewers secretly hope you won’t bitmask yourself into a corner—unless you WANT style points.
• Time & space: Worst-case is O(k × n × 2ⁿ). It’s fine if you don’t invite too many numbers to the party.

🧠 Interviewer Brain-Tattoo

“Always check if your buckets are leak-proof before you pour in the numbers, or you’ll be debugging your own tears by 2AM.”