1760. Minimum Limit of Balls in a Bag - Detailed Explanation

Complexity: The binary search approach is much more efficient. Each feasibility check is O(n) (we iterate through the bags to sum splits), and we perform O(log M) such checks, where M is the max balls in a bag. So overall time complexity is about O(n · log M). With n up to 10^5 and M up to 10^9, this is manageable (log2(10^9) ≈ 30, so roughly 30 * 10^5 operations in worst case). Space complexity is O(1) since we only use a few variables for counting and the binary search bounds.

Step-by-Step Walkthrough of the Solution

To summarize the strategy, we narrow down the penalty limit using binary search, adjusting based on feasibility:

1. Initial Bounds: The smallest possible penalty is 1 (all balls in separate bags) and the largest is max(nums) (no splits at all). We start with low = 1 and high = max(nums).

2. Binary Search Loop: We repeatedly pick a midpoint mid = (low + high)//2 as a trial penalty. Then we check if this penalty is achievable with the allowed operations:

   • We compute the total splits required to enforce that no bag has more than mid balls. This is done by summing (balls_i - 1)//mid for each bag i. (This formula effectively calculates ceil(balls_i / mid) - 1 .)
   • If the required splits ≤ maxOperations, it means mid is a feasible penalty – we can achieve this bag size limit. So we try to lower the penalty and search in the lower half (high = mid).
   • If the required splits > maxOperations, then mid is not feasible – we don’t have enough operations to reach that small of a max bag size. So we must allow a higher penalty and search in the upper half (low = mid + 1).

3. Convergence: Eventually, low and high converge to the smallest feasible penalty, at which point the loop ends. That value is our answer.

Let’s visualize this with a simple case for clarity: nums = [9], maxOperations = 2 (Example 1). Initially low=1, high=9.

• Try mid=5: We need (9-1)//5 = 1 split for the bag of 9 (split 9 into, say, [5,4]). 1 split ≤ 2, so penalty 5 is doable. We tighten upper bound to 5.
• Try mid=3: Now high=5, low=1 ⇒ mid=3. Splits needed = (9-1)//3 = 2 (split 9 into [3,6], then [3,3,3] after two splits). 2 splits ≤ 2, so penalty 3 is feasible. Update high = 3.
• Try mid=2: Now high=3, low=1 ⇒ mid=2. Splits needed = (9-1)//2 = 4. This means even after 2 splits, we would still have a bag larger than 2 (indeed, from [9] → [5,4] → [5,2,2], we’d need more splits). 4 > 2, not feasible. So we raise low to 3.
• Now low=3, high=3, the loop ends. The answer is 3.

This matches the example result and shows how the algorithm honed in on the correct penalty by testing only a few candidates instead of all values. During the process, we distributed balls by splitting as needed for each candidate penalty and adjusted our search range based on whether the distribution fit within the allowed operations.

Common Mistakes and Pitfalls

• Attempting a Greedy Splitting Strategy: A naive instinct might be to always split the largest bag first until you run out of operations, hoping this yields the minimal possible maximum. However, this greedy approach doesn’t always work optimally for this problem. The distribution of splits among bags matters. For example, if you have multiple large bags, you might need to distribute your splits among them rather than focus only on the single largest bag. The correct approach is to consider a target penalty and ensure all bags meet that threshold, rather than splitting one bag at a time without a global plan. (In fact, trying to simulate greedy splitting is essentially what the feasibility check does for a given penalty, but doing it without a guiding penalty can lead to suboptimal or complex logic.)

• Off-by-One Errors in Calculating Splits: A frequent bug is computing the number of splits for a bag incorrectly. Remember that if a bag has x balls and we want each bag to have at most p balls: the number of splits required is ceil(x/p) - 1. Using integer arithmetic, this is (x - 1) // p . If you use x // p (floor division) without the adjustment, you’ll underestimate splits when x is exactly divisible by p. For example, if x=6 and p=3, 6 // 3 = 2 (which might misleadingly suggest 2 splits needed) but actually a bag of 6 only needs 1 split to become two bags of 3 and 3. Using (6-1)//3 = 1 correctly gives the needed one split. Be careful with this calculation.

• Not Recognizing the Monotonic Pattern: Some might try a linear scan of possible penalties or other complex strategies. It’s important to see that once a penalty is achievable, any larger penalty is also achievable (and vice versa for failing). Missing this insight can lead to inefficient solutions. Make sure to leverage the binary search pattern for problems like this.

• Ignoring Edge Cases: Forgetting to handle extreme scenarios, such as when no splits are needed or when splitting down to 1-ball bags, can cause wrong answers or runtime errors. Always consider the boundaries (penalty = 1 and penalty = max(nums)) in your reasoning.

Edge Cases to Consider

• No Operations Allowed: If maxOperations = 0, you cannot split any bag. The answer in this case is simply the size of the largest bag (since you’re stuck with the initial configuration). The algorithm should handle this by immediately returning max(nums) when maxOperations is 0.

• All Bags Already Small: If every bag already has a very small number of balls (e.g. 1 or 2 balls each) and/or maxOperations is high, the answer might just be the current max. For example, nums=[1,2,2] with any number of operations – the penalty starts at 2 and you can’t improve it by splitting 1’s and 2’s (splitting a 2 would actually increase the count of bags but the max would remain 2). Ensure the solution doesn’t try unnecessary splits in such cases (the formula naturally yields 0 required splits for those).

• Extremely Large Single Bag: If one bag has a very large number of balls and you have many operations, the optimal strategy may be to split that bag almost completely. For instance, nums=[10^9] and maxOperations = 10^9: you could split into all 1’s using all operations, achieving a penalty of 1. The binary search should gracefully handle searching down to 1 in such cases.

• Multiple Large Bags vs Limited Operations: If there are multiple bags with the maximum value and not enough operations to split all of them, the answer will remain that maximum value. For example, nums=[5,5,5] with maxOperations=2. Even though you can split two of the bags (reducing their sizes), one bag of 5 will remain un-split, so the penalty can’t go below 5. The algorithm’s feasibility check inherently covers this, since the total required splits would exceed maxOperations for any penalty < 5.

• Very Large nums.length: The solution should handle up to 100k bags. Our approach iterates through all bags for each check, which is fine, but ensure no step has worse complexity than O(n). We do not store huge additional structures, so memory isn’t a big problem (aside from input storage).

Alternative Variations and Related Problems

This problem is an example of using binary search on the answer (also known as binary search on a monotonic condition). Once you recognize this pattern, many seemingly different problems can be solved similarly by choosing a “guess” and checking a condition. Here are a few related problems and variations:

• LeetCode 875: Koko Eating Bananas – Another problem where you find the minimum eating speed so that Koko can finish all bananas in H hours. The structure is analogous: choosing a speed and checking if it’s sufficient (monotonic feasibility check), and using binary search to find the minimum feasible speed.

• LeetCode 1011: Capacity To Ship Packages Within D Days – You must find the smallest ship capacity to transport all packages in D days. You can binary search the capacity and check if a given capacity can schedule the shipments within D days.

• LeetCode 1482: Minimum Number of Days to Make m Bouquets – You have flowers blooming over days and need to make bouquets. You binary search on days and check if by that day you can form the required bouquets. It’s a similar feasibility check pattern (if it’s possible by day X, it’s possible by any later day, etc.).

• LeetCode 1552: Magnetic Force Between Two Balls – This involves placing balls in baskets such that the minimum distance between any two balls is maximized. You guess a distance and check if you can place balls with at least that spacing. It’s a binary search on the distance with a greedy check, again a similar pattern of monotonic feasibility.

• LeetCode 410: Split Array Largest Sum – Although a different scenario (splitting an array into subarrays to minimize the largest sum), it also can be solved by binary searching the answer (the maximum subarray sum allowed) and checking feasibility by greedily forming subarrays. This is conceptually similar in that splitting and checking a threshold is involved.

• Other variations: Generally, any problem that asks for a minimum possible maximum (or maximum possible minimum) and provides a way to test a candidate solution is a good candidate for binary search on the answer. Practice identifying the monotonic property (if one guess works/fails, what about higher or lower guesses?) and writing a check function. This will help in problems like allocating resources, minimizing maximum load, or meeting deadlines with variable speed/capacity.