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

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Problem Statement

You are given an array nums where the i-th element is the number of balls in the i-th bag. You can perform at most maxOperations operations, where in each operation you split one bag into two bags (each new bag must get a positive number of balls). For example, a bag of 5 balls could be split into bags of [1,4] or [2,3]. Your penalty is defined as the maximum number of balls in any bag, and you want to minimize this value after performing up to maxOperations splits. The task is to find the minimum possible penalty.

Example 1:
Input: nums = [9], maxOperations = 2
Output: 3
Explanation: Start with one bag of 9. Use 1st operation to split 9 into [6,3]. Use 2nd operation to split the bag of 6 into [3,3]. The bags become [3,3,3]. The largest bag now has 3 balls, so the penalty is 3.

Example 2:
Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation: We can achieve a penalty of 2 with 4 splits. For instance: split the bag of 8 into [4,4], then split each of those 4’s into [2,2] (using 2 more operations), and also split the original bag of 4 into [2,2] (4th operation). The final bags could be [2,2,2,2,2,2,2,2], all of size 2. The largest bag has 2 balls, so penalty = 2.

Example 3:
Input: nums = [7,17], maxOperations = 2
Output: 7
Explanation: With 2 splits, we can reduce the bag of 17 into bags of [7,7,3] (split 17 into 7 and 10, then split 10 into 7 and 3). The other bag has 7. Now the largest bag contains 7 balls, so the penalty can be 7. We cannot achieve a penalty of 6 or less because that would require more than 2 splits (for instance, making all bags ≤ 6 would demand a third split).

Constraints: 1 <= nums.length <= 10^5; 1 <= maxOperations, nums[i] <= 10^9. We expect an integer result representing the minimum achievable penalty. The solution should be efficient given the large input size.

Hints

Think in terms of “feasible penalty”: Suppose you fix a value X as the allowed maximum balls per bag (a potential penalty). Can you check if it’s possible to make every bag contain at most X balls with at most maxOperations splits? Try simulating the splitting process for a given X. If you can determine feasibility for any chosen X, it might guide you to the answer.
Use the search space wisely: Note that if a certain penalty value X is achievable, then any larger penalty (higher X) would also be achievable (since allowing bigger bags only makes it easier). Conversely, if X is not achievable, any smaller penalty would also fail (smaller limit requires even more splits). This monotonic behavior hints that you can apply binary search on the penalty value range instead of checking each value one by one.

Approaches

Brute Force Approach (Simulation of Penalties):
- How it works: A straightforward idea is to try every possible penalty limit from 1 up to the largest initial bag, and check if that limit can be achieved with the available operations. For each candidate penalty value p, we simulate splitting: for each bag in nums, calculate how many splits would be needed to ensure it has at most p balls. If the total splits required ≤ maxOperations, then p is a feasible penalty. We choose the smallest p that is feasible. This approach is easy to understand but not efficient for large inputs (the range of penalties can be up to the max number of balls, which is up to 10^9). However, it’s useful for conceptual understanding and for very small cases.
- Step-by-step example: Imagine nums = [3, 11] and maxOperations = 3. We’ll test increasing penalty limits:
  - Try p = 5. Bag of 3 needs 0 splits (already ≤5). Bag of 11 needs to be split: we can divide 11 into [5,6] (1 split), but 6 is still >5, so split 6 into [5,1] (2nd split). Now all bags are [3,5,5,1] and the largest has 5. We used 2 operations (≤3 allowed), so penalty 5 is feasible. We record 5 as a possible answer, but we will check smaller penalties to see if we can do better.
  - Try p = 4. Bag of 3 needs 0 splits. Bag of 11: split into [4,7] (1 split), then 7 into [4,3] (2nd split). Now bags [3,4,4,3], largest has 4 balls, and we used 2 splits (≤3). Penalty 4 is feasible.
  - Try p = 3. Bag of 3 needs 0 splits. Bag of 11: split into [3,8] (1), 8 into [3,5] (2), 5 into [3,2] (3rd split). After 3 operations, we have [3,3,3,2], largest bag = 3, and we've exactly used 3 ops. Penalty 3 is still feasible.
  - Try p = 2. Bag of 3: split into [2,1] (1 split). Bag of 11: we would need to split 11 into parts of size ≤2. 11 divided into parts of 2 would require many splits (in fact, at least 5 splits because 11→[2,9], 9→[2,7], 7→[2,5], 5→[2,3], 3→[2,1]). The total needed exceeds 3 operations. So penalty 2 is not feasible.
  From this simulation, the smallest feasible penalty was 3. We would output 3 for this example. This brute force checked each penalty in increasing order until it found a feasible one.

Python Implementation:

Python3

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Java Implementation:

Java

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Complexity: This brute force approach is extremely slow in the worst case. In the worst scenario, we might try every penalty from 1 up to max(nums). If M = max(nums) and n = len(nums), the time complexity is about O(n * M). With M up to 10^9 and n up to 10^5, this is completely impractical. (Even if we break out early when feasible, the upper bound remains too large.) The brute force is only viable for very small inputs or as a thought experiment. We need a more efficient strategy, which leads us to using binary search.

Optimized Approach Using Binary Search:
- Key Idea: Instead of testing every possible penalty, we leverage the monotonic nature of the problem (if a penalty is achievable, higher penalties are also achievable, and if it’s not achievable, lower ones aren’t either). This means we can apply binary search on the range of possible penalties. The lowest possible penalty is 1 (if we could split everything into single-ball bags), and the highest possible penalty is max(nums) (if we do no operations at all). We binary-search within this range to find the minimum feasible penalty.
- Feasibility Check: For a given candidate penalty p (midpoint of our search range), we need to check if it’s possible to make all bags ≤ p using at most maxOperations splits. We do this by summing the required splits for each bag:
  - A bag with x balls can be split into ceiling(x/p) bags to ensure each has ≤ p balls. The number of splits needed for that bag is ceil(x/p) - 1. This formula comes from dividing x into chunks of size at most p. Another way to compute this is (x - 1) // p (integer division after subtracting 1). For example, if x=8 and p=3, ceil(8/3) = 3, so 2 splits are needed (indeed 8 → split into 3 and 5 → split 5 into 3 and 2, total 2 splits). If x <= p, then 0 splits are needed.
  - We sum the splits needed for all bags. Let this sum be totalSplits. If totalSplits <= maxOperations, then penalty p is feasible (we can achieve that max bag size with the given operations). If totalSplits > maxOperations, then p is too low (not feasible).
- Binary Search Procedure:
  1. Initialize low = 1 and high = max(nums) as the bounds for penalty.
  2. While low < high:
    - Compute mid = (low + high) // 2 (this is a candidate penalty to test).
    - Perform the feasibility check for mid.
    - If mid is achievable (feasible), that means we might be able to do even better (try a smaller penalty), so set high = mid.
    - If mid is not achievable (needs too many splits), we must allow a higher penalty, so set low = mid + 1.
  3. When the loop finishes, low will have converged to the smallest feasible penalty. We return low.
  This approach drastically reduces the number of trials by halving the search space each time, instead of linear scanning.
- Step-by-step example (binary search in action): Let’s walk through the example nums = [2,4,8,2], maxOperations = 4 (from Example 2) using the binary search method:
  - Start with low = 1 and high = 8 (since the largest bag has 8 balls). The search space for penalty is [1, 8].
  - Mid = 4: We check if a penalty of 4 is feasible.
    For each bag, splits needed = (balls - 1) // 4:
    - Bag of 2: (2-1)//4 = 0 splits (already ≤4)
    - Bag of 4: (4-1)//4 = 0 splits (exactly 4, no split needed)
    - Bag of 8: (8-1)//4 = 7//4 = 1 split needed (8 can be split into two bags of 4 and 4)
    - Bag of 2: 0 splits
      Total splits needed = 1. We have up to 4 ops, so 1 ≤ 4, feasible. Because 4 works, we try to go lower – set high = 4. (Now range is [1, 4].)
  - Mid = 2: Now check if penalty 2 is feasible.
    Calculate splits (balls - 1)//2 for each bag:
    - 2 → (2-1)//2 = 0 splits
    - 4 → (4-1)//2 = 3//2 = 1 split (4 → split into 2 and 2)
    - 8 → (8-1)//2 = 7//2 = 3 splits (8 → split into 4+4, then each 4 → split into 2+2; that’s 3 splits in total for the 8-ball bag)
    - 2 → 0 splits
      Total splits = 0+1+3+0 = 4, which is exactly equal to maxOperations. That means it’s just barely feasible (we can achieve all bags ≤2 with 4 splits, as demonstrated in the example). So penalty 2 works; update high = 2. (Now range [1, 2].)
  - Mid = 1: Check penalty 1.
    Compute splits (balls - 1)//1 for each bag:
    - 2 → (2-1)//1 = 1 split (2 → split into 1 and 1)
    - 4 → (4-1)//1 = 3 splits (4 → split into 1,1,1,1 requires 3 splits)
    - 8 → (8-1)//1 = 7 splits (8 → split into eight 1’s requires 7 splits)
    - 2 → 1 split
      Total splits = 1+3+7+1 = 12, which far exceeds 4. Not feasible at all. So penalty 1 fails; we set low = mid + 1 = 2. (Range becomes [2, 2].)
  - Now low == high == 2, so the binary search ends. The algorithm returns 2, which matches the expected minimum penalty. We found the answer by checking only a few candidate values (in this case, 3 values: 4, 2, 1) instead of every number from 1 to 8.

Python Implementation:

Python3

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Java Implementation:

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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:

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).
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).
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).

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.

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

Problem Statement

Hints

Approaches

Step-by-Step Walkthrough of the Solution

Common Mistakes and Pitfalls

Edge Cases to Consider

Alternative Variations and Related Problems