Pseudo-code:

text
max_val = max(nums)
for candidate in 1 to max_val:
    ops_needed = 0
    for bag in nums:
        ops_needed += (bag - 1) / candidate
    if ops_needed <= maxOperations:
        return candidate

Why this fails: The maximum value in nums can be up to $10^9$. A linear scan from 1 to $10^9$ results in a time complexity of $O(N \cdot \max(nums))$, where $N$ is the array length. With $N=10^5$ and values up to $10^9$, this leads to approximately $10^{14}$ operations, causing a Time Limit Exceeded (TLE) error. We need a logarithmic approach to search the solution space.

Core Insight for Minimum Limit of Balls in a Bag

The key intuition is to invert the problem. Instead of asking "What is the minimum penalty?", we ask "Given a specific penalty $K$, can we achieve it within maxOperations?"

If we decide that no bag should contain more than $K$ balls, we can calculate exactly how many split operations are required for each existing bag.

• For a bag of size $S$, we need to split it into enough pieces such that each piece has size $\le K$.
• The number of pieces required is $\lceil S / K \rceil$.
• Since each split operation increases the number of bags by 1 (e.g., 1 bag becomes 2, then 3), the number of operations required to get $P$ pieces is $P - 1$.
• Therefore, the operations needed for a bag of size $S$ is $\lceil S / K \rceil - 1$. Using integer arithmetic, this is calculated as (S - 1) / K.

By summing these required operations for all bags, we get the total cost to achieve penalty $K$. If total_cost <= maxOperations, then $K$ is feasible.

Since the feasibility function is monotonic (larger $K$ requires fewer operations), we can binary search for the smallest feasible $K$.

Visual Description: Imagine the search space as a horizontal line representing all possible maximum bag sizes, ranging from 1 to the largest initial bag.

• The left side (small sizes) requires many operations (likely infeasible).
• The right side (large sizes) requires few operations (feasible).
• There is a specific point where the status flips from "Infeasible" to "Feasible".
• Our algorithm probes the midpoint. If it's feasible, we discard the right half (we want smaller). If it's infeasible, we discard the left half (we need larger). This logarithmic reduction quickly isolates the boundary.

Execution Flow

Let's trace Example 1: nums = [9], maxOperations = 2.

1. Initialization:

   • left = 1, right = 9.

2. Iteration 1:

   • mid = (1 + 9) / 2 = 5.
   • Check feasibility for limit 5:
     • Bag 9: (9 - 1) / 5 = 1 operation (split 9 -> 4, 5).
     • Total ops = 1.
   • $1 \le 2$ (maxOperations). Feasible.
   • Try smaller: right = 5.

3. Iteration 2:

   • mid = (1 + 5) / 2 = 3.
   • Check feasibility for limit 3:
     • Bag 9: (9 - 1) / 3 = 2 operations (split 9 -> 3, 3, 3).
     • Total ops = 2.
   • $2 \le 2$. Feasible.
   • Try smaller: right = 3.

4. Iteration 3:

   • mid = (1 + 3) / 2 = 2.
   • Check feasibility for limit 2:
     • Bag 9: (9 - 1) / 2 = 4 operations.
     • Total ops = 4.
   • $4 > 2$. Not feasible.
   • Must be larger: left = 2 + 1 = 3.

5. Termination:

   • left is 3, right is 3. Loop terminates.
   • Return left (3).

Proof of Correctness

The correctness relies on the monotonicity of the condition function. Let $f(x)$ be the minimum operations required to ensure all bags are size $\le x$.

• As $x$ increases (allowing larger bags), $f(x)$ decreases or stays the same.
• As $x$ decreases (forcing smaller bags), $f(x)$ increases.

We are looking for the smallest $x$ such that $f(x) \le \text{maxOperations}$. Since $f(x)$ is non-increasing, the boolean predicate $P(x): f(x) \le \text{maxOperations}$ transitions from False to True exactly once as $x$ increases. Binary search is guaranteed to find this transition point in logarithmic time relative to the range size.

The formula for operations, (bag - 1) / limit, correctly computes $\lceil \frac{\text{bag}}{\text{limit}} \rceil - 1$, which is the minimal number of cuts required to partition bag into chunks of size at most limit.

Reference Implementation

C++ Solution for LeetCode 1760

cpp
class Solution {
public:
    int minimumSize(vector<int>& nums, int maxOperations) {
        int left = 1;
        int right = 0;
        
        // The search space upper bound is the largest bag
        for (int num : nums) {
            right = max(right, num);
        }
        
        int result = right;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            // Check if it's possible to achieve max bag size 'mid'
            // within maxOperations
            long long opsNeeded = 0;
            for (int num : nums) {
                if (num > mid) {
                    // Equivalent to ceil(num / mid) - 1
                    opsNeeded += (num - 1) / mid;
                }
            }
            
            if (opsNeeded <= maxOperations) {
                result = mid;
                right = mid - 1; // Try to achieve a smaller max size
            } else {
                left = mid + 1; // Impossible, need larger max size
            }
        }
        
        return result;
    }
};

Java Solution for LeetCode 1760

java
class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int left = 1;
        int right = 0;
        
        // Establish upper bound
        for (int num : nums) {
            right = Math.max(right, num);
        }
        
        int result = right;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            if (canAchieve(mid, nums, maxOperations)) {
                result = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        
        return result;
    }
    
    private boolean canAchieve(int maxBagSize, int[] nums, int maxOperations) {
        long opsNeeded = 0;
        for (int num : nums) {
            if (num > maxBagSize) {
                // Determine splits needed. 
                // (num - 1) / maxBagSize is integer math for ceil(num/maxBagSize) - 1
                opsNeeded += (num - 1) / maxBagSize;
            }
            // Optimization: Early exit if we exceed limits
            if (opsNeeded > maxOperations) return false;
        }
        return true;
    }
}

Python Solution for LeetCode 1760

python
class Solution:
    def minimumSize(self, nums: List[int], maxOperations: int) -> int:
        left, right = 1, max(nums)
        
        def can_achieve(target_size):
            ops_needed = 0
            for num in nums:
                if num > target_size:
                    # Calculate required splits
                    # Equivalent to math.ceil(num / target_size) - 1
                    ops_needed += (num - 1) // target_size
            return ops_needed <= maxOperations
            
        while left < right:
            mid = (left + right) // 2
            if can_achieve(mid):
                right = mid
            else:
                left = mid + 1
                
        return left

Q: Why does my solution fail on large inputs even with Binary Search?

• Mistake: Using a 32-bit integer (int) to accumulate opsNeeded.
• Why: If you have many large bags and try to reduce them to a small size (e.g., 1), the total operations required can exceed $2^{31}-1$.
• Consequence: Integer overflow leads to incorrect feasibility checks. Always use long long (C++) or long (Java) for the accumulator.

Q: Why is my split calculation wrong?

• Mistake: Using num / mid instead of (num - 1) / mid.
• Why: If a bag is exactly size 6 and the target is 3, 6/3 = 2. But you only need 1 split to get [3, 3]. The formula (6-1)/3 = 5/3 = 1 correctly accounts for exact multiples.
• Consequence: You overestimate the operations needed, potentially returning a suboptimal answer.