Solution Approach

The solution implements a binary search to find the minimum possible maximum bag size. This is a classic "find first true" binary search pattern.

Binary Search Setup:

• Left boundary left = 1: The minimum possible bag size is 1 ball
• Right boundary right = max(nums): If we don't perform any operations, the maximum is the largest original bag
• first_true_index = -1: Tracks the minimum achievable maximum bag size

Feasibility Check can_achieve_max_size(max_size): This function determines if we can achieve a maximum bag size of max_size within maxOperations:

• For each bag with x balls, we calculate how many splits are needed: (x - 1) // max_size
• This formula works because to split x balls into bags of at most max_size balls, we need ceil(x/max_size) - 1 operations
• Sum all required operations and check if the total is ≤ maxOperations
• Returns True if achievable (feasible), False otherwise

Binary Search Template: The algorithm follows the standard "find first true" template:

left, right = 1, max(nums)
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if can_achieve_max_size(mid):
        first_true_index = mid
        right = mid - 1  # Try to find smaller valid size
    else:
        left = mid + 1   # Need larger size

return first_true_index

The key characteristics of this template:

1. Uses while left <= right (not left < right)
2. When feasible, records first_true_index = mid and searches left with right = mid - 1
3. When not feasible, searches right with left = mid + 1
4. Returns first_true_index which holds the smallest valid maximum size

Time Complexity: O(n * log(max(nums))) where n is the length of the array. We perform binary search over the range [1, max(nums)], and for each iteration, we check all n bags.

Space Complexity: O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with nums = [7, 17] and maxOperations = 2.

Step 1: Initialize Binary Search

• left = 1 (minimum possible bag size)
• right = max([7, 17]) = 17 (maximum without any operations)
• first_true_index = -1 (tracks the answer)

Step 2: Binary Search Process with Template

Iteration 1: left=1, right=17, mid=9

• Check feasible(9):
  • Bag with 7 balls: (7-1)//9 = 0 operations
  • Bag with 17 balls: (17-1)//9 = 1 operation
  • Total: 0 + 1 = 1 ≤ 2 → feasible(9) = True
• Since feasible: first_true_index = 9, right = 9 - 1 = 8

Iteration 2: left=1, right=8, mid=4

• Check feasible(4):
  • Bag with 7 balls: (7-1)//4 = 1 operation
  • Bag with 17 balls: (17-1)//4 = 4 operations
  • Total: 1 + 4 = 5 > 2 → feasible(4) = False
• Since not feasible: left = 4 + 1 = 5

Iteration 3: left=5, right=8, mid=6

• Check feasible(6):
  • Bag with 7 balls: (7-1)//6 = 1 operation
  • Bag with 17 balls: (17-1)//6 = 2 operations
  • Total: 1 + 2 = 3 > 2 → feasible(6) = False
• Since not feasible: left = 6 + 1 = 7

Iteration 4: left=7, right=8, mid=7

• Check feasible(7):
  • Bag with 7 balls: (7-1)//7 = 0 operations
  • Bag with 17 balls: (17-1)//7 = 2 operations
  • Total: 0 + 2 = 2 ≤ 2 → feasible(7) = True
• Since feasible: first_true_index = 7, right = 7 - 1 = 6

Iteration 5: left=7, right=6

• left > right, loop terminates

Result: Return first_true_index = 7

The minimum possible penalty is 7. We split the 17-ball bag twice to get [7, 7, 3], combined with the original 7-ball bag gives us [7, 7, 7, 3]. The maximum is 7.

Time and Space Complexity

The time complexity is O(n × log M), where n is the length of the array nums and M is the maximum value in the array nums.

This complexity arises from:

• Binary search over the range [1, max(nums)], which has M elements, taking O(log M) iterations
• For each binary search iteration, the check function is called, which iterates through all n elements in nums to compute sum((x - 1) // mx for x in nums), taking O(n) time
• Total: O(n) × O(log M) = O(n × log M)

The space complexity is O(1).

The algorithm only uses a constant amount of extra space for variables in the check function and the binary search operation. The range object in Python is a lazy iterator that doesn't create the entire list in memory, so it doesn't contribute to space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

Pitfall: Using while left < right with right = mid instead of the correct template:

# WRONG template variant
while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid
    else:
        left = mid + 1
return left

Correct template:

left, right = 1, max(nums)
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index

Key differences:

• Uses while left <= right (not left < right)
• Tracks the answer in first_true_index
• Uses right = mid - 1 when feasible (not right = mid)

2. Incorrect Split Calculation Formula

Pitfall: Using the wrong formula to calculate the number of operations needed:

• x // mx (forgetting to subtract 1)
• (x + mx - 1) // mx (this gives the number of resulting bags, not operations)
• x // mx - 1 (off by one when x is divisible by mx)

Why the formula (x - 1) // mx works:

• To split x balls into bags of at most mx balls each, you need ceil(x/mx) total bags
• Since you start with 1 bag, you need ceil(x/mx) - 1 operations
• The formula (x - 1) // mx is equivalent to ceil(x/mx) - 1

Example:

• With 9 balls and max size 3: (9-1)//3 = 8//3 = 2 operations needed
• This creates 3 bags of 3 balls each (9 → 6+3 → 3+3+3)

3. Binary Search Boundary Issues

Pitfall: Setting incorrect search boundaries:

• Starting with left = 0 instead of left = 1 (bag size must be positive)
• Not initializing first_true_index = -1 to handle edge cases

Correct initialization:

left, right = 1, max(nums)
first_true_index = -1

4. Integer Overflow in Other Languages

Pitfall: When translating to languages like C++ or Java, the sum of operations might overflow for large inputs.

Solution: Use appropriate data types:

long long totalOperations = 0;  // Use long long instead of int
for (int num : nums) {
    totalOperations += (num - 1) / maxSize;
}

5. Misunderstanding the Feasible Function

Pitfall: The feasible function should return True when the condition is satisfied (we CAN achieve the target with given operations), creating a pattern like:

false, false, false, true, true, true, ...

The binary search finds the first true (minimum feasible max size).