Solution Approach

The solution uses binary search to find the smallest valid divisor. Here's how the implementation works:

Binary Search Setup:

• Left boundary: l = 1 (smallest possible divisor)
• Right boundary: r = max(nums) (any divisor larger than this won't reduce the sum further)

The Search Process:

1. Calculate the middle point: mid = (l + r) // 2
2. For the current mid, calculate the sum of all divisions: s = sum(ceil(x/mid) for x in nums)
3. Compare the sum with threshold:
   • If s ≤ threshold: This divisor works, but there might be a smaller one. Update r = mid
   • If s > threshold: We need a larger divisor to reduce the sum. Update l = mid + 1
4. Repeat until l meets r

Calculating Ceiling Division: The code uses a clever trick to calculate ceil(x/v) without using the ceiling function:

• Formula: (x + v - 1) // v
• This works because adding v - 1 to the numerator ensures rounding up for any remainder

The Given Solution's Approach: The provided code uses Python's bisect_left with a custom key function:

def f(v: int) -> bool:
    v += 1
    return sum((x + v - 1) // v for x in nums) <= threshold

return bisect_left(range(max(nums)), True, key=f) + 1

This function f(v) returns True if divisor v+1 produces a sum ≤ threshold. The bisect_left finds the leftmost position where True would fit in the boolean sequence, which corresponds to our smallest valid divisor. The +1 adjustments handle the 0-indexed nature of the range.

Time Complexity: O(n * log(max(nums))) where n is the length of the array Space Complexity: O(1) excluding the input array

Example Walkthrough

Let's walk through a small example to illustrate the binary search approach.

Given: nums = [2, 7, 11] and threshold = 10

Goal: Find the smallest divisor such that ceil(2/d) + ceil(7/d) + ceil(11/d) ≤ 10

Step 1: Initialize binary search boundaries

• Left boundary: l = 1
• Right boundary: r = max(nums) = 11
• Search range: [1, 11]

Step 2: First iteration

• mid = (1 + 11) // 2 = 6
• Calculate sum with divisor 6:
  • ceil(2/6) + ceil(7/6) + ceil(11/6) = 1 + 2 + 2 = 5
• Since 5 ≤ 10, divisor 6 works! But can we find a smaller one?
• Update: r = 6, new range: [1, 6]

Step 3: Second iteration

• mid = (1 + 6) // 2 = 3
• Calculate sum with divisor 3:
  • ceil(2/3) + ceil(7/3) + ceil(11/3) = 1 + 3 + 4 = 8
• Since 8 ≤ 10, divisor 3 works! Continue searching for smaller.
• Update: r = 3, new range: [1, 3]

Step 4: Third iteration

• mid = (1 + 3) // 2 = 2
• Calculate sum with divisor 2:
  • ceil(2/2) + ceil(7/2) + ceil(11/2) = 1 + 4 + 6 = 11
• Since 11 > 10, divisor 2 is too small (sum exceeds threshold).
• Update: l = 3, new range: [3, 3]

Step 5: Convergence

• l = r = 3, so we've found our answer!

Answer: The smallest divisor is 3, which gives us a sum of 8 ≤ 10.

Verification of neighboring values:

• Divisor 2: sum = 11 (too large, exceeds threshold)
• Divisor 3: sum = 8 (valid, within threshold) ✓
• Divisor 4: sum = 6 (valid, but not the smallest)

The binary search efficiently found the turning point where the divisor becomes just large enough to keep the sum within the threshold.

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 is because:

• The binary search using bisect_left performs O(log M) iterations over the range [0, max(nums)).
• In each iteration, the function f is called, which computes the sum of ceiling divisions for all elements in nums, taking O(n) time.
• Therefore, the total time complexity is O(n × log M).

The space complexity is O(1). The algorithm only uses a constant amount of extra space for variables like v and the sum calculation, regardless of the input size. The range(max(nums)) in bisect_left is treated as a virtual range and doesn't create an actual list in memory.

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

The most critical mistake is using while left < right with right = mid instead of the standard template. This problem requires finding the first true index (smallest valid divisor).

Wrong template:

while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid  # Wrong pattern
    else:
        left = mid + 1
return left

Correct template:

first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1  # Correct: search left for smaller
    else:
        left = mid + 1
return first_true_index

2. Incorrect Ceiling Division Formula

Many developers try to use floating-point division with math.ceil, which can lead to precision issues.

Problematic approach:

import math
total_sum = sum(math.ceil(num / divisor) for num in nums)
# Can have floating-point precision issues

Better approach:

# Integer-only arithmetic: ceil(x/d) = (x + d - 1) // d
total_sum = sum((num + divisor - 1) // divisor for num in nums)

3. Wrong Binary Search Boundaries

Setting the right boundary incorrectly can miss valid solutions or cause unnecessary iterations.

Wrong:

left, right = 1, min(nums)  # Too small - may miss valid divisors
left, right = 1, sum(nums)  # Too large - wastes iterations

Correct:

# Right boundary should be max(nums) since any divisor larger
# than the maximum element won't reduce the sum further
left, right = 1, max(nums)

4. Not Tracking the Answer Separately

Returning left directly instead of tracking first_true_index can cause issues in edge cases.

Wrong:

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid - 1
    else:
        left = mid + 1
return left  # May be incorrect if no valid answer exists

Correct:

first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid  # Track the answer
        right = mid - 1
    else:
        left = mid + 1
return first_true_index