Understanding the Problem
We are given an array of positive integers and a threshold
. Our goal is to find the smallest positive divisor such that when each number in the array is divided by this divisor and rounded up (using the ceiling function), the total sum of these results is less than or equal to the threshold.
Let’s look at an example to make it clear. Suppose:
nums = [1, 2, 5, 9]
threshold = 6
If we try divisor = 5, the results of ceiling division are:
- ceil(1 / 5) = 1
- ceil(2 / 5) = 1
- ceil(5 / 5) = 1
- ceil(9 / 5) = 2
The total sum is 1 + 1 + 1 + 2 = 5
, which is less than or equal to the threshold 6. Hence, 5 is a valid divisor. But is it the smallest one? That’s what we need to figure out.
Step-by-Step Solution with Example
Step 1: Understand ceiling division
For any number x
and divisor d
, ceil(x/d)
means rounding the result up. For example, ceil(9/4) = 3
, not 2. This helps ensure that the result is always at least as large as the exact division, rounded up.
Step 2: Define the problem as a search problem
We can treat the search for the smallest divisor as a range of numbers from 1 to the max(nums)
. For each value in this range, we calculate the total of all ceil(num / divisor)
and check if it’s less than or equal to the threshold.
Step 3: Use binary search to optimize
Trying each number one-by-one would be slow. Instead, we use binary search:
- Start with low = 1 and high = max(nums)
- At each step, pick
mid = (low + high) / 2
and calculate the total sum using this mid
as the divisor
- If the sum is within the threshold, try smaller divisors (high = mid - 1)
- Otherwise, try larger divisors (low = mid + 1)
Step 4: Walk through the example
With nums = [1, 2, 5, 9]
and threshold = 6
:
- Start with low = 1, high = 9
- mid = 5 ➝ total = 5 (valid) ➝ try smaller (high = 4)
- mid = 2 ➝ total = ceil(1/2)+ceil(2/2)+ceil(5/2)+ceil(9/2) = 1+1+3+5 = 10 ➝ too big ➝ try larger (low = 3)
- mid = 3 ➝ total = 1+1+2+3 = 7 ➝ too big ➝ try larger (low = 4)
- mid = 4 ➝ total = 1+1+2+3 = 7 ➝ too big ➝ try larger (low = 5)
At the end, we settle on divisor = 5 as the smallest valid divisor.
Edge Cases
- Empty array: There are no elements to divide. Return
-1
.
- Threshold is 0 or negative: It’s impossible to satisfy the requirement. Return
-1
.
- All numbers are 1: Any divisor ≥ 1 results in a sum equal to the length of the array.
- Very large numbers: Binary search ensures that we only do log(max) checks, so it stays efficient.
Finally
This problem is a great example of turning a mathematical challenge into an efficient algorithm. By using binary search over possible divisors, we quickly narrow down to the smallest one that meets the threshold. This approach handles large inputs well and avoids brute-force inefficiencies.
Always remember to consider edge cases and validate inputs before applying the main logic. And when you see a problem asking for “minimum x that satisfies a condition,” binary search is often your best friend.
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