To solve this problem, we need to understand what it means to divide all numbers by a divisor and take the ceiling of each result. The goal is to find the smallest such divisor that keeps the sum of all these rounded-up divisions within the specified threshold
.
Understanding the Problem
For example, if the array is [1, 2, 5, 9]
and we divide each number by 5
, we get:
ceil(1/5) = 1
ceil(2/5) = 1
ceil(5/5) = 1
ceil(9/5) = 2
So the sum is 1 + 1 + 1 + 2 = 5
, which is within the threshold of 6. The challenge is to find the smallest such divisor that still keeps this sum within limits.
Different Situations to Consider
- If the divisor is small (e.g., 1): Then each number is divided by 1, so the sum of the array stays the same (no reduction).
- If the divisor is large (e.g., more than the max value in array): Then all divisions will round up to 1, which may help reduce the total sum.
The key is to find a balance — a divisor large enough to reduce the sum, but as small as possible.
What Happens in Edge Cases?
- If the array is empty, we can't divide anything. So we return
-1
.
- If the
threshold
is zero or negative, there's no way to achieve a valid result, so again return -1
.
How Do We Search Efficiently?
Instead of checking each divisor one by one, we can use binary search from 1
to max(nums)
. For each candidate divisor, we calculate the sum of ceilings, and decide if we need a larger or smaller divisor. This drastically reduces the number of checks and makes it efficient.
At the end of this process, we’ll have the smallest divisor that satisfies the condition. If no such divisor is found (which happens only if input is invalid), we return -1
.
This method ensures optimal performance and is suitable even when the input array is very large.