# How to Find the Smallest Divisor Given a Threshold in Java ## The challenge

Given an array of integers `nums` and an integer `threshold`, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to `threshold`.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

Example 1:

```Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1.
If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2). ```

Example 2:

```Input: nums = [2,3,5,7,11], threshold = 11
Output: 3```

Example 3:

```Input: nums = , threshold = 5
Output: 4```

Constraints:

• `1 <= nums.length <= 5 * 10^4`
• `1 <= nums[i] <= 10^6`
• `nums.length <= threshold <= 10^6`

## The solution in Java code

Option 1:

```.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}```class Solution {
public long computeSum(int[] nums, int right) {
long sum = 0;
for (int n : nums) {
sum += n / right + (n % right == 0 ? 0 : 1);
}
return sum;
}

public int smallestDivisor(int[] nums, int threshold) {
int left = 1, right = 2;
while (computeSum(nums, right) > threshold) {
left = right;
right <<= 1;
}
while (left <= right) {
int pivot = left + ((right - left) >> 1);
long num = computeSum(nums, pivot);

if (num > thrPeshold) left = pivot + 1;
else right = pivot - 1;
}
return left;
}
}
```Code language: Java (java)```

There may be a simpler way to do this, let’s explore further:

``````class Solution {
public int smallestDivisor(int[] nums, int threshold)  {
int m = 0;
int l = 1;
int r = max(nums);

while (l <= r) {
if (sum(nums, threshold, m = (l + r) / 2))
l = m + 1;
else r = m - 1;
}
return l;
}

private static int max(int[] nums) {
int max = 0;

for (int n : nums) {
max = Math.max( max, n );
}
return max;
}

private static boolean sum(int[] nums, int t, int d) {
int sum = 0;

for (int n : nums) {
sum += (n - 1) / d + 1;
}
return sum > t;
}
}
```Code language: Java (java)```
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