Skip to content

1283. Find the Smallest Divisor Given a Threshold

Description

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

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

The test cases are generated so 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 of 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

Input: nums = [44,22,33,11,1], threshold = 5
Output: 44

 

Constraints:

  • 1 <= nums.length <= 5 * 104
  • 1 <= nums[i] <= 106
  • nums.length <= threshold <= 106

Solutions

Thinking

A larger divisor only decreases the sum of ceilings, so the predicate is monotone. \(n \le 5\times 10^4\) forbids trying every divisor. We binary-search the smallest \(v\) in \([1,\max nums]\) such that \(\sum \lceil nums_i/v \rceil \le threshold\). Each check is one linear scan.

Notice that for number \(v\), if the sum of results of dividing each number in \(nums\) by \(v\) is less than or equal to \(threshold\), then all values greater than \(v\) satisfy the condition. There is a monotonicity, so we can use binary search to find the smallest \(v\) that satisfies the condition.

We define the left boundary of the binary search \(l=1\), \(r=\max(nums)\). Each time we take \(mid=(l+r)/2\), calculate the sum of the results of dividing each number in \(nums\) by \(mid\) \(s\), if \(s\) is less than or equal to \(threshold\), then it means that \(mid\) satisfies the condition, we will update \(r\) to \(mid\), otherwise we will update \(l\) to \(mid+1\).

Finally, return \(l\).

The time complexity is \(O(n \times \log M)\), where \(n\) is the length of the array \(nums\) and \(M\) is the maximum value in the array \(nums\). The space complexity is \(O(1)\).

1
2
3
4
5
6
7
class Solution:
    def smallestDivisor(self, nums: List[int], threshold: int) -> int:
        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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
class Solution {
    public int smallestDivisor(int[] nums, int threshold) {
        int l = 1, r = 1000000;
        while (l < r) {
            int mid = (l + r) >> 1;
            int s = 0;
            for (int x : nums) {
                s += (x + mid - 1) / mid;
            }
            if (s <= threshold) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution {
public:
    int smallestDivisor(vector<int>& nums, int threshold) {
        int l = 1;
        int r = *max_element(nums.begin(), nums.end());
        while (l < r) {
            int mid = (l + r) >> 1;
            int s = 0;
            for (int x : nums) {
                s += (x + mid - 1) / mid;
            }
            if (s <= threshold) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
func smallestDivisor(nums []int, threshold int) int {
    return sort.Search(slices.Max(nums), func(v int) bool {
        v++
        s := 0
        for _, x := range nums {
            s += (x + v - 1) / v
        }
        return s <= threshold
    }) + 1
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
function smallestDivisor(nums: number[], threshold: number): number {
    let l = 1;
    let r = Math.max(...nums);
    while (l < r) {
        const mid = (l + r) >> 1;
        const s = nums.reduce((acc, x) => acc + Math.ceil(x / mid), 0);
        if (s <= threshold) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
impl Solution {
    pub fn smallest_divisor(nums: Vec<i32>, threshold: i32) -> i32 {
        let mut l = 1;
        let mut r = *nums.iter().max().unwrap();
        while l < r {
            let mid = (l + r) / 2;
            let s: i32 = nums.iter().map(|&x| (x + mid - 1) / mid).sum();
            if s <= threshold {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
/**
 * @param {number[]} nums
 * @param {number} threshold
 * @return {number}
 */
var smallestDivisor = function (nums, threshold) {
    let l = 1;
    let r = Math.max(...nums);
    while (l < r) {
        const mid = (l + r) >> 1;
        const s = nums.reduce((acc, x) => acc + Math.ceil(x / mid), 0);
        if (s <= threshold) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
public class Solution {
    public int SmallestDivisor(int[] nums, int threshold) {
        int l = 1;
        int r = nums.Max();
        while (l < r) {
            int mid = (l + r) >> 1;
            int s = 0;
            foreach (int x in nums) {
                s += (x + mid - 1) / mid;
            }
            if (s <= threshold) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

Comments