1283. Find the Smallest Divisor Given a Threshold

Method 1

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

Method 1

【O(N * log(M)) time | O(1) space】

package Leetcode.BinarySearch.FindMinimum;

/**
 * @author zhengxingxing
 * @date 2025/04/28
 */
public class FindTheSmallestDivisorGivenAThreshold {

    public static int smallestDivisor(int[] nums, int threshold) {
        // Find maximum number in array to set binary search upper bound
        int maxNum = 0;
        for (int num : nums) {
            maxNum = Math.max(maxNum, num);
        }

        // Perform binary search to find smallest valid divisor
        // Left boundary: 1 (minimum possible divisor)
        // Right boundary: maxNum (maximum needed divisor)
        int left = 1, right = maxNum;
        while (left < right) {
            // Calculate middle point avoiding overflow
            int mid = left + (right - left) / 2;

            // If current divisor produces sum <= threshold,
            // try to find smaller divisor in left half
            if (calculateSum(nums, mid) <= threshold) {
                right = mid;
            } else {
                // If sum > threshold, need larger divisor
                left = mid + 1;
            }
        }
        return left;
    }

    private static int calculateSum(int[] nums, int divisor) {
        int sum = 0;
        for (int num : nums) {
            // Calculate ceiling division using integer arithmetic
            // Formula: ceil(num/divisor) = (num + divisor - 1) / divisor
            sum += (num + divisor - 1) / divisor;
        }
        return sum;
    }

    public static void main(String[] args) {
        // Test Case 1: Basic example with small numbers
        int[] nums1 = {1, 2, 5, 9};
        System.out.println("Test Case 1: " + smallestDivisor(nums1, 6));  // Expected: 5

        // Test Case 2: Example with prime numbers
        int[] nums2 = {2, 3, 5, 7, 11};
        System.out.println("Test Case 2: " + smallestDivisor(nums2, 11));  // Expected: 3

        // Test Case 3: Single element array
        int[] nums3 = {19};
        System.out.println("Test Case 3: " + smallestDivisor(nums3, 5));  // Expected: 4
    }
}

Method 1

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