LeetCode #1283 — MEDIUM

Find the Smallest Divisor Given a Threshold

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[1,2,5,9]
6

Example 2

[44,22,33,11,1]
5

Core Insight

What unlocks the optimal approach

Examine every possible number for solution. Choose the largest of them.
Use binary search to reduce the time complexity.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1283: Find the Smallest Divisor Given a Threshold
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;
    }
}

// Accepted solution for LeetCode #1283: Find the Smallest Divisor Given a Threshold
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
}

# Accepted solution for LeetCode #1283: Find the Smallest Divisor Given a Threshold
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

// Accepted solution for LeetCode #1283: Find the Smallest Divisor Given a Threshold
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
    }
}

// Accepted solution for LeetCode #1283: Find the Smallest Divisor Given a Threshold
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.