What is the time complexity of Find the Smallest Divisor Given a Threshold?

The optimal solution has a time complexity of O(nlog(m)).

What data structures are used to solve Find the Smallest Divisor Given a Threshold?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find the Smallest Divisor Given a Threshold?

Insight 1: The problem can be solved using Binary Search because the result of dividing each element by the divisor and summing them up is monotonically decreasing as the divisor increases. This monotonicity property allows us to efficiently search for the smallest divisor. Insight 2: The minimum possible divisor is 1, and the maximum possible divisor is the largest number in the `nums` array. These boundaries define the search space for the binary search.

Which companies ask LeetCode 1283. Find the Smallest Divisor Given a Threshold in interviews?

This problem is frequently asked at Agoda, DE Shaw, Expedia, Millennium, PayPal and 1 more companies.

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Find the Smallest Divisor Given a Threshold - LeetCode 1283 Solution

Find the Smallest Divisor Given a Threshold - Complete Solution Guide

Find the Smallest Divisor Given a Threshold is LeetCode problem 1283, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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], t

Detailed Explanation

The problem requires finding the smallest positive integer divisor that, when used to divide each number in a given array `nums`, results in a sum of the rounded-up quotients that is less than or equal to a given `threshold`. In simpler terms, we need to find the smallest divisor such that the sum of `ceil(num / divisor)` for all `num` in `nums` is at most `threshold`. The division result is always rounded up to the nearest integer.

Solution Approach

The solution uses binary search to find the smallest divisor. It starts with a search range from 1 (minimum divisor) to the maximum value in the input array `nums` (maximum divisor). In each iteration of the binary search, the midpoint `mid` is chosen as a potential divisor. The sum of the rounded-up quotients (using the formula `(num + mid - 1) // mid`) is calculated for all numbers in `nums`. If this sum is less than or equal to the `threshold`, it means we can potentially find a smaller divisor, so the upper bound of the search range is updated to `mid`. Otherwise, the sum exceeds the threshold, indicating that the divisor is too small, so the lower bound of the search range is updated to `mid + 1`. This process continues until the lower and upper bounds converge, at which point the lower bound represents the smallest divisor that satisfies the given condition.

Step-by-Step Algorithm

Step 1: Initialize `low` to 1 (minimum possible divisor) and `high` to the maximum element in `nums` (maximum possible divisor).
Step 2: While `low` is less than `high`, calculate the middle value `mid` as `(low + high) // 2`.
Step 3: Calculate `current_sum` by iterating through `nums` and summing the rounded-up quotients `(num + mid - 1) // mid`.
Step 4: If `current_sum` is less than or equal to `threshold`, update `high` to `mid` to search for a potentially smaller divisor in the lower half.
Step 5: Otherwise, if `current_sum` is greater than `threshold`, update `low` to `mid + 1` to search for a larger divisor in the upper half.
Step 6: After the loop finishes (when `low` is equal to `high`), `low` (or `high`) will contain the smallest divisor that satisfies the condition. Return `low`.

Key Insights

Insight 1: The problem can be solved using Binary Search because the result of dividing each element by the divisor and summing them up is monotonically decreasing as the divisor increases. This monotonicity property allows us to efficiently search for the smallest divisor.
Insight 2: The minimum possible divisor is 1, and the maximum possible divisor is the largest number in the `nums` array. These boundaries define the search space for the binary search.
Insight 3: The ceiling operation `ceil(num / divisor)` can be efficiently calculated using `(num + divisor - 1) // divisor` using integer division, avoiding floating-point calculations which could introduce inaccuracies.

Complexity Analysis

Time Complexity: O(nlog(m))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Agoda, DE Shaw, Expedia, Millennium, PayPal, ZScaler.

Find the Smallest Divisor Given a Threshold | LeetCode 1283

Medium

Array Binary Search

Time: O(nlog(m))Space: O(1)

Solution Code

Available in:

class Solution {
    public int smallestDivisor(int[] nums, int threshold) {
        int low = 1, high = 0;
        for (int num : nums) {
            high = Math.max(high, num);
        }

        while (low < high) {
            int mid = (low + high) / 2;
            
            int current_sum = 0;
            for (int num : nums) {
                current_sum += (num + mid - 1) / mid;
            }
            
            if (current_sum <= threshold) {
                high = mid;
            } else {
                low = mid + 1;
            }
        
        return low;
    }
}

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Problem Info

Description

Problem 1283: Find the Smallest Divisor Given a Threshold - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

AgodaDE ShawExpediaMillenniumPayPalZScaler

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:1283

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize `low` to 1 (minimum possible divisor) and `high` to the maximum element in `nums` (maximum possible divisor).
2
Step 2: While `low` is less than `high`, calculate the middle value `mid` as `(low + high) // 2`.
3
Step 3: Calculate `current_sum` by iterating through `nums` and summing the rounded-up quotients `(num + mid - 1) // mid`.
4
Step 4: If `current_sum` is less than or equal to `threshold`, update `high` to `mid` to search for a potentially smaller divisor in the lower half.
5
Step 5: Otherwise, if `current_sum` is greater than `threshold`, update `low` to `mid + 1` to search for a larger divisor in the upper half.
6
Step 6: After the loop finishes (when `low` is equal to `high`), `low` (or `high`) will contain the smallest divisor that satisfies the condition. Return `low`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(nlog(m))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the binary search which has O(log(m)) iterations and in each iteration a linear scan through the input array is performed with cost O(n). The space complexity is constant because we only use a fixed number of variables irrespective of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Linear Search

Iterate through all possible divisors starting from 1 until a divisor is found that satisfies the condition. The search stops when the sum of rounded-up quotients is less than or equal to the threshold.

Time: O(n*m) where n is the size of the input array and m is the maximum element in the array. We iterate through the divisor values from 1 to the largest value in the input array. For each divisor value, we sum the result of the rounded divisions.Space: O(1)

Trade-offs: The linear search is much simpler to implement but has a significantly worse time complexity compared to the binary search. It is only practical for very small input arrays and thresholds. The binary search approach is preferred for larger inputs.

✨

Key Insights

1Insight 1: The problem can be solved using Binary Search because the result of dividing each element by the divisor and summing them up is monotonically decreasing as the divisor increases. This monotonicity property allows us to efficiently search for the smallest divisor.
2Insight 2: The minimum possible divisor is 1, and the maximum possible divisor is the largest number in the `nums` array. These boundaries define the search space for the binary search.
3Insight 3: The ceiling operation `ceil(num / divisor)` can be efficiently calculated using `(num + divisor - 1) // divisor` using integer division, avoiding floating-point calculations which could introduce inaccuracies.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: `nums` contains only one element. The binary search will still converge to the correct divisor. For instance, if `nums = [10]` and `threshold = 3`, the algorithm correctly finds the divisor to be 4 (since 10/4 rounded up is 3).
Edge case 2: The `threshold` is very large, larger than the sum of the elements in `nums`. The algorithm will still converge to the smallest possible divisor, which is 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not rounding up the division result correctly. Integer division in most languages truncates towards zero, so one must use the `(num + divisor - 1) // divisor` formula to achieve proper rounding up.

✗Common mistake 2: Incorrectly initializing the `low` and `high` values for the binary search. `low` must be 1, and `high` must be the maximum value in `nums` to ensure the entire search space is covered.

✗Common mistake 3: Not handling the case where the threshold is very large which leads to divisor equal to 1.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling: Suppose we have a set of tasks with varying workloads (represented by `nums`) and a constraint on the maximum number of machines we can use (represented by `threshold`). We need to find the smallest workload divisor such that each machine's load is balanced and the number of machines doesn't exceed the threshold. The divisor then defines the granularity of the tasks to be assigned to a single machine.

🏦Real-world scenario 2: Resource allocation: Imagine allocating bandwidth to different users. The `nums` array represents the data requirements of each user, and the `threshold` represents the maximum allowable delay. The smallest divisor represents the minimum bandwidth that can be allocated to each user such that the total delay remains within the acceptable threshold.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 24, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find the Smallest Divisor Given a Threshold"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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