Allocate Minimum Pages Binary Search on Answer in DSA Rust - Build from Scratch

Allocate Minimum Pages using Binary Search on Answer

📖 Scenario: You are a librarian who needs to allocate books to students. Each book has a certain number of pages. You want to distribute the books so that the student who reads the most pages reads as few pages as possible.This helps keep reading fair and balanced among students.

🎯 Goal: Build a program that finds the minimum possible maximum number of pages assigned to any student using binary search on the answer.

📋 What You'll Learn

Create a vector called books with these exact page counts: 12, 34, 67, 90

Create a variable called students and set it to 2

Write a function called can_allocate that takes books, students, and a max_pages limit, and returns true if allocation is possible without exceeding max_pages per student

Use binary search between the maximum single book pages and the sum of all pages to find the minimum maximum pages

Print the minimum maximum pages after allocation

💡 Why This Matters

🌍 Real World

This problem models fair distribution of workload or resources, such as assigning tasks or dividing files among workers.

💼 Career

Understanding binary search on answer is useful for optimization problems in software engineering and algorithm design roles.

1

Create the books vector

Create a vector called books with these exact page counts: 12, 34, 67, 90

DSA Rust

# Create the books vector with exact page counts
# Your code here

Hint

Use vec![12, 34, 67, 90] to create the vector.

2

Set the number of students

Create a variable called students and set it to 2

DSA Rust

let books = vec![12, 34, 67, 90];
# Create the students variable and set it to 2
# Your code here

Hint

Use let students = 2; to create the variable.

3

Write the allocation check function

Write a function called can_allocate that takes books: &Vec<i32>, students: i32, and max_pages: i32, and returns bool. It should return true if books can be allocated to students without any student reading more than max_pages pages.

DSA Rust

let books = vec![12, 34, 67, 90];
let students = 2;

// Write the can_allocate function below
# Your code here

Hint

Check if adding a book exceeds max_pages. If yes, assign to next student. Return false if students needed exceed given students.

4

Find and print the minimum maximum pages

Use binary search between the maximum pages in books and the sum of all pages in books to find the minimum maximum pages that can be allocated to students. Print the result using println!("{}", result).

DSA Rust

let books = vec![12, 34, 67, 90];
let students = 2;

fn can_allocate(books: &Vec, students: i32, max_pages: i32) -> bool {
    let mut required_students = 1;
    let mut current_sum = 0;
    for &pages in books.iter() {
        if pages > max_pages {
            return false;
        }
        if current_sum + pages > max_pages {
            required_students += 1;
            current_sum = pages;
            if required_students > students {
                return false;
            }
        } else {
            current_sum += pages;
        }
    }
    true
}

// Use binary search to find minimum max pages and print it
# Your code here

Hint

Use binary search between max single book pages and total pages. Update result when allocation is possible.

Practice

1. What is the main goal of the Allocate Minimum Pages problem using binary search on the answer?

easy

A. To minimize the maximum number of pages assigned to a student

B. To maximize the total number of pages assigned

C. To find the average number of pages per student

D. To assign pages randomly among students

Solution

1. Step 1: Understand the problem objective

   The problem aims to distribute book pages among students such that the maximum pages assigned to any student is as small as possible.

2. Step 2: Identify the goal of binary search on answer

   Binary search is used to find this minimum possible maximum pages value efficiently.

3. Final Answer:

   To minimize the maximum number of pages assigned to a student -> Option A

4. Quick Check:

   Goal = minimize max pages [OK]

Hint: Minimize max pages per student using binary search [OK]

Common Mistakes:

• Thinking the goal is to maximize pages
• Confusing average pages with maximum pages
• Assigning pages randomly without constraints

2. Which of the following Rust function signatures correctly represents the binary search on answer approach for the Allocate Minimum Pages problem?

easy

A. fn allocate_pages(pages: &[u32], students: usize) -> u32

B. fn allocate_pages(pages: &[u32], students: u32) -> u32

C. fn allocate_pages(pages: Vec<u32>, students: u32) -> usize

D. fn allocate_pages(pages: Vec<u32>, students: usize) -> usize

Solution

1. Step 1: Analyze parameter types for pages and students

   Pages should be a slice reference &[u32] for read-only access, and students count is a u32 as a positive integer.

2. Step 2: Check return type for minimum maximum pages

   The function returns a u32 representing the minimum maximum pages allocated.

3. Final Answer:

   fn allocate_pages(pages: &[u32], students: u32) -> u32 -> Option B

4. Quick Check:

   Signature matches input slice and u32 output [OK]

Hint: Use slice &[u32] and u32 for students and return [OK]

Common Mistakes:

• Using Vec instead of slice for input when not modifying
• Mixing usize and u32 inconsistently
• Returning usize instead of u32 for pages

3. Given the pages array [10, 20, 30, 40] and 2 students, what is the minimum maximum pages allocated after running the binary search algorithm?

medium

A. 60

B. 50

C. 70

D. 40

Solution

1. Step 1: Calculate search range for binary search

   Minimum pages = max page = 40, maximum pages = sum = 100.

2. Step 2: Apply binary search to find minimum max pages

   Check mid = 70: possible to allocate? Yes (10+20+30=60 <=70, 40 alone). Try lower mid = 60: possible? Yes (10+20+30=60, 40 alone). Try mid = 50: possible? No (10+20=30, next 30 alone, 40 alone means 3 students needed). So minimum max pages = 60.

3. Final Answer:

   60 -> Option A

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Check feasibility at mid, adjust search range [OK]

Common Mistakes:

• Choosing sum of all pages as answer
• Ignoring max single page as lower bound
• Miscounting number of students needed

4. Identify the bug in this Rust code snippet for checking feasibility in Allocate Minimum Pages problem:

fn is_possible(pages: &[u32], students: u32, max_pages: u32) -> bool {
    let mut count = 1;
    let mut total = 0;
    for &page in pages {
        total += page;
        if total > max_pages {
            count += 1;
            total = page;
        }
    }
    count <= students
}

medium

A. It increments count before adding page to total

B. It resets total incorrectly after exceeding max_pages

C. It does not check if a single page is greater than max_pages

D. It uses u32 instead of usize for count

Solution

1. Step 1: Understand feasibility check logic

   The function sums pages until exceeding max_pages, then assigns to next student.

2. Step 2: Identify missing check for single page size

   If any single page is larger than max_pages, allocation is impossible, but code does not check this explicitly.

3. Final Answer:

   It does not check if a single page is greater than max_pages -> Option C

4. Quick Check:

   Check single page > max_pages missing [OK]

Hint: Always check if any single page exceeds max_pages [OK]

Common Mistakes:

• Ignoring single page size check
• Resetting total before incrementing count
• Using wrong variable types causing overflow

5. You have pages array [12, 34, 67, 90] and 2 students. Using binary search on answer, what is the minimum maximum pages allocation? Also, explain why the naive approach of assigning equal pages to each student might fail.

hard

A. 67; Naive approach fails due to uneven page sizes

B. 124; Naive approach always works for equal distribution

C. 90; Naive approach is optimal for sorted pages

D. 113; Naive approach ignores page order and max page size

Solution

1. Step 1: Calculate binary search range and feasibility

   Max single page = 90, sum = 203. Check mid = 113: allocate (12+34+67=113) to first student, 90 to second student. Possible. Try lower mid = 101: (12+34=46, next 67 alone, 90 alone) needs 3 students, not possible. So minimum max pages = 113.

2. Step 2: Explain naive approach failure

   Naive approach divides total pages (203) by students (2) = 101.5, but cannot split pages arbitrarily or ignore max single page size, so it fails to find feasible allocation.

3. Final Answer:

   113; Naive approach ignores page order and max page size -> Option D

4. Quick Check:

   Minimum max pages = 113; naive fails due to constraints [OK]

Hint: Check max single page and feasibility, naive ignores constraints [OK]

Common Mistakes:

• Assuming equal division always works
• Ignoring max single page size
• Not testing feasibility with actual page grouping