Allocate Minimum Pages Binary Search on Answer in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Minimum Pages Allocation for given books and students

What is the output of the Rust function that finds the minimum number of pages allocated to a student when books are divided optimally among students?

DSA Rust

fn is_possible(books: &[i32], students: i32, max_pages: i32) -> bool {
    let mut required_students = 1;
    let mut current_sum = 0;
    for &pages in books {
        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
}

fn allocate_min_pages(books: &[i32], students: i32) -> i32 {
    let mut low = *books.iter().max().unwrap();
    let mut high: i32 = books.iter().sum();
    let mut result = high;
    while low <= high {
        let mid = low + (high - low) / 2;
        if is_possible(books, students, mid) {
            result = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    result
}

fn main() {
    let books = vec![12, 34, 67, 90];
    let students = 2;
    let result = allocate_min_pages(&books, students);
    println!("{}", result);
}

A 113

B 157

C 90

D 67

💡 Hint

Think about how binary search narrows down the maximum pages a student can get.

The minimum maximum pages allocated when dividing books [12, 34, 67, 90] among 2 students is 113. One optimal division is [12, 34, 67] and [90], where the maximum pages assigned to a student is 113.

🧠 Conceptual

intermediate

Understanding the role of 'is_possible' function in Allocate Minimum Pages

What does the 'is_possible' function check in the Allocate Minimum Pages problem?

A Whether the total pages of all books is less than max_pages

B Whether the number of students is greater than the number of books

C Whether books can be allocated to students such that no student reads more than max_pages

D Whether the books are sorted in ascending order

💡 Hint

Focus on the condition checked inside the loop for each book.

The 'is_possible' function checks if it is possible to allocate books to the given number of students such that no student reads more than max_pages pages.

🔧 Debug

advanced

Identify the error in this Rust code snippet for Allocate Minimum Pages

What error will this Rust code produce when compiled?

DSA Rust

fn allocate_min_pages(books: &[i32], students: i32) -> i32 {
    let mut low = 0;
    let mut high = books.iter().sum();
    let mut result = high;
    while low < high {
        let mid = (low + high) / 2;
        if is_possible(books, students, mid) {
            result = mid;
            high = mid;
        } else {
            low = mid + 1;
        }
    }
    result
}

fn is_possible(books: &[i32], students: i32, max_pages: i32) -> bool {
    let mut required_students = 1;
    let mut current_sum = 0;
    for &pages in books {
        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
}

A Infinite loop because high is never decreased properly

B Compilation error due to missing semicolon

C Runtime panic due to integer overflow

D No error, code runs correctly

💡 Hint

Check the binary search loop condition and how high is updated.

The loop condition 'while low < high' with 'high = mid;' can cause infinite loop when low == mid, because high does not decrease below mid.

🚀 Application

advanced

Applying Allocate Minimum Pages to a new scenario

Given books with pages [10, 20, 30, 40, 50] and 3 students, what is the minimum maximum pages allocated to a student?

A 90

B 70

C 60

D 50

💡 Hint

Try dividing books into 3 parts minimizing the maximum sum.

One optimal division is [10, 20, 30], [40], [50]. The maximum pages assigned is 70.

🧠 Conceptual

expert

Why binary search is used in Allocate Minimum Pages problem?

Why is binary search an efficient approach to solve the Allocate Minimum Pages problem?

A Because it tries all possible allocations exhaustively

B Because it sorts the books before allocation

C Because it uses recursion to allocate pages

D Because the answer lies within a numeric range and feasibility can be checked quickly

💡 Hint

Think about how the search space is defined and how feasibility is tested.

Binary search efficiently narrows down the minimum maximum pages by checking feasibility in a numeric range without trying all allocations.

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