Allocate Minimum Pages Binary Search on Answer in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum number of pages in any single book.

DSA Rust

let max_pages = *books.iter().[1]().unwrap();

Drag options to blanks, or click blank then click option'

Amax

Bmin

Csum

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using min() instead of max()

Using sum() which returns total pages

Using count() which returns number of books

We use iter().max() to find the maximum pages in any book, which is the lower bound for allocation.

2fill in blank

medium

Complete the code to check if allocation is possible with given max pages per student.

DSA Rust

if allocate(books, students, [1]) {

Drag options to blanks, or click blank then click option'

Amin_pages

Bmax_pages

Cmid

Dtotal_pages

Attempts:

3 left

💡 Hint

Common Mistakes

Using max_pages or min_pages directly instead of mid

Using total_pages which is sum of all pages

We check allocation feasibility with mid as the maximum pages allowed per student.

3fill in blank

hard

Fix the error in the allocation function to correctly count students needed.

DSA Rust

if current_sum + book > [1] {
    students_needed += 1;
    current_sum = book;
} else {
    current_sum += book;
}

Drag options to blanks, or click blank then click option'

Atotal_pages

Bmax_pages

Cmin_pages

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with max_pages or total_pages instead of mid

Not resetting current_sum correctly

We compare current sum plus book pages to mid, the max pages allowed per student.

4fill in blank

hard

Fill both blanks to update binary search bounds correctly after allocation check.

DSA Rust

if allocate(books, students, mid) {
    result = mid;
    [1] = mid + 1;
} else {
    [2] = mid - 1;
}

Drag options to blanks, or click blank then click option'

Alow

Bhigh

Cmid

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping low and high updates

Updating mid instead of low or high

If allocation is possible, move low up to mid+1 to try for a better answer; else move high down to mid-1.

5fill in blank

hard

Fill both blanks to complete the binary search loop for minimum pages allocation.

DSA Rust

let mut low = [1];
let mut high = [2];
let mut result = high;
while low <= high {
    let mid = (low + high) / 2;
    if allocate(books, students, mid) {
        result = mid;
        low = mid + 1;
    } else {
        high = mid - 1;
    }
}
result

Drag options to blanks, or click blank then click option'

Amax_pages

Btotal_pages

Cmin_pages

Dstudents

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to 0 instead of max_pages

Setting high to max_pages instead of total_pages

Initializing result incorrectly

low starts at max_pages (maximum pages in any single book), high at total_pages (sum of all pages), result initialized to high.

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