Allocate Minimum Pages Binary Search on Answer in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Allocate Minimum Pages problem?

To divide books among students so that the maximum number of pages assigned to a student is minimized.

intermediate

Why do we use binary search on the answer in this problem?

Because the answer (minimum maximum pages) lies between the maximum pages of a single book and the sum of all pages, allowing efficient searching in this range.

beginner

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

It checks if it is possible to allocate books to students such that no student reads more than a given maximum number of pages.

beginner

In the binary search approach, what do the 'low' and 'high' pointers represent?

'Low' represents the maximum pages in a single book, and 'high' represents the sum of all pages in all books.

intermediate

What is the time complexity of the Allocate Minimum Pages solution using binary search on answer?

O(N * log S), where N is the number of books and S is the sum of all pages.

What is the minimum possible value for the maximum pages assigned to a student?

A The maximum pages in a single book

B Zero

C The sum of all pages

D The average pages per student

The minimum maximum pages cannot be less than the largest single book because that book must be assigned whole.

What does the binary search 'mid' value represent in this problem?

A A candidate maximum pages per student

B Number of students

C Total pages of all books

D Number of books

The 'mid' is a guess for the maximum pages a student can read; we check feasibility with this value.

If the allocation is not possible with 'mid' pages, what should we do?

A Return mid as answer

B Increase 'low' to mid + 1

C Decrease 'high' to mid - 1

D Stop the algorithm

If allocation fails, we need to allow more pages per student, so increase 'low'.

What is the role of the 'is_possible' function in the binary search?

A To sort the books

B To calculate total pages

C To count number of books

D To check if current mid can be a valid maximum pages allocation

It tests if the current guess for max pages per student is feasible.

What happens when 'low' exceeds 'high' in the binary search?

A We decrease the number of books

B We reset the search

C The search ends and the answer is found

D We increase the number of students

When low > high, the binary search completes and the minimum maximum pages is found.

Explain how binary search is applied to find the minimum maximum pages in the Allocate Minimum Pages problem.

Think about how you guess and check the maximum pages a student can read.

Describe the purpose and working of the 'is_possible' function in the Allocate Minimum Pages problem.

It tests if a given maximum pages limit works for the allocation.

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