Allocate Minimum Pages Binary Search on Answer in DSA Rust - Time & Space Complexity

Time Complexity: Allocate Minimum Pages Binary Search on Answer

O(n * log m)

Understanding Time Complexity

We want to find how fast the binary search approach works to allocate minimum pages among students.

Specifically, how the number of operations grows as the number of books or pages increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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
}
    

This code tries to find the smallest maximum number of pages assigned to any student by using binary search on the answer range.

Identify Repeating Operations

• Primary operation: The binary search loop runs over the range of possible page sums.
• How many times: Binary search runs about log of the sum of pages times.
• Inside each binary search step: The is_possible function traverses all books once (linear scan).
• Dominant operation: The linear scan inside is_possible repeated for each binary search step.

How Execution Grows With Input

As the number of books (n) increases, the linear scan inside is_possible grows linearly.

The binary search runs about log of the total pages range, which depends on the sum of pages.

Input Size (n)	Approx. Operations
10 books	~10 * log(sum of pages)
100 books	~100 * log(sum of pages)
1000 books	~1000 * log(sum of pages)

Pattern observation: Operations grow linearly with number of books and logarithmically with the range of pages.

Final Time Complexity

Time Complexity: O(n * log m)

This means the time grows linearly with the number of books and logarithmically with the total pages range.

Common Mistake

[X] Wrong: "The binary search alone makes this O(log n) time."

[OK] Correct: Each binary search step requires scanning all books, so the linear scan dominates and multiplies the log factor.

Interview Connect

Understanding this time complexity helps you explain how binary search can be applied on answers, a common pattern in coding problems.

It shows how combining binary search with linear checks balances speed and correctness.

Self-Check

What if we replaced the linear scan in is_possible with a more complex check? How would that affect the overall time complexity?

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