Allocate Minimum Pages Binary Search on Answer in DSA Rust - Execution Trace

Concept Flow - Allocate Minimum Pages Binary Search on Answer

Set low = max(pages array)

↓

Set high = sum(pages array)

↓

While low <= high

↓

Calculate mid = (low + high) / 2

↓

Check if allocation possible with mid

↓

Update answer

↓

Decrease high

↓

Return answer

We use binary search on the range of possible maximum pages to allocate, checking feasibility each time, narrowing down to the minimum maximum pages.

Execution Sample

DSA Rust

fn is_possible(pages: &[i32], students: i32, max_pages: i32) -> bool {
    let mut required = 1;
    let mut current_sum = 0;
    for &page in pages {
        if page > max_pages {
            return false;
        }
        if current_sum + page > max_pages {
            required += 1;
            current_sum = page;
            if required > students {
                return false;
            }
        } else {
            current_sum += page;
        }
    }
    true
}

This function checks if it is possible to allocate books so that no student reads more than max_pages.

Execution Table

Step	Operation	mid (max pages)	Allocation Possible?	Update low/high	Answer	Visual State (Students allocation)
1	Initialize low and high	low=40, high=100	-	-	-	Pages array: [10, 20, 30, 40]
2	Calculate mid	mid=70	-	-	-	-
3	Check allocation with mid=70	70	Yes	high=mid-1=69	answer=70	Student1: [10,20,30] sum=60; Student2: [40] sum=40
4	Calculate mid	mid=54	-	-	-	-
5	Check allocation with mid=54	54	No	low=mid+1=55	answer=70	Student1: [10,20] sum=30; Student2: [30] sum=30; Student3: [40] sum=40 (3 students needed)
6	Calculate mid	mid=62	-	-	-	-
7	Check allocation with mid=62	62	Yes	high=mid-1=61	answer=62	Student1: [10,20,30] sum=60; Student2: [40] sum=40
8	Calculate mid	mid=58	-	-	-	-
9	Check allocation with mid=58	58	No	low=mid+1=59	answer=62	Student1: [10,20] sum=30; Student2: [30] sum=30; Student3: [40] sum=40 (3 students needed)
10	Calculate mid	mid=60	-	-	-	-
11	Check allocation with mid=60	60	Yes	high=mid-1=59	answer=60	Student1: [10,20,30] sum=60; Student2: [40] sum=40
12	Loop ends	-	-	-	Final answer=60	Final allocation as above

💡 Loop ends when low (60) > high (59), final answer is 60 which is minimum max pages.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 9	After Step 11	Final
low	40	40	55	55	59	59	60
high	100	69	69	61	61	59	59
mid	-	70	54	62	58	60	-
answer	-	70	70	62	62	60	60
required_students	-	2	3	2	3	2	-

Key Moments - 3 Insights

Why do we set low to max(pages) at the start?

Because no student can be assigned less than the largest single book, so low must start at least max(pages). See Step 1 in execution_table.

Why does allocation fail when required students exceed given students?

Because it means with current max_pages limit, we need more students than available. This is shown in Steps 5 and 9 where allocation needs 3 students but only 2 are available.

Why do we update high when allocation is possible?

Because we try to find a smaller maximum pages allocation, so we reduce high to mid-1 to search lower values. See Steps 3, 7, and 11.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'answer' after Step 7?

A 62

B 70

C 60

D 54

💡 Hint

Check the 'Answer' column in execution_table row for Step 7.

At which step does the allocation first become impossible with the mid value?

A Step 3

B Step 5

C Step 7

D Step 11

💡 Hint

Look at 'Allocation Possible?' column in execution_table to find first 'No'.

If the number of students increased to 3, how would the final answer change?

A It would decrease

B It would increase

C It would stay the same

D Cannot determine

💡 Hint

More students allow smaller max pages per student; see variable_tracker for required_students.

Concept Snapshot

Allocate Minimum Pages using Binary Search:
- Set low = max single book pages
- Set high = sum of all pages
- While low <= high:
  - mid = (low + high) / 2
  - Check if allocation possible with mid
  - If yes, update answer and high = mid - 1
  - Else, low = mid + 1
- Return answer as minimum max pages

Full Transcript

This concept uses binary search on the answer space to allocate books to students such that the maximum pages assigned to any student is minimized. We start with low as the largest single book pages and high as the sum of all pages. We repeatedly check if allocation is possible with mid value. If yes, we try to find a smaller max by reducing high; if no, we increase low. The process continues until low exceeds high, and the last successful mid is the answer.

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