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

Recall & Review

beginner

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

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

intermediate

Why do we use binary search in the Allocate Minimum Pages problem?

Because the answer (minimum maximum pages) lies between the maximum single book pages and the sum of all pages, binary search helps efficiently find the minimum feasible maximum pages.

beginner

What does the 'isPossible' function check in this problem?

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

beginner

In the binary search approach, what are the initial low and high values?

Low is the maximum number of pages in a single book, and high is 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 does the 'high' value represent in the binary search for Allocate Minimum Pages?

A The maximum pages in a single book

B The number of students

C The minimum pages any student can read

D The sum of all pages in all books

The 'high' value is set to the sum of all pages because the maximum pages assigned to a student cannot exceed this.

What is the purpose of the 'isPossible' function?

A To check if a student can read all books alone

B To check if a given maximum page limit per student is feasible

C To find the maximum pages in a book

D To calculate the total pages

It verifies if books can be allocated so that no student reads more than the given page limit.

If the number of students is more than the number of books, what is the minimum maximum pages?

A Maximum pages in a single book

B Number of students

C Zero

D Sum of all pages

Each student can get at most one book, so the minimum maximum pages is the largest single book.

What is the main advantage of using binary search on the answer in this problem?

A It reduces the search space for the answer efficiently

B It sorts the books

C It counts the number of students

D It finds the smallest book

Binary search narrows down the minimum maximum pages quickly by checking feasibility.

Which of these is NOT a step in the binary search solution?

A Calculate sum of all pages

B Set low to max pages in a book

C Sort the books in descending order

D Check feasibility with mid value

Sorting is not required; allocation depends on page sums and feasibility.

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

Think about narrowing down the possible maximum pages by checking if allocation is possible.

Describe the role of the feasibility function in the Allocate Minimum Pages problem and how it works.

Focus on how the function simulates allocation under a page limit.

Practice

1. What is the main goal of the Allocate Minimum Pages problem using binary search on 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 asks to allocate books to students so 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 the minimum possible maximum pages that can be assigned to a student.

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: Focus on minimizing the largest pages assigned [OK]

Common Mistakes:

• Thinking the goal is to maximize pages
• Confusing average pages with maximum pages
• Assuming random allocation is acceptable

2. Which of the following is the correct way to initialize the binary search boundaries for the Allocate Minimum Pages problem in Java?

easy

A. int low = 0; int high = max pages in a single book;

B. int low = 0; int high = sum of all pages;

C. int low = max pages in a single book; int high = sum of all pages;

D. int low = sum of all pages; int high = max pages in a single book;

Solution

1. Step 1: Determine the minimum possible max pages

   The minimum max pages must be at least the largest single book's pages, because no book can be split.

2. Step 2: Determine the maximum possible max pages

   The maximum max pages is the sum of all pages if one student gets all books.

3. Final Answer:

   int low = max pages in a single book; int high = sum of all pages; -> Option C

4. Quick Check:

   low = max book pages, high = total pages [OK]

Hint: Low = max single book, High = sum all pages [OK]

Common Mistakes:

• Setting low to 0 which is invalid
• Swapping low and high values
• Using average pages as boundaries

3. Consider the books with pages [10, 20, 30, 40] and 2 students. What is the minimum maximum pages allocated to a student?

medium

A. 40

B. 70

C. 50

D. 60

Solution

1. Step 1: Check possible max pages allocations

   Try to split books so that max pages per student is minimized. Possible splits: [10,20,30] and [40] max=60; or [10,20] and [30,40] max=70.

2. Step 2: Choose minimum max pages

   Between 60 and 70, 60 is smaller and valid allocation.

3. Final Answer:

   60 -> Option D

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Try splits and pick smallest max sum [OK]

Common Mistakes:

• Choosing sum of all pages as answer
• Ignoring largest single book pages
• Incorrectly splitting books

4. Given the following code snippet for checking feasibility in the Allocate Minimum Pages problem, what is the bug?

boolean isPossible(int[] pages, int students, int mid) {
    int count = 1, sum = 0;
    for (int page : pages) {
        sum += page;
        if (sum > mid) {
            count++;
            sum = 0;
        }
    }
    return count <= students;
}

medium

A. sum should be reset to current page, not zero, after count increment

B. count should be reset to zero inside the loop

C. The loop should start from index 1, not zero

D. The function should return count >= students

Solution

1. Step 1: Analyze sum reset logic

   When sum exceeds mid, count increments but sum resets to zero, losing the current page's value.

2. Step 2: Correct sum reset to current page

   Sum should reset to current page's value to include it in the new student's allocation.

3. Final Answer:

   sum should be reset to current page, not zero, after count increment -> Option A

4. Quick Check:

   Reset sum = current page [OK]

Hint: Reset sum to current page, not zero [OK]

Common Mistakes:

• Resetting sum to zero loses current page
• Returning wrong comparison for count and students
• Incorrect loop indexing

5. You have books with pages [12, 34, 67, 90] and 2 students. Using binary search on answer, what is the minimum maximum pages that can be allocated to a student? Show the reasoning.

hard

A. 90

B. 113

C. 101

D. 67

Solution

1. Step 1: Set binary search boundaries

   Low = max single book pages = 90, High = sum = 12+34+67+90=203.

2. Step 2: Check mid and feasibility

   Mid = (90+203)/2=146. Check if allocation possible with max 146 pages per student: [12,34,67] sum=113 <=146, next student gets [90]. Possible with 2 students.

3. Step 3: Narrow search

   Try lower mid: (90+146)/2=118. Check allocation: [12,34,67] sum=113 <=118, next student [90]. Possible.

4. Step 4: Try even lower mid

   Mid = (90+118)/2=104. Allocation: [12,34,67] sum=113 >104, so split: [12,34] sum=46 <=104, next student [67,90] sum=157 >104 not possible.

5. Step 5: Increase low to 105 and continue

   Mid = (105+118)/2=111. Allocation: [12,34,67] sum=113 >111 no, split: [12,34] sum=46 <=111, next student [67,90] sum=157 >111 no.

6. Step 6: Mid = (112+118)/2=115

   Allocation: [12,34,67] sum=113 <=115 yes, next student [90]. Possible.

7. Step 7: Narrow down to minimum feasible

   Try mid= (112+114)/2=113 allocation possible, mid=112 allocation not possible.

8. Final Answer:

   113 -> Option B

9. Quick Check:

   Minimum max pages = 113 [OK]

Hint: Use binary search between max book and total pages [OK]

Common Mistakes:

• Choosing max single book pages as answer
• Not checking feasibility correctly
• Stopping binary search too early