Allocate Minimum Pages Binary Search on Answer in DSA Java - Build from Scratch

Allocate Minimum Pages using Binary Search on Answer

📖 Scenario: You are a librarian who needs to allocate books to students. Each book has a certain number of pages. You want to distribute the books so that the student who gets the most pages has as few pages as possible. Books must be allocated in order and each student gets at least one book.

🎯 Goal: Build a program that finds the minimum possible maximum number of pages assigned to a student using binary search on the answer.

📋 What You'll Learn

Create an integer array called books with these exact values: 12, 34, 67, 90

Create an integer variable called students and set it to 2

Write a function called isPossible that takes books, students, and a maximum page limit mid and returns true if allocation is possible without exceeding mid

Use binary search between the maximum single book pages and the sum of all pages to find the minimum maximum pages

Print the minimum maximum pages after allocation

💡 Why This Matters

🌍 Real World

This problem models real-world scenarios like task distribution, workload balancing, and resource allocation where fairness and limits are important.

💼 Career

Understanding binary search on answer and allocation problems is useful for software engineers working on optimization, scheduling, and system design.

1

Create the books array

Create an integer array called books with these exact values: 12, 34, 67, 90

DSA Java

// Create the books array with the given values
// Your code here

Hint

Use int[] books = {12, 34, 67, 90}; to create the array.

2

Set the number of students

Create an integer variable called students and set it to 2

DSA Java

int[] books = {12, 34, 67, 90};
// Create the students variable and set it to 2
// Your code here

Hint

Use int students = 2; to create the variable.

3

Write the isPossible function

Write a function called isPossible that takes int[] books, int students, and int mid and returns true if it is possible to allocate books so that no student gets more than mid pages. Otherwise, return false.

DSA Java

int[] books = {12, 34, 67, 90};
int students = 2;

// Write the isPossible function below
// Your code here

Hint

Check if any single book has more pages than mid. If yes, return false. Otherwise, count how many students are needed if no student gets more than mid pages.

4

Find and print the minimum maximum pages

Use binary search between the maximum pages in books and the sum of all pages in books to find the minimum maximum pages that can be allocated to students. Print the result using System.out.println.

DSA Java

int[] books = {12, 34, 67, 90};
int students = 2;

public static boolean isPossible(int[] books, int students, int mid) {
    int studentCount = 1;
    int pageSum = 0;
    for (int pages : books) {
        if (pages > mid) {
            return false;
        }
        if (pageSum + pages > mid) {
            studentCount++;
            pageSum = pages;
            if (studentCount > students) {
                return false;
            }
        } else {
            pageSum += pages;
        }
    }
    return true;
}

// Use binary search to find minimum maximum pages and print the result
// Your code here

Hint

Use binary search between 0 and sum of all pages. Use isPossible to check feasibility. Print the final minimum maximum pages.

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