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

Allocate Minimum Pages using Binary Search on Answer

📖 Scenario: Imagine you are a librarian who needs to allocate books to students. Each student must get consecutive books, and you want to minimize the maximum number of pages assigned to any student.This is a common problem in resource allocation where you want to balance the load fairly.

🎯 Goal: You will write a program to find the minimum possible maximum number of pages that can be assigned to a student using binary search on the answer.

📋 What You'll Learn

Create a list called books with the exact values [100, 200, 300, 400]

Create a variable called students and set it to 2

Write a function called is_possible that takes mid and checks if allocation is possible with max pages mid

Use binary search between low and high to find the minimum max pages

Print the final minimum maximum pages

💡 Why This Matters

🌍 Real World

This technique is useful in scheduling tasks, dividing workloads, and resource allocation where fairness and limits matter.

💼 Career

Binary search on answer is a common pattern in coding interviews and software engineering roles involving optimization.

1

Create the list of books

Create a list called books with these exact values: [100, 200, 300, 400]

DSA Python

# Create the list of books
# Your code here

Hint

Use square brackets and separate numbers by commas.

2

Set the number of students

Create a variable called students and set it to 2

DSA Python

books = [100, 200, 300, 400]
# Set the number of students
# Your code here

Hint

Just assign the number 2 to the variable students.

3

Write the function to check allocation possibility

Write a function called is_possible that takes a parameter mid and returns True if it is possible to allocate books to students such that no student reads more than mid pages. Use variables current_sum and count_students inside the function.

DSA Python

books = [100, 200, 300, 400]
students = 2

# Define the is_possible function
# Your code here

Hint

Check if adding a book exceeds mid. If yes, allocate to next student.

4

Find and print the minimum maximum pages

Use binary search with variables low set to the maximum pages in books and high set to the sum of books. Use a while loop to find the minimum maximum pages. Print the final answer using print(min_pages).

DSA Python

books = [100, 200, 300, 400]
students = 2

def is_possible(mid):
    current_sum = 0
    count_students = 1
    for pages in books:
        if pages > mid:
            return False
        if current_sum + pages > mid:
            count_students += 1
            current_sum = pages
            if count_students > students:
                return False
        else:
            current_sum += pages
    return True

# Use binary search to find minimum max pages
# Your code here

Hint

Start low at the largest book pages and high at total pages. Narrow down using binary search.

Practice

1. What is the main goal of the Allocate Minimum Pages problem when using binary search on the answer?

easy

A. To find the minimum maximum number of pages assigned to a student

B. To maximize the total pages assigned to all students

C. To find the maximum number of students that can be assigned books

D. To minimize the total number of books assigned

Solution

1. Step 1: Understand the problem goal

   The problem aims to allocate books to students such that the maximum pages assigned to any student is minimized.

2. Step 2: Clarify the binary search target

   Binary search is used on the range of possible maximum pages to find the smallest feasible maximum.

3. Final Answer:

   To find the minimum maximum number of pages assigned to a student -> Option A

4. Quick Check:

   Minimum max pages = goal [OK]

Hint: Focus on minimizing the largest pages any student gets [OK]

Common Mistakes:

• Confusing total pages with maximum pages per student
• Trying to maximize pages instead of minimizing
• Ignoring the binary search on answer approach

2. Which of the following is the correct way to initialize the binary search boundaries for the Allocate Minimum Pages problem given an array pages of book pages?

easy

A. low = sum(pages), high = max(pages)

B. low = max(pages), high = sum(pages)

C. low = 0, high = max(pages)

D. low = 0, high = sum(pages) * 2

Solution

1. Step 1: Determine minimum and maximum possible max pages

   The minimum max pages is at least the largest single book (max of pages), and the maximum is the sum of all pages if one student gets all books.

2. Step 2: Set binary search boundaries accordingly

   Set low = max(pages) and high = sum(pages) to cover all feasible max page values.

3. Final Answer:

   low = max(pages), high = sum(pages) -> Option B

4. Quick Check:

   Boundaries = max and sum [OK]

Hint: Low starts at max single book, high at total pages sum [OK]

Common Mistakes:

• Setting low to 0 which is invalid
• Swapping low and high values
• Using arbitrary high value instead of sum

3. What will be the output of the following code snippet?

def is_possible(pages, students, mid):
    count = 1
    total = 0
    for p in pages:
        if total + p > mid:
            count += 1
            total = p
            if count > students:
                return False
        else:
            total += p
    return True

pages = [10, 20, 30, 40]
students = 2
mid = 50
print(is_possible(pages, students, mid))

medium

A. True

B. None

C. SyntaxError

D. False

Solution

1. Step 1: Trace the allocation with mid = 50

   pages=[10,20,30,40], students=2: count=1, total=0 - p=10: total=10 <=50 - p=20: total=30 <=50 - p=30: 30+30=60 >50 -> count=2, total=30 - p=40: 30+40=70 >50 -> count=3 >2 -> return False

2. Step 2: Check if count exceeds students

   Count reaches 3 > 2, so return False.

3. Final Answer:

   False -> Option D

4. Quick Check:

   Allocation not possible with mid=50 = False [OK]

Hint: Check if books fit in mid pages per student [OK]

Common Mistakes:

• Miscounting students needed
• Adding pages incorrectly
• Confusing > and >= in condition

4. Identify the bug in the following code snippet for the Allocate Minimum Pages problem:

def allocate_pages(pages, students):
    low = max(pages)
    high = sum(pages)
    result = 0
    while low <= high:
        mid = (low + high) // 2
        if is_possible(pages, students, mid):
            result = mid
            low = mid + 1
        else:
            high = mid - 1
    return result

medium

A. The high pointer should move to mid - 1 when allocation is possible

B. The low pointer should move to mid + 1 when allocation is possible

C. The low pointer should move to mid when allocation is possible

D. The low pointer should move to mid - 1 when allocation is not possible

Solution

1. Step 1: Understand binary search update rules

   If allocation is possible at mid, we try to find smaller max pages by moving high to mid - 1. If not possible, we increase low to mid + 1.

2. Step 2: Identify incorrect pointer update

   The code incorrectly moves low to mid + 1 when allocation is possible instead of high to mid - 1, which searches larger values instead of smaller.

3. Final Answer:

   The high pointer should move to mid - 1 when allocation is possible -> Option A

4. Quick Check:

   Move high down when allocation succeeds [OK]

Hint: Move low up only when allocation fails, else move high down [OK]

Common Mistakes:

• Swapping low and high updates
• Not updating pointers correctly causing infinite loop
• Returning wrong result due to pointer errors

5. Given the books with pages [12, 34, 67, 90] and 3 students, what is the minimum possible maximum number of pages assigned to a student using binary search on answer?

hard

A. 67

B. 113

C. 90

D. 34

Solution

1. Step 1: Set binary search boundaries

   low = max(pages) = 90, high = sum(pages) = 203.

2. Step 2: Check feasibility at mid values

   Try mid = (90+203)//2 = 146: count=1, total=0 - p=12: total=12 <=146 - p=34: total=46 <=146 - p=67: total=113 <=146 - p=90: 113+90=203 >146 -> count=2, total=90 count=2 <=3, possible -> try smaller max Try mid = (90+145)//2 = 117: count=1, total=0 - p=12: total=12 <=117 - p=34: total=46 <=117 - p=67: total=113 <=117 - p=90: 113+90=203 >117 -> count=2, total=90 count=2 <=3, possible -> try smaller max Try mid = (90+116)//2 = 103: count=1, total=0 - p=12: total=12 <=103 - p=34: total=46 <=103 - p=67: 46+67=113 >103 -> count=2, total=67 - p=90: 67+90=157 >103 -> count=3, total=90 count=3 <=3, possible -> try smaller max Try mid = (90+102)//2 = 96: count=1, total=0 - p=12: total=12 <=96 - p=34: total=46 <=96 - p=67: 46+67=113 >96 -> count=2, total=67 - p=90: 67+90=157 >96 -> count=3, total=90 count=3 <=3, possible -> try smaller max Try mid = (90+95)//2 = 92: count=1, total=0 - p=12: total=12 <=92 - p=34: total=46 <=92 - p=67: 46+67=113 >92 -> count=2, total=67 - p=90: 67+90=157 >92 -> count=3, total=90 count=3 <=3, possible -> try smaller max Try mid = (90+91)//2 = 90: count=1, total=0 - p=12: total=12 <=90 - p=34: total=46 <=90 - p=67: 46+67=113 >90 -> count=2, total=67 - p=90: 67+90=157 >90 -> count=3, total=90 count=3 <=3, possible Try mid = (90+89)//2 = 89 (low > high, stop) Minimum max pages = 90

3. Final Answer:

   90 -> Option C

4. Quick Check:

   Minimum max pages = 90 [OK]

Hint: Binary search between max single book and total pages sum [OK]

Common Mistakes:

• Choosing max single book as answer without checking feasibility
• Not updating low and high correctly during binary search
• Misallocating books to students during checks