Allocate Minimum Pages Binary Search on Answer in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the low boundary for binary search as the maximum pages in any single book.

DSA C#

int low = [1];

Drag options to blanks, or click blank then click option'

Abooks.Min()

Bbooks.Max()

Cbooks.Sum()

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to 0 or sum of all pages instead of max pages.

Using Min() instead of Max() for low boundary.

The minimum number of pages allocated to a student cannot be less than the maximum pages in a single book.

2fill in blank

medium

Complete the code to initialize the high boundary for binary search as the sum of all pages in the books.

DSA C#

int high = [1];

Drag options to blanks, or click blank then click option'

Abooks.Min()

Bbooks.Max()

C0

Dbooks.Sum()

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to max pages instead of sum.

Using Min() or 0 for high boundary.

The maximum number of pages allocated to a student can be the sum of all pages if one student reads all books.

3fill in blank

hard

Fix the error in the condition to check if the current mid value can be a valid maximum page allocation.

DSA C#

if (IsPossible(books, students, [1]))

Drag options to blanks, or click blank then click option'

Ahigh

Blow

Cmid

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using low or high instead of mid in the feasibility check.

Using sum which is not defined in this context.

We check feasibility using the mid value as the candidate maximum pages.

4fill in blank

hard

Fill both blanks to update the binary search boundaries correctly after checking feasibility.

DSA C#

if (IsPossible(books, students, mid)) {
    result = mid;
    [1] = mid [2] 1;
} else {
    low = mid + 1;
}

Drag options to blanks, or click blank then click option'

Alow

Bhigh

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Updating low instead of high when mid is possible.

Using '+' instead of '-' in the update.

If mid is possible, we try to find a smaller maximum by moving high down to mid - 1.

5fill in blank

hard

Fill all three blanks to complete the IsPossible function that checks if books can be allocated within maxPages to given students.

DSA C#

bool IsPossible(int[] books, int students, int maxPages) {
    int count = 1;
    int currentSum = 0;
    foreach (int pages in books) {
        if (pages > maxPages) return false;
        if (currentSum + pages > [1]) {
            count++;
            currentSum = pages;
            if (count > [2]) return false;
        } else {
            currentSum += [3];
        }
    }
    return true;
}

Drag options to blanks, or click blank then click option'

AmaxPages

Bstudents

Cpages

DcurrentSum

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variables in comparisons or additions.

Not resetting currentSum correctly after incrementing count.

We check if adding pages exceeds maxPages, increment student count, and add pages to currentSum otherwise.

Practice

1. What is the main goal of the Allocate Minimum Pages problem when 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 to students

Solution

1. Step 1: Understand the problem goal

   The problem wants to split book pages among students so that the student with the most pages has as few pages as possible.

2. Step 2: Identify the binary search target

   Binary search is used to find the minimum possible maximum pages assigned to any 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 any student gets [OK]

Common Mistakes:

• Thinking it maximizes total pages
• Confusing average with maximum
• Assuming random assignment

2. Which of the following is the correct way to initialize the binary search range for the minimum pages problem in C#?

easy

A. int low = 0, high = pages.Max();

B. int low = 0, high = pages.Sum();

C. int low = pages.Max(), high = pages.Sum();

D. int low = pages.Sum(), high = pages.Max();

Solution

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

   The minimum max pages must be at least the largest single book (pages.Max()), and the maximum is the sum of all pages (pages.Sum()).

2. Step 2: Set binary search range accordingly

   Set low = pages.Max() and high = pages.Sum() to cover all possible max page values.

3. Final Answer:

   int low = pages.Max(), high = pages.Sum(); -> Option C

4. Quick Check:

   low = max page, high = sum pages [OK]

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

Common Mistakes:

• Starting low at 0 (too low)
• Swapping low and high
• Using max as low and max as high

3. Given the array pages = [10, 20, 30, 40] and students = 2, what is the minimum maximum pages assigned after running the binary search solution?

medium

A. 70

B. 50

C. 40

D. 60

Solution

1. Step 1: Understand the problem with given input

   We want to split [10,20,30,40] into 2 students minimizing the max pages assigned.

2. Step 2: Check possible splits and their max pages

   Possible splits: [10,20,30] & [40] max=60; [10,20] & [30,40] max=70; [10] & [20,30,40] max=90; [10,20,30,40] & [] invalid. The minimum max is 60 or 50? Actually, [10,20,30] & [40] max=60, but [10,20] & [30,40] max=70. Check if 50 is possible: [10,20] & [30,40] max=70 >50 no; [10,20,30] & [40] max=60 >50 no; So 50 is not possible. But 60 is possible. So minimum max pages is 60.

3. Final Answer:

   60 -> Option D

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Try splitting pages to minimize largest sum per student [OK]

Common Mistakes:

• Choosing max single book (40) as answer
• Choosing total sum (100) as answer
• Not checking all splits

4. Consider this C# code snippet for checking feasibility in the minimum pages problem:

bool IsPossible(int[] pages, int students, int mid) {
    int count = 1, sum = 0;
    foreach (int page in pages) {
        if (sum + page > mid) {
            count++;
            sum = page;
        } else {
            sum += page;
        }
    }
    return count <= students;
}

What is the bug in this code?

medium

A. It should return count >= students instead of count <= students

B. It does not handle the case when a single book has more pages than mid

C. It increments count incorrectly inside the loop

D. It resets sum to zero instead of page after incrementing count

Solution

1. Step 1: Analyze the feasibility check logic

   The function checks if pages can be allocated so no student gets more than mid pages.

2. Step 2: Identify missing check for single book pages

   If any single book has pages > mid, allocation is impossible, but code does not check this, causing wrong results.

3. Final Answer:

   It does not handle the case when a single book has more pages than mid -> Option B

4. Quick Check:

   Check single book > mid to avoid wrong feasibility [OK]

Hint: Always check if any single book exceeds mid [OK]

Common Mistakes:

• Ignoring single book size check
• Wrong comparison operator in return
• Resetting sum incorrectly

5. You have pages = [12, 34, 67, 90] and students = 2. Using binary search on answer, what is the minimum maximum pages that can be allocated? Also, explain why the answer is not the maximum single book pages.

hard

A. 113, because splitting as [12,34,67] and [90] minimizes max pages

B. 90, because the largest single book sets the minimum max pages

C. 67, because the average pages per student is less than 70

D. 123, because total pages must be divided equally

Solution

1. Step 1: Calculate total and max pages

   Total pages = 12+34+67+90 = 203; max single book = 90.

2. Step 2: Check possible splits for 2 students

   Splits: [12,34,67] & [90] max=113; [12,34] & [67,90] max=157; [12] & [34,67,90] max=191. Minimum max pages is 113.

3. Step 3: Explain why answer is not max single book

   Because 90 alone is too small to cover all pages in 2 students without exceeding max pages, the minimum max pages must be higher than 90.

4. Final Answer:

   113, because splitting as [12,34,67] and [90] minimizes max pages -> Option A

5. Quick Check:

   Minimum max pages = 113 > max single book 90 [OK]

Hint: Check splits; answer ≥ max single book, often higher [OK]

Common Mistakes:

• Choosing max single book as answer
• Ignoring total pages sum
• Assuming equal division always works