Allocate Minimum Pages Binary Search on Answer in DSA Java - 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 a single book.

DSA Java

int low = [1]; // max pages in a single book

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

Apages.length

BArrays.stream(pages).max().getAsInt()

C0

DArrays.stream(pages).sum()

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using pages.length which is the number of books, not pages.

The low boundary is the maximum pages in any single book because no student can be assigned less than that.

2fill in blank

medium

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

DSA Java

int high = [1]; // sum of all pages

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

AArrays.stream(pages).sum()

Bpages.length

CArrays.stream(pages).max().getAsInt()

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using max pages instead of sum for high boundary.

Using pages.length which is count of books, not pages.

The high boundary is the total pages of all books, representing the maximum pages one student can read if assigned all books.

3fill in blank

hard

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

DSA Java

if (isPossible(pages, students, [1])) {

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

Amid

Bstudents

Chigh

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Passing low or high instead of mid to isPossible.

Passing students count instead of pages limit.

We check if the mid value can be a valid maximum pages allocation for the students.

4fill in blank

hard

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

DSA Java

if (isPossible(pages, students, mid)) {
    result = mid;
    [1] = mid - 1;
} else {
    [2] = mid + 1;
}

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

Alow

Bhigh

Cresult

Dstudents

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping low and high updates.

Updating result instead of low or high.

If mid is possible, we update result to mid and try for a lower value by setting high = mid - 1. Otherwise, we increase low to mid + 1.

5fill in blank

hard

Fill all three blanks to complete the isPossible function that checks if allocation is feasible with max pages limit.

DSA Java

int studentCount = 1;
int currentSum = 0;
for (int page : pages) {
    if (currentSum + page > [1]) {
        studentCount++;
        currentSum = page;
        if (studentCount > [2]) {
            return [3];
        }
    } else {
        currentSum += page;
    }
}
return true;

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

AmaxPages

Bstudents

Cfalse

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Returning true instead of false when students exceed limit.

Using wrong variable names for max pages or students.

We check if assigning pages exceeds maxPages, increase student count, and if students exceed limit, return false.

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