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

Recall & Review

beginner

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

To divide an array of book pages among students such that the maximum 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, allowing us to efficiently search this range.

intermediate

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

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

beginner

In the binary search approach, what do the 'low' and 'high' pointers represent?

'Low' is the maximum 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 problem 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 is the initial lower bound in binary search for Allocate Minimum Pages?

A Number of students

B Maximum pages in a single book

C Zero

D Sum of all pages

The minimum maximum pages cannot be less than the largest single book.

What does the 'isPossible' function return if allocation exceeds the number of students?

A False

B True

C Number of pages

D Number of books

If allocation requires more students than available, it returns false.

Which technique is used to find the minimum maximum pages?

A Binary search on answer

B Depth-first search

C Linear search

D Greedy sorting

Binary search on the answer range efficiently finds the minimum maximum pages.

If the mid value in binary search is feasible, what should we do next?

A Stop searching

B Search in higher half

C Search in lower half

D Increase number of students

We try to find a smaller maximum by searching the lower half.

What is the maximum number of students allowed in the problem?

A Equal to number of books

B Always 1

C Sum of pages

D Given input value

The number of students is given as input and is fixed.

Explain how binary search is applied to solve the Allocate Minimum Pages problem.

Think about searching for the best maximum pages value.

Describe the role of the 'isPossible' function in the allocation process.

It tests if a given max pages limit works.

Practice

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

easy

A. To maximize the total number of pages assigned

B. To find the average number of pages per student

C. To minimize the maximum number of pages assigned to any student

D. To assign pages randomly to students

Solution

1. Step 1: Understand the problem goal

   The problem wants to divide 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 any student -> Option C

4. Quick Check:

   Goal = minimize max pages per student [OK]

Hint: Focus on minimizing the max pages per student, not total pages [OK]

Common Mistakes:

• Confusing total pages with maximum pages per student
• Thinking binary search finds average pages
• Assuming random assignment solves the problem

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

easy

A. int low = max_page_count, high = sum_of_all_pages;

B. int low = 0, high = sum_of_all_pages;

C. int low = 0, high = max_page_count;

D. int low = sum_of_all_pages, high = max_page_count;

Solution

1. Step 1: Determine the search boundaries

   The minimum possible max pages is at least the largest single book (max_page_count), and the maximum possible is the sum of all pages.

2. Step 2: Set binary search range accordingly

   So, low = max_page_count and high = sum_of_all_pages is correct initialization.

3. Final Answer:

   int low = max_page_count, high = sum_of_all_pages; -> Option A

4. Quick Check:

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

Hint: Start low at max single book pages, high at total pages sum [OK]

Common Mistakes:

• Starting low at 0 instead of max book pages
• Swapping low and high values
• Using arbitrary values outside page counts

3. Given the array of pages [10, 20, 30, 40] and 2 students, what is the minimum maximum pages assigned to a student?

medium

A. 30

B. 50

C. 40

D. 60

Solution

1. Step 1: Check possible max pages assignments

   Try to split books into 2 parts so max pages per student is minimized. Possible splits: [10,20,30] and [40] max 60; [10,20] and [30,40] max 70; [10,20,30,40] all to one student max 100.

2. Step 2: Use binary search or reasoning

   Splitting as [10,20] and [30,40] gives max 70, but [10,20,30] and [40] gives max 60. Check if 50 is possible: [10,20] and [30,40] max 70 too high, but [10,20,30] is 60. So 50 is not possible because 40 > 50. So minimum max pages is 60.

3. Final Answer:

   60 -> Option D

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Try splitting books to minimize max pages per student [OK]

Common Mistakes:

• Choosing max single book as answer without checking splits
• Assuming equal split always works
• Ignoring that a student must get continuous books

4. Identify the error in this C code snippet for checking feasibility in Allocate Minimum Pages problem:

int isPossible(int arr[], int n, int m, int mid) {
    int studentCount = 1, pageSum = 0;
    for (int i = 0; i < n; i++) {
        if (pageSum + arr[i] <= mid) {
            pageSum += arr[i];
        } else {
            studentCount++;
            pageSum = arr[i];
        }
    }
    return studentCount <= m;
}

medium

A. It does not check if a single book has more pages than mid

B. It increments studentCount incorrectly

C. It should return studentCount >= m instead

D. It does not initialize pageSum to arr[0]

Solution

1. Step 1: Analyze feasibility check logic

   The function tries to assign books to students without exceeding mid pages per student.

2. Step 2: Identify missing check

   If a single book has pages more than mid, the function still tries to assign it, which is invalid. This case is not handled.

3. Final Answer:

   It does not check if a single book has more pages than mid -> Option A

4. Quick Check:

   Check single book pages <= mid [OK]

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

Common Mistakes:

• Ignoring single book size check
• Returning wrong comparison for studentCount
• Incorrect initialization of pageSum

5. You have books with pages [12, 34, 67, 90] and 2 students. Using binary search on answer, what is the minimum maximum pages a student can get? Explain the steps.

hard

A. 101

B. 113

C. 90

D. 67

Solution

1. Step 1: Set binary search range

   Minimum max pages = max single book = 90, maximum = sum = 12+34+67+90=203.

2. Step 2: Apply binary search and feasibility check

   Mid = (90+203)/2=146. Check if possible with max 146 pages: assign [12,34,67] sum=113 to student1, [90] to student2. Possible. Try lower mid.

3. Step 3: Narrow down mid

   Mid = (90+146)/2=118. Check with 118: assign [12,34,67] sum=113 student1, [90] student2. Possible. Try lower mid.

4. Step 4: Check mid=104

   Assign [12,34] sum=46 student1, [67,90] sum=157 student2 exceeds 104. Not possible.

5. Step 5: Check mid=111

   Assign [12,34,67] sum=113 >111 no. Assign [12,34] sum=46 student1, [67,90] sum=157 no. Not possible.

6. Step 6: Check mid=113

   Assign [12,34,67] sum=113 student1, [90] student2. Possible.

7. Step 7: Final minimum max pages

   Minimum max pages = 113.

8. Final Answer:

   113 -> Option B

9. Quick Check:

   Minimum max pages = 113 [OK]

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

Common Mistakes:

• Choosing max single book as answer without checking splits
• Not testing feasibility properly
• Confusing sum and max in binary search range