Why Allocate Minimum Pages Binary Search on Answer in DSA C?

The Big Idea

What if you could find the fairest way to share work without trying every option?

The Scenario

Imagine you have a stack of books with different numbers of pages. You want to divide these books among a few students so that the student who gets the most pages has as few pages as possible. If you try to do this by guessing and checking every possible way manually, it will take forever and be very confusing.

The Problem

Manually trying all ways to split books is slow and tiring. You might miss better ways or make mistakes. It's like trying to find a needle in a haystack by looking at every straw one by one.

The Solution

Using binary search on the answer lets you quickly guess a maximum pages limit and check if the books can be divided under that limit. By adjusting guesses smartly, you find the best minimum maximum pages without checking every possibility.

Before vs After

✗ Before

int minPages(int books[], int n, int students) {
    int result = INT_MAX;
    for (int maxPages = 0; maxPages <= sumOfPages(books, n); maxPages++) {
        if (canAllocate(books, n, students, maxPages)) {
            result = maxPages;
            break;
        }
    }
    return result;
}

✓ After

int minPages(int books[], int n, int students) {
    int low = maxPage(books, n);
    int high = sumOfPages(books, n);
    int result = high;
    while (low <= high) {
        int mid = (low + high) / 2;
        if (canAllocate(books, n, students, mid)) {
            result = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return result;
}

What It Enables

This method lets you quickly find the fairest way to split work or resources, saving time and effort.

Real Life Example

Dividing chapters of a textbook among students so that no one has too much reading, or assigning tasks to workers so the busiest worker has the least load possible.

Key Takeaways

Manual checking is slow and error-prone.

Binary search on answer guesses limits smartly to find the best solution fast.

Useful for fair division problems where you minimize the maximum load.

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