Allocate Minimum Pages Binary Search on Answer in DSA C# - Time & Space Complexity

Time Complexity: Allocate Minimum Pages Binary Search on Answer

O(n * log S)

Understanding Time Complexity

We want to find how long it takes to find the minimum maximum pages allocated using binary search on the answer space.

How does the number of steps grow as the number of books and pages increase?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

int AllocateMinimumPages(int[] books, int students) {
    int low = 0, high = 0;
    foreach (int pages in books) high += pages;
    int result = high;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (IsPossible(books, students, mid)) {
            result = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return result;
}
    

This code tries to find the minimum maximum pages a student can get by using binary search on the possible answer range.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The IsPossible function scans all books once to check feasibility.
• How many times: This check runs inside a binary search loop over the answer range, which repeats about log(sum of pages) times.

How Execution Grows With Input

As the number of books (n) grows, the scan inside IsPossible grows linearly. The binary search cuts the answer range roughly in half each time.

Input Size (n)	Approx. Operations
10 books	About 10 * log(total pages) checks
100 books	About 100 * log(total pages) checks
1000 books	About 1000 * log(total pages) checks

Pattern observation: Operations grow linearly with number of books and logarithmically with total pages.

Final Time Complexity

Time Complexity: O(n * log S)

This means the time grows linearly with the number of books and logarithmically with the sum of all pages.

Common Mistake

[X] Wrong: "The binary search runs in O(log n) where n is the number of books."

[OK] Correct: The binary search is on the range of possible page sums, not the number of books. The number of books affects the linear scan inside each check.

Interview Connect

Understanding this time complexity helps you explain how binary search can be applied beyond sorted arrays, showing problem-solving skills that interviewers appreciate.

Self-Check

"What if we replaced the linear scan in IsPossible with a more complex check? How would that affect the time complexity?"

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