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

Time Complexity: Allocate Minimum Pages Binary Search on Answer

O(n * log m)

Understanding Time Complexity

We want to find how fast the binary search approach works to allocate minimum pages among students.

Specifically, how the number of operations grows as the number of books or pages increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public boolean isPossible(int[] books, int students, int mid) {
    int requiredStudents = 1, currentSum = 0;
    for (int pages : books) {
        if (pages > mid) return false;
        if (currentSum + pages > mid) {
            requiredStudents++;
            currentSum = pages;
        } else {
            currentSum += pages;
        }
    }
    return requiredStudents <= students;
}

public int allocatePages(int[] books, int students) {
    int start = 0, end = 0;
    for (int pages : books) end += pages;
    int result = end;
    while (start <= end) {
        int mid = start + (end - start) / 2;
        if (isPossible(books, students, mid)) {
            result = mid;
            end = mid - 1;
        } else {
            start = mid + 1;
        }
    }
    return result;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop inside isPossible that checks if allocation is possible.
• How many times: This check runs for each mid value in the binary search, which runs about log of the sum of pages times.

How Execution Grows With Input

As the number of books (n) increases, the for-loop inside isPossible runs n times for each binary search step.

Input Size (n)	Approx. Operations
10	10 * log(sum of pages)
100	100 * log(sum of pages)
1000	1000 * log(sum of pages)

Pattern observation: The operations grow linearly with the number of books and logarithmically with the total pages.

Final Time Complexity

Time Complexity: O(n * log m)

This means the time grows linearly with the number of books and logarithmically with the total pages to allocate.

Common Mistake

[X] Wrong: "The binary search alone makes this O(log n) time."

[OK] Correct: Each binary search step runs a full check over all books, so the loop over books dominates and multiplies the log factor.

Interview Connect

Understanding this time complexity helps you explain how binary search can be applied on answers, a common pattern in allocation and partition problems.

Self-Check

"What if we used a linear search instead of binary search on the answer range? How would the time complexity change?"

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