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DSA Javascriptprogramming~5 mins

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

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Time Complexity: Allocate Minimum Pages Binary Search on Answer
O(n x log S)
Understanding Time Complexity

We want to find how fast the binary search approach works when allocating minimum pages to students.

How does the number of operations grow as the number of books or pages increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


function isPossible(books, students, maxPages) {
  let requiredStudents = 1;
  let currentSum = 0;
  for (let pages of books) {
    if (pages > maxPages) return false;
    if (currentSum + pages > maxPages) {
      requiredStudents++;
      currentSum = pages;
    } else {
      currentSum += pages;
    }
  }
  return requiredStudents <= students;
}

function allocateMinimumPages(books, students) {
  let start = 0;
  let end = books.reduce((a, b) => a + b, 0);
  let result = end;
  while (start <= end) {
    let mid = Math.floor((start + end) / 2);
    if (isPossible(books, students, mid)) {
      result = mid;
      end = mid - 1;
    } else {
      start = mid + 1;
    }
  }
  return result;
}

This code tries to find the smallest maximum number of pages assigned to any student by using binary search on the answer range.

Identify Repeating Operations
  • Primary operation: The while loop performing binary search on the range of possible page sums.
  • How many times: The binary search loop runs about log(sum of pages) times.
  • Secondary operation: The for loop inside isPossible that checks if allocation is possible for a given maxPages.
  • How many times: This loop runs once per binary search iteration, iterating over all books, so n times per check.
How Execution Grows With Input

As the number of books n grows, the inner check runs n times each binary search step. The binary search runs about log(S) times, where S is the sum of all pages.

Input Size (n)Approx. Operations
10 (books)10 x log(S) (e.g., 10 x 7 = 70)
100 (books)100 x log(S) (e.g., 100 x 14 = 1400)
1000 (books)1000 x log(S) (e.g., 1000 x 20 = 20000)

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

Final Time Complexity

Time Complexity: O(n x log S)

This means the time grows linearly with the number of books and logarithmically with the total pages, making it efficient for large inputs.

Common Mistake

[X] Wrong: "The binary search runs in O(log n) because it depends on the number of books."

[OK] Correct: The binary search runs on the range of possible page sums, not the number of books, so it depends on the sum of pages, not just n.

Interview Connect

Understanding this time complexity helps you explain how binary search can be applied beyond simple arrays, showing your problem-solving skills with searching answer spaces efficiently.

Self-Check

What if we replaced the linear check in isPossible with a more complex operation? How would that affect the overall time complexity?

Practice

(1/5)
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 student with the least pages
D. To sort the pages in ascending order

Solution

  1. Step 1: Understand the problem goal

    The problem aims to divide books 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 on the maximum pages assigned to any student to find the minimum possible maximum.

  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 sorting with allocation
  • Assuming equal pages per student
2. Which of the following is the correct way to initialize the binary search boundaries for the Allocate Minimum Pages problem in JavaScript?
easy
A. let low = 0, high = sum of all pages;
B. let low = max pages in a single book, high = sum of all pages;
C. let low = 0, high = max pages in a single book;
D. let low = sum of all pages, high = max pages in a single book;

Solution

  1. Step 1: Determine low boundary

    Low should be at least the maximum pages in a single book because no student can get less than that.

  2. Step 2: Determine high boundary

    High is the sum of all pages, representing the case where one student reads all books.

  3. Final Answer:

    let low = max pages in a single book, high = sum of all pages; -> Option B

  4. Quick Check:

    low = max book pages, high = total pages [OK]
Hint: Low = max single book pages, High = total pages sum [OK]
Common Mistakes:
  • Setting low to 0 which is too low
  • Swapping low and high values
  • Using arbitrary boundaries
3. Given books with pages [10, 20, 30, 40] and 2 students, what is the minimum maximum pages assigned to a student using binary search on answer?
medium
A. 40
B. 50
C. 60
D. 70

Solution

  1. Step 1: Calculate boundaries

    Max single book pages = 40, sum = 100. So low=40, high=100.

  2. Step 2: Apply binary search and check feasibility

    Mid = 70: Can we allocate so max pages ≤ 70? Yes: [10,20,30] and [40]. Try lower mid.

  3. Step 3: Check mid = 55

    Allocate: [10,20,30] sum=60 > 55, so no. Increase low to 56.

  4. Step 4: Check mid = 60

    Allocate: [10,20,30] sum=60 ≤ 60, [40] next. Possible, try lower.

  5. Step 5: Check mid = 50

    Allocate: [10,20] sum=30 ≤ 50, next [30,40] sum=70 > 50 no. So 50 not possible.

  6. Final Answer:

    60 -> Option C

  7. Quick Check:

    Minimum max pages = 60 [OK]
Hint: Check mid feasibility by simulating allocation [OK]
Common Mistakes:
  • Choosing sum/2 without checking feasibility
  • Ignoring max single book pages
  • Misallocating books to students
4. Identify the bug in this JavaScript snippet for checking feasibility in Allocate Minimum Pages problem: 
function isPossible(books, students, maxPages) {
  let count = 1, sum = 0;
  for (let i = 0; i < books.length; i++) {
    sum += books[i];
    if (sum > maxPages) {
      count++;
      sum = books[i];
    }
  }
  return count <= students;
}
medium
A. It does not check if a single book has pages more than maxPages
B. It does not reset sum correctly after exceeding maxPages
C. It increments count even when sum equals maxPages
D. It uses a for loop instead of while loop

Solution

  1. Step 1: Analyze sum accumulation

    The code adds book pages to sum and resets sum when sum > maxPages, increasing count.

  2. Step 2: Check for single book pages exceeding maxPages

    There is no check if any single book's pages exceed maxPages, which makes allocation impossible.

  3. Final Answer:

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

  4. Quick Check:

    Missing single book page check [OK]
Hint: Always check single book pages ≤ maxPages first [OK]
Common Mistakes:
  • Ignoring single book page size check
  • Resetting sum incorrectly
  • Confusing loop types
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? Also, explain why allocating [12, 34, 67] to one student and [90] to another is optimal.
hard
A. Minimum max pages is 101; allocation is not optimal because 67+34=101 < 113
B. Minimum max pages is 90; allocation is optimal
C. Minimum max pages is 113; allocation is not optimal because 12+34+67=113 is too high
D. Minimum max pages is 113; allocation is optimal

Solution

  1. Step 1: Calculate boundaries and try allocations

    Max single book = 90, sum = 203. Low=90, High=203.

  2. Step 2: Check mid = 146

    Allocate: [12,34,67] sum=113 ≤146, [90] sum=90 ≤146. Possible, try lower.

  3. Step 3: Check mid = 113

    Allocate: [12,34,67] sum=113 ≤113, [90] sum=90 ≤113. Possible, try lower.

  4. Step 4: Check mid = 101

    Allocate: [12,34,67] sum=113 >101 no. Try [12,34] sum=46 ≤101, next [67,90] sum=157 >101 no. Not possible.

  5. Step 5: Conclusion

    Minimum max pages = 113. Allocation [12,34,67] and [90] achieves max(113,90)=113, which is the minimum possible maximum, so it is optimal.

  6. Final Answer:

    Minimum max pages is 113; allocation is optimal -> Option D

  7. Quick Check:

    Minimum max pages = 113, allocation optimal [OK]
Hint: Try mid values and simulate allocation carefully [OK]
Common Mistakes:
  • Assuming 90 is minimum max pages
  • Ignoring sum of allocated books
  • Not testing all allocation splits
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