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

Time Complexity: Allocate Minimum Pages Binary Search on Answer

O(n x log m)

Understanding Time Complexity

We want to find how fast the binary search approach works to allocate minimum pages among 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: number[], students: number, maxPages: number): boolean {
  let requiredStudents = 1;
  let currentSum = 0;
  for (const pages of books) {
    if (pages > maxPages) return false;
    if (currentSum + pages > maxPages) {
      requiredStudents++;
      currentSum = pages;
      if (requiredStudents > students) return false;
    } else {
      currentSum += pages;
    }
  }
  return true;
}

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

This code uses binary search on the answer range to find the minimum maximum pages allocated to any student.

Identify Repeating Operations

Look at the loops and checks that repeat.

• Primary operation: The for-loop inside isPossible that checks if allocation is possible.
• How many times: This loop runs once for each book, so n times per check.
• Binary search loop: The while loop in allocateMinimumPages runs about log(sum of pages) times.

How Execution Grows With Input

Each binary search step checks all books once.

Input Size (n)	Approx. Operations
10 books	10 x log(total pages)
100 books	100 x log(total pages)
1000 books	1000 x log(total pages)

As books increase, operations grow linearly with n, multiplied by the logarithm of total pages.

Final Time Complexity

Time Complexity: O(n x log m)

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 alone makes this O(log n) time."

[OK] Correct: Each binary search step runs a full check over all books, so the total time depends on both the number of books and the binary search steps.

Interview Connect

Understanding this time complexity helps you explain how binary search can be used 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 assign pages randomly among students

B. To maximize the total number of pages assigned

C. To find the average number of pages per student

D. To minimize the maximum number of pages assigned to a student

Solution

1. Step 1: Understand the problem objective

   The problem asks to allocate books to students such that the maximum pages assigned to any student is minimized.

2. Step 2: Identify the binary search approach

   We use binary search on the range of possible maximum pages to find the minimum feasible maximum.

3. Final Answer:

   To minimize the maximum number of pages assigned to a student -> Option D

4. Quick Check:

   Goal = minimize max pages [OK]

Hint: Minimize max pages per student using binary search [OK]

Common Mistakes:

• Trying to maximize pages instead of minimizing
• Ignoring the maximum pages constraint
• Confusing total pages with max pages per student

2. Which of the following is the correct TypeScript syntax to declare the binary search boundaries for the Allocate Minimum Pages problem given an array pages?

easy

A. let low = pages.length; let high = Math.max(...pages);

B. let low = Math.min(...pages); let high = pages.length;

C. let low = 0; let high = pages.reduce((a, b) => a + b, 0);

D. let low = 0; let high = Math.min(...pages);

Solution

1. Step 1: Determine low and high boundaries

   Low is minimum possible max pages (0 or max single book pages), high is sum of all pages (max total pages).

2. Step 2: Check correct syntax for sum

   Using reduce sums all pages correctly for high boundary.

3. Final Answer:

   let low = 0; let high = pages.reduce((a, b) => a + b, 0); -> Option C

4. Quick Check:

   low = 0, high = sum pages [OK]

Hint: Sum array for high, zero or max page for low [OK]

Common Mistakes:

• Setting low as pages.length instead of page count
• Using min pages as high boundary
• Incorrect sum calculation syntax

3. Consider the following TypeScript code snippet for checking feasibility in Allocate Minimum Pages problem:

function isPossible(pages: number[], students: number, maxPages: number): boolean {
  let requiredStudents = 1;
  let currentSum = 0;
  for (let page of pages) {
    if (page > maxPages) return false;
    if (currentSum + page > maxPages) {
      requiredStudents++;
      currentSum = page;
    } else {
      currentSum += page;
    }
  }
  return requiredStudents <= students;
}

What is the output of isPossible([10, 20, 30, 40], 2, 50)?

medium

A. false

B. SyntaxError

C. true

D. RuntimeError

Solution

1. Step 1: Trace allocation with maxPages = 50

   Assign pages: 10+20=30 (still <=50), add 30 would exceed 50, so new student starts with 30, then 40 exceeds 50 if added, so new student starts with 40.

2. Step 2: Count required students

   We used 3 students but only 2 allowed, so requiredStudents=3 > 2 means return false. But check carefully: 10+20=30, next 30 alone fits in second student, 40 alone needs third student.

3. Final Answer:

   false -> Option A

4. Quick Check:

   Required students = 3 > 2, so false [OK]

Hint: Count students needed, compare with allowed [OK]

Common Mistakes:

• Ignoring that a single book can exceed maxPages
• Miscounting students when sum exceeds maxPages
• Assuming sum of all pages fits in one student

4. The following TypeScript function attempts to find the minimum maximum pages but has a bug. Identify the bug.

function findMinPages(pages: number[], students: number): number {
  let low = Math.min(...pages);
  let high = pages.reduce((a, b) => a + b, 0);
  let result = high;
  while (low <= high) {
    let mid = Math.floor((low + high) / 2);
    if (isPossible(pages, students, mid)) {
      result = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  return result;
}

medium

A. High should be min page, not sum of pages

B. Low should start from max page, not min page

C. Mid calculation causes infinite loop

D. Result variable is not updated

Solution

1. Step 1: Analyze low boundary initialization

   Low should be the maximum single book pages, because no student can get less than the largest book.

2. Step 2: Identify impact of wrong low

   Starting low at minimum page can cause incorrect feasibility checks and longer search.

3. Final Answer:

   Low should start from max page, not min page -> Option B

4. Quick Check:

   Low = max page [OK]

Hint: Low boundary = max single book pages [OK]

Common Mistakes:

• Setting low to min page instead of max page
• Confusing high boundary with low
• Not updating result inside loop

5. Given books with pages [12, 34, 67, 90] and 3 students, what is the minimum maximum pages allocated to any student using binary search on answer?

hard

A. 90

B. 113

C. 67

D. 34

Solution

1. Step 1: Set search boundaries

   Low = max pages = 90, High = sum pages = 12+34+67+90=203.

2. Step 2: Binary search feasibility checks

   Mid = (90+203)//2=146. Check if possible with maxPages=146: allocate [12+34+67=113] to student1, 90 to student2, only 2 students needed <=3, possible. So high=145.

3. Step 3: Narrow down to minimum max pages

   Continuing: mid=103 possible [12+34=46,67,90] 3 students. mid=96 possible. mid=92 possible. mid=90: [12+34=46], [67], [90] all <=90, 3 students <=3 feasible, high=89. Loop ends with result=90.

4. Step 4: Re-evaluate minimum max pages

   Check if 113 is feasible with 3 students: [12+34+67=113], [90] 2 students needed <=3, so 113 is feasible but not minimum.

5. Step 5: Check if 90 is feasible

   Allocations: [12+34=46], [67], [90] 3 students needed <=3, feasible.

6. Final Answer:

   90 -> Option A

7. Quick Check:

   Minimum max pages = 90 [OK]

Hint: Binary search between max page and sum pages [OK]

Common Mistakes:

• Misallocating at max=90 thinking 4 students needed
• Miscounting students during allocation
• Confusing with 2 students (113)