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

The Big Idea

What if you could find the fairest way to split tasks without trying every possibility?

The Scenario

Imagine you have a stack of books with different numbers of pages. You want to divide these books among a few friends so that the friend who gets the most pages has as few pages as possible. Doing this by guessing and checking every possible way manually would take forever!

The Problem

Trying every way to split the books manually is slow and confusing. You might miss better ways or spend hours checking all possibilities. It's easy to make mistakes and hard to know if you found the best answer.

The Solution

Using binary search on the answer lets us quickly find the smallest maximum pages any friend has to read. Instead of guessing blindly, we check if a guess works and then narrow down the range until we find the best solution fast and without errors.

Before vs After

✗ Before

function allocateBooks(books: number[], friends: number): number {
  // Try all splits manually - very slow
  let minMax = Infinity;
  // Complex nested loops to check all divisions
  return minMax;
}

✓ After

function allocateBooks(books: number[], friends: number): number {
  let low = Math.max(...books);
  let high = books.reduce((a, b) => a + b, 0);
  while (low < high) {
    let mid = Math.floor((low + high) / 2);
    if (canAllocate(books, friends, mid)) high = mid;
    else low = mid + 1;
  }
  return low;
}

What It Enables

This method makes it easy to find the best way to split work or resources evenly and efficiently, even when there are many options.

Real Life Example

Imagine dividing chapters of a textbook among study group members so no one has too much to read. Using this method, you quickly find the fairest way to split the chapters.

Key Takeaways

Manual checking of all splits is slow and error-prone.

Binary search on the answer narrows down the best maximum pages quickly.

This approach helps balance workloads or resources efficiently.

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)