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

The Big Idea

What if you could find the perfect way to share work 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. If you try to do this by guessing and checking manually, it can take forever!

The Problem

Trying every possible way to split the books manually is slow and confusing. You might miss the best way or spend hours checking each option. It's easy to make mistakes and waste time.

The Solution

Using binary search on the answer lets you quickly find the smallest maximum number of pages any friend has to read. Instead of guessing blindly, you check smartly by narrowing down the possible answers step-by-step until you find the best one.

Before vs After

✗ Before

function allocateBooks(books, friends) {
  // Try all splits manually
  // Very slow and complex
}

✓ After

function allocateBooks(books, friends) {
  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;
}

function canAllocate(books, friends, maxPages) {
  let requiredFriends = 1;
  let currentSum = 0;
  for (let pages of books) {
    if (currentSum + pages <= maxPages) {
      currentSum += pages;
    } else {
      requiredFriends++;
      currentSum = pages;
      if (requiredFriends > friends) return false;
    }
  }
  return true;
}

What It Enables

This method lets you solve complex allocation problems quickly and correctly, even with many books and friends.

Real Life Example

Think about dividing tasks among team members so no one is overloaded. Using this approach helps balance work fairly and efficiently.

Key Takeaways

Manual guessing is slow and error-prone.

Binary search on answer narrows down the best solution fast.

It helps balance loads fairly in allocation problems.

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 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