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

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, and you want to divide them 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 and be very confusing.

The Problem

Trying every possible way to split the books manually is slow and tiring.

You might miss better ways to split or spend too much time checking all possibilities.

The Solution

Using binary search on the answer helps you quickly find the smallest maximum number of pages any friend has to read.

This method smartly guesses a number, checks if it works, and adjusts the guess until it finds the best answer.

Before vs After

✗ Before

maxPages := sum(pages)
for maxPages > minPages {
  // try all splits manually
}

✓ After

low, high := max(pages), sum(pages)
for low <= high {
  mid := (low + high) / 2
  if canAllocate(pages, mid, friends) {
    high = mid - 1
  } else {
    low = mid + 1
  }
}

What It Enables

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

Real Life Example

Dividing tasks among team members so no one is overloaded, or splitting files across servers to balance load.

Key Takeaways

Manual checking is slow and error-prone.

Binary search on answer narrows down the best maximum quickly.

It helps balance loads fairly and efficiently.

Practice

1. What is the main goal of the Allocate Minimum Pages problem using binary search on answer?

easy

A. To maximize the total number of pages assigned

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

C. To find the student with the least pages

D. To count the total number of pages in all books

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 B

4. Quick Check:

   Goal = minimize max pages [OK]

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

Common Mistakes:

• Confusing minimizing max pages with maximizing total pages
• Thinking the problem counts students instead of pages
• Ignoring the binary search on answer approach

2. Which of the following is the correct way to initialize the binary search boundaries for the Allocate Minimum Pages problem in Go?

easy

A. low := 0; high := sum of all pages

B. low := 0; high := max pages in a single book

C. low := max pages in a single book; high := sum of all pages

D. low := number of students; high := number of books

Solution

1. Step 1: Determine minimum and maximum boundaries

   The minimum possible max pages is the largest single book pages, because no book can be split.

2. Step 2: Set high boundary as total pages

   The maximum possible max pages is the sum of all pages if one student reads all books.

3. Final Answer:

   low := max pages in a single book; high := sum of all pages -> Option C

4. Quick Check:

   Boundaries = max single book to total pages [OK]

Hint: Low = max single book pages, High = sum of all pages [OK]

Common Mistakes:

• Setting low to 0 which is invalid
• Setting high less than sum of pages
• Confusing number of students with page boundaries

3. Given books with pages [10, 20, 30, 40] and 2 students, what is the minimum maximum pages allocated per student?

medium

A. 40

B. 70

C. 50

D. 60

Solution

1. Step 1: Check possible max pages allocations

   Try max pages = 50: allocation could be [10,20,30] and [40], but first student has 60 pages, which is >50, so invalid.

2. Step 2: Try max pages = 60

   Allocation: [10,20,30] = 60 pages, [40] = 40 pages. Both <= 60, valid allocation.

3. Final Answer:

   60 -> Option D

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Try mid values to check feasibility [OK]

Common Mistakes:

• Choosing max single book pages as answer without checking allocation
• Assuming equal split always works
• Not verifying allocation feasibility

4. Identify the bug in this Go code snippet for checking allocation feasibility:

func isPossible(books []int, students int, mid int) bool {
    count := 1
    sum := 0
    for _, pages := range books {
        if pages > mid {
            return false
        }
        sum += pages
        if sum > mid {
            count++
            sum = pages
        }
    }
    return count <= students
}

medium

A. No bug, code is correct

B. count should increment before sum addition

C. sum should reset to 0 instead of pages after count++

D. Function should return true if count >= students

Solution

1. Step 1: Analyze sum reset logic

   When sum exceeds mid, count increments and sum resets to current pages, which is correct to start new allocation.

2. Step 2: Check return condition

   Returns true if count of students needed is less than or equal to available students, which is correct.

3. Final Answer:

   No bug, code is correct -> Option A

4. Quick Check:

   Sum reset and count logic correct [OK]

Hint: Reset sum to current pages after exceeding mid [OK]

Common Mistakes:

• Resetting sum to 0 instead of current pages
• Returning true when count exceeds students
• Incrementing count too early

5. You have books with pages [12, 34, 67, 90] and 2 students. Using binary search on answer, what is the minimum maximum pages allocated? Also, explain why allocating [12, 34] to one student and [67, 90] to another is not optimal.

hard

A. 113; Because max pages is 157 which is more than minimum possible

B. 67; Because 67 is the median of pages

C. 90; Because 90 is the largest single book pages

D. 113; Because max pages is 90 which is less than 113

Solution

1. Step 1: Calculate sum and max single book pages

   Sum = 12+34+67+90 = 203; max single book = 90; low=90, high=203.

2. Step 2: Check allocation with mid = (90+203)/2 = 146

   Try allocation: [12,34,67] = 113 pages, [90] = 90 pages; max = 113 <= 146, feasible.

3. Step 3: Try smaller mid to find minimum max pages

   Try mid=113: allocation same as above, feasible. Try mid=112: allocation fails.

4. Step 4: Explain why [12,34] and [67,90] is not optimal

   Allocation sums: 46 and 157; max is 157 which is greater than minimum max pages 113.

5. Final Answer:

   113; Because max pages is 157 which is more than minimum possible -> Option A

6. Quick Check:

   Minimum max pages = 113; given allocation max = 157 [OK]

Hint: Check allocations sums to find minimal max pages [OK]

Common Mistakes:

• Assuming largest single book pages is answer
• Not verifying allocation sums
• Confusing median with allocation max pages