Why Allocate Minimum Pages Binary Search on Answer in DSA C++?

The Big Idea

What if you could find the perfect fair split without trying every possibility?

The Scenario

Imagine you have a stack of books and a few friends. You want to divide the books so each friend gets some books to read, but you want to make sure the friend who gets the most pages has as few pages as possible. Doing this by guessing and checking each way manually is like trying to split a big pizza into equal slices without a knife.

The Problem

Trying every possible way to split the books is slow and tiring. You might miss the best way or spend hours checking all options. It's easy to make mistakes and hard to know if you found the best split.

The Solution

Using binary search on the answer means guessing a number of pages and checking if it's possible to split the books so no friend reads more than that. If it's possible, try a smaller number; if not, try a bigger number. This way, you quickly find the smallest maximum pages anyone has to read.

Before vs After

✗ Before

int minPages = INT_MAX;
for (int guess = maxPage; guess <= totalPages; guess++) {
  if (canSplit(books, guess, friends)) {
    minPages = guess;
    break;
  }
}

✓ After

int low = maxPage, high = totalPages, result = totalPages;
while (low <= high) {
  int mid = low + (high - low) / 2;
  if (canSplit(books, mid, friends)) {
    result = mid;
    high = mid - 1;
  } else {
    low = mid + 1;
  }
}

What It Enables

This method lets you quickly find the best way to divide work or resources fairly and efficiently.

Real Life Example

Dividing chapters of a textbook among students so that no one has to read too much, ensuring everyone finishes around the same time.

Key Takeaways

Manual checking is slow and error-prone.

Binary search on answer guesses and checks efficiently.

It finds the smallest maximum workload quickly.

Practice

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

easy

A. To assign pages randomly to 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 wants to divide book pages among students so that the student with the most pages has as few pages as possible.

2. Step 2: Recognize the use of binary search on answer

   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 per student [OK]

Hint: Focus on minimizing the largest pages assigned [OK]

Common Mistakes:

• Thinking it maximizes total pages
• Confusing average with maximum
• Ignoring the binary search approach

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

easy

A. int low = max pages in a single book, high = sum of all pages;

B. int low = 0, high = sum of all pages;

C. int low = 0, high = max pages in a single book;

D. int low = sum of all pages, high = max pages in a single book;

Solution

1. Step 1: Determine the minimum and maximum boundaries

   The minimum possible max pages is the largest single book (no student can get less than that), and the maximum is the sum of all pages (one student gets all books).

2. Step 2: Set binary search boundaries accordingly

   So, low = max pages in a single book, high = sum of all pages.

3. Final Answer:

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

4. Quick Check:

   low = max single book, high = total pages [OK]

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

Common Mistakes:

• Setting low to 0 which is invalid
• Swapping low and high values
• Using average pages as boundary

3. What will be the output of the following C++ code snippet for the Allocate Minimum Pages problem?

int pages[] = {10, 20, 30, 40};
int n = 4, students = 2;

// Function to check feasibility
bool isPossible(int mid) {
    int studentCount = 1, pageSum = 0;
    for (int i = 0; i < n; i++) {
        if (pages[i] > mid) return false;
        if (pageSum + pages[i] > mid) {
            studentCount++;
            pageSum = pages[i];
            if (studentCount > students) return false;
        } else {
            pageSum += pages[i];
        }
    }
    return true;
}

int low = 40, high = 100, result = high;
while (low <= high) {
    int mid = (low + high) / 2;
    if (isPossible(mid)) {
        result = mid;
        high = mid - 1;
    } else {
        low = mid + 1;
    }
}
std::cout << result << std::endl;

medium

A. 50

B. 70

C. 60

D. 40

Solution

1. Step 1: Understand the feasibility check

   The function checks if pages can be allocated so no student gets more than mid pages.

2. Step 2: Binary search to find minimum max pages

   Try mid values between 40 and 100. The minimum feasible max pages is 60 (assign {10,20,30} to student 1 and {40} to student 2).

3. Final Answer:

   60 -> Option C

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Check feasibility by simulating allocation [OK]

Common Mistakes:

• Not updating student count correctly
• Choosing mid incorrectly
• Ignoring single book page size

4. Identify the bug in the following C++ code snippet for the Allocate Minimum Pages problem:

bool isPossible(int mid, int pages[], int n, int students) {
    int studentCount = 1, pageSum = 0;
    for (int i = 0; i < n; i++) {
        if (pageSum + pages[i] <= mid) {
            pageSum += pages[i];
        } else {
            studentCount++;
            pageSum = pages[i];
        }
    }
    return studentCount <= students;
}

medium

A. Missing check if a single book has pages more than mid

B. Incorrect increment of studentCount

C. Page sum should reset to zero instead of pages[i]

D. Loop should run from i = 1 instead of 0

Solution

1. Step 1: Analyze the feasibility function

   The function tries to allocate books without exceeding mid pages per student.

2. Step 2: Identify missing validation

   It does not check if any single book has pages more than mid, which makes allocation impossible.

3. Final Answer:

   Missing check if a single book has pages more than mid -> Option A

4. Quick Check:

   Check single book pages <= mid [OK]

Hint: Always check single book pages against mid [OK]

Common Mistakes:

• Ignoring single book size check
• Resetting pageSum incorrectly
• Wrong loop start index

5. Given the books with pages {12, 34, 67, 90} and 2 students, what is the minimum maximum number of pages a student can get when using binary search on answer to allocate books optimally?

hard

A. 90

B. 113

C. 101

D. 67

Solution

1. Step 1: Set binary search boundaries

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

2. Step 2: Use binary search to find minimum max pages

   Check mid values. The minimum feasible max pages is 113 (assign {12,34,67}=113 to one student and {90} to the other).

3. Final Answer:

   113 -> Option B

4. Quick Check:

   Minimum max pages = 113 [OK]

Hint: Binary search between max single book and total pages [OK]

Common Mistakes:

• Choosing max single book as answer
• Choosing sum of pages as answer
• Not testing feasibility properly