Allocate Minimum Pages to Students
Binary Search Optimization

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Problem Statement

You are given a list of N books where each book has a certain number of pages, and you have to allocate these books to M students.

Your goal is to distribute the books among students in such a way that:

Each student gets at least one book.
Each book is allocated to exactly one student.
Books allocated to a student must be in a contiguous block (i.e., no skipping).

The task is to minimize the maximum number of pages assigned to a student across all students.

If it is not possible to allocate the books as per the rules, return -1.

Examples❯

Books (Pages)	Students	Output	Description
[10, 20, 30, 40]	2	60	Allocate [10, 20, 30] to one student and [40] to another → max pages = 60
[12, 34, 67, 90]	2	113	Best allocation: [12, 34, 67] and [90] → max = 113
[5, 17, 100, 11]	4	100	Each student gets one book → max = 100
[10, 20, 30, 40]	5	-1	More students than books → allocation not possible
[100]	1	100	Only one book and one student → give full book
[50, 60]	1	110	One student must take all books → total = 110
[]	1	-1	Empty book list → cannot allocate
[15, 25, 35]	0	-1	No students to allocate → invalid input

Solution❯

This problem is about fairly distributing work (in the form of book pages) among a group (students), such that no single person is overloaded compared to others.

The key constraint is that each student must receive a contiguous block of books. You cannot give a student book 1 and book 3 while skipping book 2.

Let's explore different scenarios to understand what the correct answer should look like:

Case 1: More students than books — This is not possible. Every student must get at least one book, and if books are fewer than students, we return -1.
Case 2: Equal number of students and books — This means each student will get exactly one book. The answer will be the maximum number of pages among the books because that’s the worst-case individual load.
Case 3: Only one student — The only student must read all the books. So the answer will be the total number of pages of all books.
Case 4: Normal case (fewer students than books) — This is where optimization is needed. We try different ways to divide books into blocks such that the student with the highest load has to read the least possible number of pages. This is done by exploring different values for the "maximum pages" and checking if the books can be divided accordingly.
Case 5: Empty book list — There's nothing to allocate, so we return -1.
Case 6: Zero students — No one is there to read the books. This is invalid input, so return -1.

The solution uses a smart way to search the answer using Binary Search instead of trying all combinations. It searches between the range max(arr) (because no student can be assigned fewer pages than the largest book) and sum(arr) (the case where one student gets everything). For each mid-value in this range, we check: is it possible to allocate books so that no student gets more than mid pages? If yes, we try to find a smaller answer. If not, we look for a higher value.

This process gives us the minimum possible value for the maximum pages a student has to read in a fair allocation.

Visualization

Algorithm Steps

Set low = max(arr) and high = sum(arr).
While low ≤ high, do:
→ Calculate mid = (low + high) / 2.
→ Check if it's possible to allocate books so that no student has more than mid pages using a helper function.
→ If possible, store the result and try to reduce it: high = mid - 1.
→ If not possible, increase the range: low = mid + 1.
Return the minimum value found that satisfies the condition.

Code

Java

Python

JavaScript

C++

Kotlin

Swift

Php

public class BookAllocator {
  public static boolean isPossible(int[] arr, int m, int maxPages) {
    int students = 1;
    int pages = 0;
    for (int i = 0; i < arr.length; i++) {
      if (arr[i] > maxPages) return false;
      if (pages + arr[i] > maxPages) {
        students++;
        pages = arr[i];
      } else {
        pages += arr[i];
      }
    }
    return students <= m;
  }

  public static int findMinimumPages(int[] arr, int m) {
    if (m > arr.length) return -1;
    int low = arr[0], high = 0;
    for (int pages : arr) {
      low = Math.max(low, pages);
      high += pages;
    }

    int result = -1;
    while (low <= high) {
      int mid = low + (high - low) / 2;
      if (isPossible(arr, m, mid)) {
        result = mid;
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    return result;
  }

  public static void main(String[] args) {
    int[] arr = {12, 34, 67, 90};
    int students = 2;
    int minPages = findMinimumPages(arr, students);
    System.out.println("Minimum Maximum Pages: " + minPages);
  }
}

Time Complexity

Case	Time Complexity	Explanation
Best Case	O(n)	Occurs if binary search hits the correct answer in the first try, but validation still needs to check all books.
Average Case	O(n * log(sum - max))	Binary search over range from max(arr) to sum(arr), and for each guess, we iterate through n books.
Worst Case	O(n * log(sum - max))	Even in the worst case, we perform log iterations and n-time validation per iteration.

Space Complexity

O(1)

Explanation: Only a fixed number of variables are used. No extra space is needed.

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Arrays32
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Arrays - Binary Search28
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Strings18
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Matrix3
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Matrix - Binary Search5
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Allocate Minimum Pages to Students
Binary Search Optimization

Problem Statement

Examples❯

Solution❯

Visualization

Algorithm Steps

Code

Time Complexity

Space Complexity

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Allocate Minimum Pages to StudentsBinary Search Optimization

Problem Statement

Examples❯

Solution❯

Visualization

Algorithm Steps

Code

Time Complexity

Space Complexity

Welcome to ProgramGuru

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Allocate Minimum Pages to Students
Binary Search Optimization