Day 7 - Allocate Minimum Pages.md

Difficulty	Source	Tags
Medium	160 Days of Problem Solving	Searching Divide and Conquer

🚀 Day 7. Allocate Minimum Pages 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

You are given an array arr[] of integers, where each element arr[i] represents the number of pages in the ith book. You also have an integer k representing the number of students. The task is to allocate books to each student such that:

1. Each student receives at least one book.
2. Each student is assigned a contiguous sequence of books.
3. No book is assigned to more than one student.

The objective is to minimize the maximum number of pages assigned to any student. In other words, find the arrangement where the student who receives the most pages has the smallest possible maximum.

Note:

• Return -1 if a valid assignment is not possible.
• Allocation must be contiguous.

🔍 Example Walkthrough:

Input:

arr[] = [12, 34, 67, 90]
k = 2

Output:

113

Explanation:

Allocation can be done in the following ways:

1. [12] and [34, 67, 90] → Maximum Pages = 191
2. [12, 34] and [67, 90] → Maximum Pages = 157
3. [12, 34, 67] and [90] → Maximum Pages = 113

The minimum of these cases is 113, which is selected as the output.

Input:

arr[] = [15, 17, 20]
k = 5

Output:

-1

Explanation:

Allocation is not possible as k > len(arr).

Input:

arr[] = [22, 23, 67]
k = 1

Output:

112

Constraints:

• $1 <= arr.size() <= 10^6$
• $1 <= arr[i] <= 10^3$
• $1 <= k <= 10^3$

🎯 My Approach:

1. Binary Search on Result:

   • The result lies between max(arr) (the maximum pages of a single book) and sum(arr) (all pages given to one student).
   • Use binary search to find the smallest valid value.

2. Validation Function:

   • Check if it is possible to allocate books such that no student receives more than mid pages:
     • Iterate through the books.
     • Add pages to the current student until it exceeds mid.
     • If it does, assign to the next student.
   • If the number of students required exceeds k, return False.

3. Edge Cases:

   • If k > len(arr), return -1 as allocation is not possible.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n * log(sum(arr) - max(arr))), where n is the size of the array. This is because we perform binary search over the range [max(arr), sum(arr)] and validate each mid-point in O(n).
• Expected Auxiliary Space Complexity: O(1), as we do not use any additional space apart from a few variables.

📝 Solution Code

Code (C++)

class Solution {
public:
    int findPages(vector<int>& arr, int k) {
        int n = arr.size();
        if (k > n) return -1; 
        int low = *max_element(arr.begin(), arr.end()), high = accumulate(arr.begin(), arr.end(), 0);
        while (low < high) {
            int mid = (low + high) / 2, students = 1, sum = 0;
            for (int pages : arr) {
                if ((sum += pages) > mid) students++, sum = pages;
            }
            if (students > k) low = mid + 1;
            else high = mid;
        }
        return low;
    }
};

Code (Java)

class Solution {
    public static int findPages(int[] arr, int k) {
        int n = arr.length;
        if (k > n) return -1; 

        int low = Arrays.stream(arr).max().getAsInt(); 
        int high = Arrays.stream(arr).sum(); 

        while (low < high) {
            int mid = (low + high) / 2;
            int students = 1, sum = 0;

            for (int pages : arr) {
                if ((sum + pages) > mid) {
                    students++;
                    sum = pages;
                } else {
                    sum += pages;
                }
            }

            if (students > k) low = mid + 1;
            else high = mid;
        }
        return low;
    }
}

Code (Python)

class Solution:
    def findPages(self, arr, k):
        n = len(arr)
        if k > n:
            return -1  

        low, high = max(arr), sum(arr)  

        while low < high:
            mid = (low + high) // 2
            students, current_sum = 1, 0

            for pages in arr:
                if current_sum + pages > mid:
                    students += 1
                    current_sum = pages
                else:
                    current_sum += pages

            if students > k:
                low = mid + 1
            else:
                high = mid

        return low

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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