Given number of pages in n different books and m students. The books are arranged in ascending order of number of pages. Every student is assigned to read some consecutive books. The task is to assign books in such a way that the maximum number of pages assigned to a student is minimum.

Input : pages[] = {12, 34, 67, 90} m = 2 Output : 113 Explanation: There are 2 number of students. Books can be distributed in following fashion : 1) [12] and [34, 67, 90] Max number of pages is allocated to student 2 with 34 + 67 + 90 = 191 pages 2) [12, 34] and [67, 90] Max number of pages is allocated to student 2 with 67 + 90 = 157 pages 3) [12, 34, 67] and [90] Max number of pages is allocated to student 1 with 12 + 34 + 67 = 113 pages Of the 3 cases, Option 3 has the minimum pages = 113.

Source : Google Interview

The idea is to use Binary Search. We fix a value for number of pages as mid of current minimum and maximum. We initialize minimum and maximum as 0 and sum-of-all-pages respectively. If a current mid can be a solution, then we search on the lower half, else we search in higher half.

Now the question arises, how to check if a mid value is feasible or not? Basically we need to check if we can assign pages to all students in a way that the maximum number doesn’t exceed current value. To do this, we sequentially assign pages to every student while current number of assigned pages doesn’t exceed the value. In this process, if number of students become more than m, then solution is not feasible. Else feasible.

Below is implementation of above idea.

## C++

// C++ program for optimal allocation of pages #include<bits/stdc++.h> using namespace std; // Utility function to check if current minimum value // is feasible or not. bool isPossible(int arr[], int n, int m, int curr_min) { int studentsRequired = 1; int curr_sum = 0; // iterate over all books for (int i = 0; i < n; i++) { // check if current number of pages are greater // than curr_min that means we will get the result // after mid no. of pages if (arr[i] > curr_min) return false; // count how many students are required // to distribute curr_min pages if (curr_sum + arr[i] > curr_min) { // increment student count studentsRequired++; // update curr_sum curr_sum = arr[i]; // if students required becomes greater // than given no. of students,return false if (studentsRequired > m) return false; } // else update curr_sum else curr_sum += arr[i]; } return true; } // function to find minimum pages int findPages(int arr[], int n, int m) { long long sum = 0; // return -1 if no. of books is less than // no. of students if (n < m) return -1; // Count total number of pages for (int i = 0; i < n; i++) sum += arr[i]; // initialize start as 0 pages and end as // total pages int start = 0, end = sum; int result = INT_MAX; // traverse until start <= end while (start <= end) { // check if it is possible to distribute // books by using mid is current minimum int mid = (start + end) / 2; if (isPossible(arr, n, m, mid)) { // if yes then find the minimum distribution result = min(result, mid); // as we are finding minimum and books // are sorted so reduce end = mid -1 // that means end = mid - 1; } else // if not possible means pages should be // increased so update start = mid + 1 start = mid + 1; } // at-last return minimum no. of pages return result; } // Drivers code int main() { //Number of pages in books int arr[] = {12, 34, 67, 90}; int n = sizeof arr / sizeof arr[0]; int m = 2; //No. of students cout << "Minimum number of pages = " << findPages(arr, n, m) << endl; return 0; }

## Java

// Java program for optimal allocation of pages public class GFG { // Utility method to check if current minimum value // is feasible or not. static boolean isPossible(int arr[], int n, int m, int curr_min) { int studentsRequired = 1; int curr_sum = 0; // iterate over all books for (int i = 0; i < n; i++) { // check if current number of pages are greater // than curr_min that means we will get the result // after mid no. of pages if (arr[i] > curr_min) return false; // count how many students are required // to distribute curr_min pages if (curr_sum + arr[i] > curr_min) { // increment student count studentsRequired++; // update curr_sum curr_sum = arr[i]; // if students required becomes greater // than given no. of students,return false if (studentsRequired > m) return false; } // else update curr_sum else curr_sum += arr[i]; } return true; } // method to find minimum pages static int findPages(int arr[], int n, int m) { long sum = 0; // return -1 if no. of books is less than // no. of students if (n < m) return -1; // Count total number of pages for (int i = 0; i < n; i++) sum += arr[i]; // initialize start as 0 pages and end as // total pages int start = 0, end = (int) sum; int result = Integer.MAX_VALUE; // traverse until start <= end while (start <= end) { // check if it is possible to distribute // books by using mid is current minimum int mid = (start + end) / 2; if (isPossible(arr, n, m, mid)) { // if yes then find the minimum distribution result = Math.min(result, mid); // as we are finding minimum and books // are sorted so reduce end = mid -1 // that means end = mid - 1; } else // if not possible means pages should be // increased so update start = mid + 1 start = mid + 1; } // at-last return minimum no. of pages return result; } // Driver Method public static void main(String[] args) { //Number of pages in books int arr[] = {12, 34, 67, 90}; int m = 2; //No. of students System.out.println("Minimum number of pages = " + findPages(arr, arr.length, m)); } }

Output:

Minimum number of pages = 113

Source: http://qa.geeksforgeeks.org/3766/allocate-the-minimum-number-of-pages-to-each-student-google

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