Allocate minimum number of pages
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.
Example :
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.
The idea is to use Binary Search. We fix a value for the 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 the current number of assigned pages doesn’t exceed the value. In this process, if the number of students becomes more than m, then the solution is not feasible. Else feasible.
Below is an 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 as 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; } |
chevron_right
filter_none
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)); } } |
chevron_right
filter_none
Python3
# Python3 program for optimal allocation of pages # Utility function to check if # current minimum value is feasible or not. def isPossible(arr, n, m, curr_min): studentsRequired = 1 curr_sum = 0 # iterate over all books for i in range(n): # 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 += 1 # 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 def findPages(arr, n, m): 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 i in range(n): sum += arr[i] # initialize start as 0 pages and # end as total pages start, end = 0, sum result = 10**9 # traverse until start <= end while (start <= end): # check if it is possible to distribute # books by using mid as current minimum 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 # Driver Code # Number of pages in books arr = [12, 34, 67, 90] n = len(arr) m = 2 # No. of students print("Minimum number of pages = ", findPages(arr, n, m)) # This code is contributed by Mohit Kumar |
chevron_right
filter_none
C#
// C# program for optimal // allocation of pages using System; class GFG { // Utility function to check // if current minimum value // is feasible or not. static 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 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 = int.MaxValue; // traverse until start <= end while (start <= end) { // check if it is possible to // distribute books by using // mid as 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; } // Drivers code static public void Main () { //Number of pages in books int []arr = {12, 34, 67, 90}; int n = arr.Length; int m = 2; // No. of students Console.WriteLine("Minimum number of pages = " + findPages(arr, n, m)); } } // This code is contributed by anuj_67. |
chevron_right
filter_none
PHP
<?php // PHP program for optimal allocation // of pages // Utility function to check if current // minimum value is feasible or not. function isPossible($arr, $n, $m, $curr_min) { $studentsRequired = 1; $curr_sum = 0; // iterate over all books for ( $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 function findPages($arr, $n, $m) { $sum = 0; // return -1 if no. of books is // less than no. of students if ($n < $m) return -1; // Count total number of pages for ($i = 0; $i < $n; $i++) $sum += $arr[$i]; // initialize start as 0 pages // and end as total pages $start = 0; $end = $sum; $result = PHP_INT_MAX; // traverse until start <= end while ($start <= $end) { // check if it is possible to // distribute books by using // mid as current minimum $mid = (int)($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; } // Driver code // Number of pages in books $arr = array(12, 34, 67, 90); $n = count($arr); $m = 2; // No. of students echo "Minimum number of pages = ", findPages($arr, $n, $m), "\n"; // This code is contributed by Sach_Code ?> |
chevron_right
filter_none
Output :
Minimum number of pages = 113
This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Recommended Posts:
- Maximum and minimum of an array using minimum number of comparisons
- Paper Cut into Minimum Number of Squares | Set 2
- Minimum number of times A has to be repeated such that B is a substring of it
- Collect all coins in minimum number of steps
- Minimum number N such that total set bits of all numbers from 1 to N is at-least X
- Find the minimum number of rectangles left after inserting one into another
- Find row with maximum and minimum number of zeroes in given Matrix
- Search an element in an unsorted array using minimum number of comparisons
- Minimum number of swaps required for arranging pairs adjacent to each other
- Minimum number of points to be removed to get remaining points on one side of axis
- Second minimum element using minimum comparisons
- Minimum value of "max + min" in a subarray
- Number of ways to divide a given number as a set of integers in decreasing order
- Middle of three using minimum comparisons
- Count the number of operations required to reduce the given number
