Allocate minimum number of pages
Given a number of pages in N different books and M students. The books are arranged in ascending order of the 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 pagesOf the 3 cases, Option 3 has the minimum pages = 113.
Approach: A Binary Search method for solving the book allocation problem:
Case 1: When no valid answer exists.
If the number of students is greater than the number of books (i.e, M > N), In this case at least 1 student will be left to which no book has been assigned.
Case 2: When a valid answer exists.
The maximum possible answer could be when there is only one student. So, all the book will be assigned to him and the result would be the sum of pages of all the books.
The minimum possible answer could be when number of student is equal to the number of book (i.e, M == N) , In this case all the students will get at most one book. So, the result would be the maximum number of pages among them (i.e, maximum(pages[])).
Hence, we can apply binary search in this given range and each time we can consider the mid value as the maximum limit of pages one can get. And check for the limit if answer is valid then update the limit accordingly.
Below is the implementation of the above idea:
- Initialise the start to maximum(pages[]) and end = sum of pages[],
- Do while start <= end
- Calculate the mid and check if mid number of pages can assign any student by satisfying the given condition such that all students will get at least one book. Follow the steps to check for validity.
- Initialise the studentsRequired = 1 and curr_sum = 0 for sum of consecutive pages of book
- Iterate over all books or say pages[]
- Add the pages to curr_sum and check curr_sum > curr_min then increment the count of studentRequired by 1.
- Check if the studentRequired > M, return false.
- Return true.
- If mid is valid then, update the result and move the end = mid – 1
- Otherwise, move the start = mid + 1
- Calculate the mid and check if mid number of pages can assign any student by satisfying the given condition such that all students will get at least one book. Follow the steps to check for validity.
- Finally, return the result.
Below is the implementation of the above approach:
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 pagesint 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; int mx = INT_MIN; // Count total number of pages for (int i = 0; i < n; i++) { sum += arr[i]; mx = max(mx, arr[i]); } // initialize start as 0 pages and end as // total pages int start = mx, 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)) { // update result to current distribution // as it's the best we have found till now. 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 codeint 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;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program for optimal allocation of pages#include <limits.h>#include <stdbool.h>#include <stdio.h>// 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 pagesint 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; int mx = arr[0] ; // Count total number of pages for (int i = 0; i < n; i++){ sum += arr[i]; mx = (arr[i] > mx ? arr[i] : mx) ; } // initialize start as 0 pages and end as // total pages int start = mx, 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)) { // update result to current distribution // as it's the best we have found till now. 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 codeint 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 printf("Minimum number of pages = %d\n", findPages(arr, n, m)); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program for optimal allocation of pagespublic 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++) { curr_sum += arr[i]; if (curr_sum > curr_min) { studentsRequired++; // increment student // count curr_sum = arr[i]; // update curr_sum } } return studentsRequired <= m; } // method to find minimum pages static int findPages(int arr[], int n, int m) { int sum = 0; // return -1 if no. of books is less than // no. of students if (n < m) return -1; int mx = arr[0] ; // Count total number of pages for (int i = 0; i < n; i++){ sum += arr[i]; mx = (arr[i] > mx ? arr[i] : mx) ; } // initialize start as arr[n-1] pages(minimum answer // possible) and end as total pages(maximum answer // possible) int start = arr[n - 1], end = 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 - start) / 2; if (isPossible(arr, n, m, mid)) { // update result to current distribution // as it's the best we have found till now. result = mid; // as we are finding minimum so, 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) { int arr[] = { 12, 34, 67, 90 }; // Number of pages in books int m = 2; // No. of students System.out.println("Minimum number of pages = " + findPages(arr, arr.length, m)); }}// This code is contributed by Aditya Kumar (adityakumar129) |
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 pagesdef 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)): # update result to current distribution # as it's the best we have found till now. 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 booksarr = [12, 34, 67, 90]n = len(arr)m = 2 # No. of studentsprint("Minimum number of pages = ", findPages(arr, n, m))# This code is contributed by Mohit Kumar |
C#
// C# program for optimal// allocation of pagesusing 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; int mx = arr[0]; // Count total number of pages for (int i = 0; i < n; i++) { sum += arr[i]; mx = (arr[i] > mx ? arr[i] : mx); } // 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)) { // update result to current distribution // as it's the best we have found till now. 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. |
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 pagesfunction 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)) { // update result to current distribution // as it's the best we have found till now $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);$m = 2; // No. of studentsecho "Minimum number of pages = ", findPages($arr, $n, $m), "\n";// This code is contributed by Sach_Code?> |
Javascript
<script>// Javascript program for optimal allocation of pages// Utility method to check if current minimum value // is feasible or not.function isPossible(arr,n,m,curr_min){ let studentsRequired = 1; let curr_sum = 0; // iterate over all books for (let 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 pagesfunction findPages(arr,n,m){ let sum = 0; int mx = arr[0] ; // return -1 if no. of books is less than // no. of students if (n < m) return -1; // Count total number of pages for (let i = 0; i < n; i++){ sum += arr[i]; mx = (arr[i] > mx ? arr[i] : mx) ; } // initialize start as 0 pages and end as // total pages let start = 0, end = sum; let result = Number.MAX_VALUE; // traverse until start <= end while (start <= end) { // check if it is possible to distribute // books by using mid as current minimum let mid =Math.floor( (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 Methodlet arr=[12, 34, 67, 90];let m = 2; //No. of studentsdocument.write("Minimum number of pages = " + findPages(arr, arr.length, m));// This code is contributed by patel2127</script> |
Output
Minimum number of pages = 113
Time Complexity: O(N*log(N)), Where N is the total number of pages in the book.
Auxiliary Space: O(1)
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