Allocate minimum number of pages

Difficulty Level : Hard
Last Updated : 12 May, 2021

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;
}

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));
    }
}

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

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.

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
?>

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 pages
function findPages(arr,n,m)
{
    let 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 (let i = 0; i < n; i++)
        sum += arr[i];
   
    // 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 Method
let arr=[12, 34, 67, 90];
let m = 2; //No. of students
document.write("Minimum number of pages = " +
                          findPages(arr, arr.length, m));
 
 
 
// This code is contributed by patel2127
</script>

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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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