# 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.

```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.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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
```

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.

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