Given an array arr[] and an integer k, where arr[i] denotes the number of pages of a book and k denotes total number of students. All the books need to be allocated to k students in contiguous manner, with each student getting at least one book.
The task is to minimize the maximum number of pages allocated to a student. If it is not possible to allocate books to all students, return -1.
Examples:
Input: arr[] = [12, 34, 67, 90], k = 2
Output: 113
Explanation: Books can be distributed in following ways:
- [12] and [34, 67, 90] - The maximum pages assigned to a student is 34 + 67 + 90 = 191.
- [12, 34] and [67, 90] - The maximum pages assigned to a student is 67 + 90 = 157.
- [12, 34, 67] and [90] - The maximum pages assigned to a student is 12 + 34 + 67 = 113.
The third combination has the minimum pages assigned to a student which is 113.
Input: arr[] = [15, 17, 20], k = 5
Output: -1
Explanation: Since there are more students than total books, it's impossible to allocate a book to each student.
Input: arr[] = [22, 23, 67], k = 1
Output: 112
Explanation: Since there is only 1 student, all books are assigned to that student. So, maximum pages assigned to a student is 22 + 23 + 67 = 112.
[Naive Approach] By Iterating Over All Possible Page Limits
The idea is to iterate over all possible page limits, or maximum pages that can be allocated to a student.
- The minimum possible page limit is the highest page count among all books, as the book with the most pages must be assigned to some student.
- The maximum possible page limit is the sum of pages of all books, It is in the case when all books are given to a single student.
To find the number of students that will be allocated books for a page limit, we start assigning books to the first student until the page limit is reached, then we move to the next student and so on. As soon as we find the first page limit with which we can allocate books to all k students, we will return it.
C++
// C++ program to find the minimum page limit by iterating
// over all possible page limits
#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
bool check(vector<int> &arr, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < arr.size(); i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
int findPages(vector<int> &arr, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.size())
return -1;
// Minimum and maximum possible page limits
int minPageLimit = *max_element(arr.begin(), arr.end());
int maxPageLimit = accumulate(arr.begin(), arr.end(), 0);
// Iterating over all possible page limits
for(int i = minPageLimit; i <= maxPageLimit; i++) {
// Return the first page limit with we can
// allocate books to all k students
if(check(arr, k, i))
return i;
}
return -1;
}
int main() {
vector<int> arr = {12, 34, 67, 90};
int k = 2;
cout << findPages(arr, k);
return 0;
}
// C program to find the minimum page limit by iterating
// over all possible page limits
#include <stdio.h>
#include <stdbool.h>
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
bool check(int arr[], int n, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < n; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
int findPages(int arr[], int n, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > n)
return -1;
// Minimum and maximum possible page limits
int minPageLimit = arr[0];
int maxPageLimit = 0;
for(int i = 0; i < n; i++) {
if(arr[i] > minPageLimit) minPageLimit = arr[i];
maxPageLimit += arr[i];
}
// Iterating over all possible page limits
for(int i = minPageLimit; i <= maxPageLimit; i++) {
// Return the first page limit with we can
// allocate books to all k students
if(check(arr, n, k, i))
return i;
}
return -1;
}
int main() {
int arr[] = {12, 34, 67, 90};
int k = 2;
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", findPages(arr, n, k));
return 0;
}
// Java program to find the minimum page limit by iterating
// over all possible page limits
import java.util.Arrays;
class GfG {
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
static boolean check(int[] arr, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < arr.length; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
static int findPages(int[] arr, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.length)
return -1;
// Minimum and maximum possible page limits
int minPageLimit = Arrays.stream(arr).max().getAsInt();
int maxPageLimit = Arrays.stream(arr).sum();
// Iterating over all possible page limits
for(int i = minPageLimit; i <= maxPageLimit; i++) {
// Return the first page limit with we can
// allocate books to all k students
if(check(arr, k, i))
return i;
}
return -1;
}
public static void main(String[] args) {
int[] arr = {12, 34, 67, 90};
int k = 2;
System.out.println(findPages(arr, k));
}
}
# Python program to find the minimum page limit by iterating
# over all possible page limits
# Function to check if books can be allocated to
# all k students without exceeding 'pageLimit'
def check(arr, k, pageLimit):
# Starting from the first student
cnt = 1
pageSum = 0
for pages in arr:
# If adding the current book exceeds the page
# limit, assign the book to the next student
if pageSum + pages > pageLimit:
cnt += 1
pageSum = pages
else:
pageSum += pages
# If books can assigned to less than k students then
# it can be assigned to exactly k students as well
return cnt <= k
def findPages(arr, k):
# If number of students are more than total books
# then allocation is not possible
if k > len(arr):
return -1
# Minimum and maximum possible page limits
minPageLimit = max(arr)
maxPageLimit = sum(arr)
# Iterating over all possible page limits
for i in range(minPageLimit, maxPageLimit + 1):
# Return the first page limit with we can
# allocate books to all k students
if check(arr, k, i):
return i
return -1
if __name__ == "__main__":
arr = [12, 34, 67, 90]
k = 2
print(findPages(arr, k))
// C# program to find the minimum page limit by iterating
// over all possible page limits
using System;
using System.Linq;
class GfG {
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
static bool check(int[] arr, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < arr.Length; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
static int findPages(int[] arr, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.Length)
return -1;
// Minimum and maximum possible page limits
int minPageLimit = arr.Max();
int maxPageLimit = arr.Sum();
// Iterating over all possible page limits
for(int i = minPageLimit; i <= maxPageLimit; i++) {
// Return the first page limit with we can
// allocate books to all k students
if(check(arr, k, i))
return i;
}
return -1;
}
static void Main() {
int[] arr = {12, 34, 67, 90};
int k = 2;
Console.WriteLine(findPages(arr, k));
}
}
// JavaScript program to find the minimum page limit by iterating
// over all possible page limits
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
function check(arr, k, pageLimit) {
// Starting from the first student
let cnt = 1;
let pageSum = 0;
for(let i = 0; i < arr.length; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
function findPages(arr, k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.length)
return -1;
// Minimum and maximum possible page limits
const minPageLimit = Math.max(...arr);
const maxPageLimit = arr.reduce((a, b) => a + b, 0);
// Iterating over all possible page limits
for(let i = minPageLimit; i <= maxPageLimit; i++) {
// Return the first page limit with we can
// allocate books to all k students
if(check(arr, k, i))
return i;
}
return -1;
}
// Driver Code
const arr = [12, 34, 67, 90];
const k = 2;
console.log(findPages(arr, k));
Time Complexity: O(n*(Sum(arr) - MAX)), where n is the total number of books, sum(arr) is the total number of pages in all the books and MAX is maximum number of pages in any book.
Auxiliary Space: O(1)
[Expected Approach] Using Binary Search
The maximum number of pages(page limit) that a student can be allocated has a monotonic property:
- If, at a page limit p, books cannot be allocated to all k students, then we need to reduce the page limit to ensure more students receive books.
- If, at a page limit p, we can allocate books to more than k students, then we need to increase the page limit so that fewer students are allocated books.
Therefore, we can apply binary search to minimize the maximum pages a student can be allocated. To check the number of students that can be allotted books for any page limit, we start assigning books to the first student until the page limit is reached, then move to the next student.
C++
// C++ program to find the minimum page limit by iterating
// over all possible page limits
#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
bool check(vector<int> &arr, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < arr.size(); i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
int findPages(vector<int> &arr, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.size())
return -1;
// Search space for Binary Search
int lo = *max_element(arr.begin(), arr.end());
int hi = accumulate(arr.begin(), arr.end(), 0);
int res = -1;
while(lo <= hi) {
int mid = lo + (hi - lo)/2;
if(check(arr, k, mid)){
res = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return res;
}
int main() {
vector<int> arr = {12, 34, 67, 90};
int k = 2;
cout << findPages(arr, k);
return 0;
}
// C program to find the minimum page limit by iterating
// over all possible page limits
#include <stdio.h>
#include <stdbool.h>
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
bool check(int arr[], int n, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < n; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
int findPages(int arr[], int n, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > n)
return -1;
// Maximum element of the array is minimum page limit
int lo = arr[0];
for(int i = 1; i < n; i++)
if(arr[i] > lo) lo = arr[i];
// Summation of all element is maximum page limit
int hi = 0;
for(int i = 0; i < n; i++)
hi += arr[i];
int res = -1;
while(lo <= hi) {
int mid = lo + (hi - lo)/2;
if(check(arr, n, k, mid)){
res = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return res;
}
int main() {
int arr[] = {12, 34, 67, 90};
int k = 2;
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", findPages(arr, n, k));
return 0;
}
// Java program to find the minimum page limit by iterating
// over all possible page limits
import java.util.Arrays;
class GfG {
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
static boolean check(int[] arr, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < arr.length; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
static int findPages(int[] arr, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.length)
return -1;
// Search space for Binary Search
int lo = Arrays.stream(arr).max().getAsInt();
int hi = Arrays.stream(arr).sum();
int res = -1;
while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
if(check(arr, k, mid)){
res = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {12, 34, 67, 90};
int k = 2;
System.out.println(findPages(arr, k));
}
}
# Python program to find the minimum page limit by iterating
# over all possible page limits
# Function to check if books can be allocated to
# all k students without exceeding 'pageLimit'
def check(arr, k, pageLimit):
# Starting from the first student
cnt = 1
pageSum = 0
for pages in arr:
# If adding the current book exceeds the page
# limit, assign the book to the next student
if pageSum + pages > pageLimit:
cnt += 1
pageSum = pages
else:
pageSum += pages
# If books can assigned to less than k students then
# it can be assigned to exactly k students as well
return cnt <= k
def findPages(arr, k):
# If number of students are more than total books
# then allocation is not possible
if k > len(arr):
return -1
# Search space for Binary Search
lo = max(arr)
hi = sum(arr)
res = -1
while lo <= hi:
mid = lo + (hi - lo) // 2
if check(arr, k, mid):
res = mid
hi = mid - 1
else:
lo = mid + 1
return res
if __name__ == "__main__":
arr = [12, 34, 67, 90]
k = 2
print(findPages(arr, k))
// C# program to find the minimum page limit by iterating
// over all possible page limits
using System;
using System.Linq;
class GfG {
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
static bool check(int[] arr, int k, int pageLimit) {
// Starting from the first student
int cnt = 1;
int pageSum = 0;
for(int i = 0; i < arr.Length; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
static int findPages(int[] arr, int k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.Length)
return -1;
// Search space for Binary Search
int lo = arr.Max();
int hi = arr.Sum();
int res = -1;
while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
if(check(arr, k, mid)){
res = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return res;
}
static void Main() {
int[] arr = {12, 34, 67, 90};
int k = 2;
Console.WriteLine(findPages(arr, k));
}
}
// JavaScript program to find the minimum page limit by iterating
// over all possible page limits
// Function to check if books can be allocated to
// all k students without exceeding 'pageLimit'
function check(arr, k, pageLimit) {
// Starting from the first student
let cnt = 1;
let pageSum = 0;
for(let i = 0; i < arr.length; i++) {
// If adding the current book exceeds the page
// limit, assign the book to the next student
if(pageSum + arr[i] > pageLimit) {
cnt++;
pageSum = arr[i];
}
else {
pageSum += arr[i];
}
}
// If books can assigned to less than k students then
// it can be assigned to exactly k students as well
return (cnt <= k);
}
function findPages(arr, k) {
// If number of students are more than total books
// then allocation is not possible
if(k > arr.length)
return -1;
// Search space for Binary Search
let lo = Math.max(...arr);
let hi = arr.reduce((a, b) => a + b, 0);
let res = -1;
while(lo <= hi) {
let mid = lo + Math.floor((hi - lo) / 2);
if(check(arr, k, mid)){
res = mid;
hi = mid - 1;
}
else {
lo = mid + 1;
}
}
return res;
}
// Driver Code
const arr = [12, 34, 67, 90];
const k = 2;
console.log(findPages(arr, k));
Time Complexity: O(n*log(Sum(arr) - MAX)), where n is the total number of books, sum(arr) is the total number of pages in all the books and MAX is maximum number of pages in any book.
Auxiliary Space: O(1)
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