What is the main goal of the Allocate Minimum Pages problem?
To divide an array of book pages among students such that the maximum pages assigned to a student is minimized.
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intermediate
Why do we use binary search on the answer in this problem?
Because the answer (minimum maximum pages) lies between the maximum single book pages and the sum of all pages, allowing efficient searching in this range.
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intermediate
What does the 'isPossible' function check in the Allocate Minimum Pages problem?
It checks if it is possible to allocate books to students such that no student reads more than a given maximum number of pages.
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beginner
In the binary search approach, what do the 'low' and 'high' pointers represent?
'Low' is the maximum pages in a single book, and 'High' is the sum of all pages in all books.
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advanced
What is the time complexity of the Allocate Minimum Pages solution using binary search on answer?
O(N * log S), where N is the number of books and S is the sum of all pages.
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What is the lower bound for the binary search in Allocate Minimum Pages?
AZero
BSum of all pages
CNumber of students
DMaximum pages in a single book
✗ Incorrect
The minimum maximum pages cannot be less than the largest single book.
What does the 'isPossible' function return if a given max page limit is too small?
AFalse
BTrue
CThrows error
DReturns number of students
✗ Incorrect
It returns false if allocation is not possible under the given max page limit.
What is the main advantage of using binary search on the answer here?
AAllocates pages randomly
BSimplifies code by removing loops
CReduces time complexity by searching in a range
DAvoids sorting the array
✗ Incorrect
Binary search efficiently narrows down the minimum maximum pages by searching in a range.
If the number of students is more than the number of books, what is the result?
AAllocation is impossible
BAllocation is always possible
CMaximum pages equals sum of all pages
DMaximum pages equals zero
✗ Incorrect
Each student must get at least one book, so if students > books, allocation is impossible.
What does the binary search 'mid' represent in this problem?
ANumber of students
BA candidate for minimum maximum pages
CNumber of books
DSum of pages
✗ Incorrect
'Mid' is the current guess for the minimum maximum pages to check feasibility.
Explain how binary search is applied to find the minimum maximum pages in the Allocate Minimum Pages problem.
Think about how you narrow down the answer range step by step.
You got /4 concepts.
Describe the role of the 'isPossible' helper function in the Allocate Minimum Pages problem.
It tests if a guess for max pages works.
You got /4 concepts.
Practice
(1/5)
1. What is the main goal of the Allocate Minimum Pages problem when using binary search on the answer?
easy
A. To minimize the maximum number of pages assigned to a student
B. To maximize the total number of pages assigned
C. To find the average number of pages per student
D. To assign pages randomly to students
Solution
Step 1: Understand the problem goal
The problem wants to split book pages among students so that the student with the most pages has as few pages as possible.
Step 2: Identify the binary search target
Binary search is used to find the minimum possible maximum pages assigned to any student.
Final Answer:
To minimize the maximum number of pages assigned to a student -> Option A
Quick Check:
Goal = minimize max pages [OK]
Hint: Focus on minimizing the largest pages any student gets [OK]
Common Mistakes:
Thinking it maximizes total pages
Confusing average with maximum
Assuming random assignment
2. Which of the following is the correct way to initialize the binary search range for the minimum pages problem in C#?
easy
A. int low = 0, high = pages.Max();
B. int low = 0, high = pages.Sum();
C. int low = pages.Max(), high = pages.Sum();
D. int low = pages.Sum(), high = pages.Max();
Solution
Step 1: Determine the minimum and maximum possible pages
The minimum max pages must be at least the largest single book (pages.Max()), and the maximum is the sum of all pages (pages.Sum()).
Step 2: Set binary search range accordingly
Set low = pages.Max() and high = pages.Sum() to cover all possible max page values.
Final Answer:
int low = pages.Max(), high = pages.Sum(); -> Option C
Quick Check:
low = max page, high = sum pages [OK]
Hint: Low starts at max single book, high at total pages [OK]
Common Mistakes:
Starting low at 0 (too low)
Swapping low and high
Using max as low and max as high
3. Given the array pages = [10, 20, 30, 40] and students = 2, what is the minimum maximum pages assigned after running the binary search solution?
medium
A. 70
B. 50
C. 40
D. 60
Solution
Step 1: Understand the problem with given input
We want to split [10,20,30,40] into 2 students minimizing the max pages assigned.
Step 2: Check possible splits and their max pages
Possible splits: [10,20,30] & [40] max=60; [10,20] & [30,40] max=70; [10] & [20,30,40] max=90; [10,20,30,40] & [] invalid. The minimum max is 60 or 50? Actually, [10,20,30] & [40] max=60, but [10,20] & [30,40] max=70. Check if 50 is possible: [10,20] & [30,40] max=70 >50 no; [10,20,30] & [40] max=60 >50 no; So 50 is not possible. But 60 is possible. So minimum max pages is 60.
Final Answer:
60 -> Option D
Quick Check:
Minimum max pages = 60 [OK]
Hint: Try splitting pages to minimize largest sum per student [OK]
Common Mistakes:
Choosing max single book (40) as answer
Choosing total sum (100) as answer
Not checking all splits
4. Consider this C# code snippet for checking feasibility in the minimum pages problem:
bool IsPossible(int[] pages, int students, int mid) {
int count = 1, sum = 0;
foreach (int page in pages) {
if (sum + page > mid) {
count++;
sum = page;
} else {
sum += page;
}
}
return count <= students;
}
What is the bug in this code?
medium
A. It should return count >= students instead of count <= students
B. It does not handle the case when a single book has more pages than mid
C. It increments count incorrectly inside the loop
D. It resets sum to zero instead of page after incrementing count
Solution
Step 1: Analyze the feasibility check logic
The function checks if pages can be allocated so no student gets more than mid pages.
Step 2: Identify missing check for single book pages
If any single book has pages > mid, allocation is impossible, but code does not check this, causing wrong results.
Final Answer:
It does not handle the case when a single book has more pages than mid -> Option B
Quick Check:
Check single book > mid to avoid wrong feasibility [OK]
Hint: Always check if any single book exceeds mid [OK]
Common Mistakes:
Ignoring single book size check
Wrong comparison operator in return
Resetting sum incorrectly
5. You have pages = [12, 34, 67, 90] and students = 2. Using binary search on answer, what is the minimum maximum pages that can be allocated? Also, explain why the answer is not the maximum single book pages.
hard
A. 113, because splitting as [12,34,67] and [90] minimizes max pages
B. 90, because the largest single book sets the minimum max pages
C. 67, because the average pages per student is less than 70
D. 123, because total pages must be divided equally
Solution
Step 1: Calculate total and max pages
Total pages = 12+34+67+90 = 203; max single book = 90.
Step 2: Check possible splits for 2 students
Splits: [12,34,67] & [90] max=113; [12,34] & [67,90] max=157; [12] & [34,67,90] max=191. Minimum max pages is 113.
Step 3: Explain why answer is not max single book
Because 90 alone is too small to cover all pages in 2 students without exceeding max pages, the minimum max pages must be higher than 90.
Final Answer:
113, because splitting as [12,34,67] and [90] minimizes max pages -> Option A
Quick Check:
Minimum max pages = 113 > max single book 90 [OK]
Hint: Check splits; answer ≥ max single book, often higher [OK]
Common Mistakes:
Choosing max single book as answer
Ignoring total pages sum
Assuming equal division always works
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