Allocate Minimum Pages Binary Search on Answer in DSA C# - Execution Trace

Concept Flow - Allocate Minimum Pages Binary Search on Answer

Set low = max pages in any book

↓

Set high = sum of all pages

↓

While low <= high

↓

Calculate mid = (low + high) / 2

↓

Check if allocation possible with mid pages

Yes No↓

high = mid-1

↓

Repeat loop

↓

Return low as minimum pages

We use binary search between max single book pages and total pages to find the minimum maximum pages per student that allows allocation.

Execution Sample

DSA C#

int AllocatePages(int[] books, int students) {
  int low = books.Max();
  int high = books.Sum();
  while (low <= high) {
    int mid = (low + high) / 2;
    if (CanAllocate(books, students, mid)) high = mid - 1;
    else low = mid + 1;
  }
  return low;
}

This code finds the minimum maximum pages per student to allocate books using binary search on answer.

Execution Table

Step	Operation	mid (max pages)	Allocation Possible?	low	high	Visual State
1	Initialize low and high	N/A	N/A	max book pages = 90	sum pages = 370	Books: [12,34,67,90,45,22]
2	Calculate mid	230	Check allocation with max 230 pages	90	370	Trying max pages = 230
3	Allocation check	230	Yes	90	370	Students needed: 2 (<=3)
4	Update high	230	N/A	90	229	high updated to 229
5	Calculate mid	159	Check allocation with max 159 pages	90	229	Trying max pages = 159
6	Allocation check	159	Yes	90	229	Students needed: 3 (<=3)
7	Update high	159	N/A	90	158	high updated to 158
8	Calculate mid	124	Check allocation with max 124 pages	90	158	Trying max pages = 124
9	Allocation check	124	No	90	158	Students needed: 4 (>3)
10	Update low	124	N/A	125	158	low updated to 125
11	Calculate mid	141	Check allocation with max 141 pages	125	158	Trying max pages = 141
12	Allocation check	141	Yes	125	158	Students needed: 3 (<=3)
13	Update high	141	N/A	125	140	high updated to 140
14	Calculate mid	132	Check allocation with max 132 pages	125	140	Trying max pages = 132
15	Allocation check	132	No	125	140	Students needed: 4 (>3)
16	Update low	132	N/A	133	140	low updated to 133
17	Calculate mid	136	Check allocation with max 136 pages	133	140	Trying max pages = 136
18	Allocation check	136	Yes	133	140	Students needed: 3 (<=3)
19	Update high	136	N/A	133	135	high updated to 135
20	Calculate mid	134	Check allocation with max 134 pages	133	135	Trying max pages = 134
21	Allocation check	134	No	133	135	Students needed: 4 (>3)
22	Update low	134	N/A	135	135	low updated to 135
23	Calculate mid	135	Check allocation with max 135 pages	135	135	Trying max pages = 135
24	Allocation check	135	Yes	135	135	Students needed: 3 (<=3)
25	Update high	135	N/A	135	134	high updated to 134
26	Exit loop	N/A	low > high	135	134	Minimum max pages = 135

💡 Loop ends when low (135) > high (134), minimum max pages found.

Variable Tracker

Variable	Start	After Step 4	After Step 7	After Step 10	After Step 13	After Step 16	After Step 19	After Step 22	Final
low	90	90	90	125	125	133	133	135	135
high	370	229	158	158	140	140	135	135	134
mid	N/A	230	159	124	141	132	136	134	135
Allocation Possible?	N/A	Yes	Yes	No	Yes	No	Yes	No	Yes

Key Moments - 3 Insights

Why do we start low with the maximum pages of a single book?

Because no student can be assigned less than the largest single book's pages, starting lower would make allocation impossible. See Step 1 in execution_table.

Why do we update high when allocation is possible with mid pages?

If allocation is possible with mid pages, we try to find a smaller maximum by reducing high to mid-1. This narrows the search space. See Steps 3 and 4.

Why do we update low when allocation is not possible with mid pages?

If allocation fails, it means mid is too small, so we increase low to mid+1 to try larger max pages. See Steps 9 and 10.

Visual Quiz - 3 Questions

Look at the execution_table at Step 9, what is the allocation result for mid = 124?

A Allocation possible with 3 students

B Allocation not possible, needs 4 students

C Allocation possible with 2 students

D Allocation not checked

💡 Hint

Check the 'Allocation Possible?' column at Step 9 in execution_table.

At which step does low first increase above 125?

A Step 10

B Step 22

C Step 16

D Step 4

💡 Hint

Look at the 'low' variable in variable_tracker after each step.

If the number of students was increased to 4, how would the final minimum max pages change?

A It would decrease

B It would increase

C It would stay the same

D Cannot determine

💡 Hint

More students means pages can be split more, so max pages per student can be smaller.

Concept Snapshot

Allocate Minimum Pages using Binary Search:
- low = max pages in one book
- high = sum of all pages
- mid = (low + high) / 2
- Check if allocation possible with mid
- If yes, high = mid - 1
- Else, low = mid + 1
- Loop until low > high
- Return low as minimum max pages

Full Transcript

This concept uses binary search on the answer space to allocate books to students such that the maximum pages assigned to any student is minimized. We start with low as the maximum pages in a single book and high as the sum of all pages. We repeatedly check the middle value mid to see if allocation is possible with that maximum. If yes, we try smaller max by moving high down. If no, we increase low. The process continues until low exceeds high, and low is the minimum maximum pages possible. The execution table shows each step with mid values, allocation checks, and updates to low and high. Variable tracker shows how low, high, mid, and allocation results change. Key moments clarify why low starts at max book pages and how updates to low and high narrow the search. The visual quiz tests understanding of allocation results and variable changes. The snapshot summarizes the binary search approach for this problem.

Practice

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

1. 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.

2. Step 2: Identify the binary search target

   Binary search is used to find the minimum possible maximum pages assigned to any student.

3. Final Answer:

   To minimize the maximum number of pages assigned to a student -> Option A

4. 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

1. 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()).

2. Step 2: Set binary search range accordingly

   Set low = pages.Max() and high = pages.Sum() to cover all possible max page values.

3. Final Answer:

   int low = pages.Max(), high = pages.Sum(); -> Option C

4. 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

1. Step 1: Understand the problem with given input

   We want to split [10,20,30,40] into 2 students minimizing the max pages assigned.

2. 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.

3. Final Answer:

   60 -> Option D

4. 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

1. Step 1: Analyze the feasibility check logic

   The function checks if pages can be allocated so no student gets more than mid pages.

2. 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.

3. Final Answer:

   It does not handle the case when a single book has more pages than mid -> Option B

4. 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

1. Step 1: Calculate total and max pages

   Total pages = 12+34+67+90 = 203; max single book = 90.

2. 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.

3. 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.

4. Final Answer:

   113, because splitting as [12,34,67] and [90] minimizes max pages -> Option A

5. 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