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

Concept Flow - Allocate Minimum Pages Binary Search on Answer

Set low = max(pages)

→Set high = sum(pages)

↓

While low <= high

↓

Calculate mid = (low + high)//2

↓

Check if allocation possible with mid

↓

Update high = mid-1

↓

Repeat loop

↓

Return low as minimum max pages

We use binary search on the range of possible max pages to find the minimum max pages that can be allocated to students.

Execution Sample

DSA Python

def is_possible(pages, students, mid):
    total = 0
    count = 1
    for p in pages:
        if p > mid:
            return False
        if total + p > mid:
            count += 1
            total = p
        else:
            total += p
    return count <= students

Check if pages can be allocated to students so that no student gets more than mid pages.

Execution Table

Step	Operation	low	high	mid	Allocation Possible?	Action	Visual State (max pages range)
1	Initialize low=max(pages)=10, high=sum(pages)=60	10	60	-	-	Start binary search	[10 ... 60]
2	Calculate mid=(10+60)//2=35	10	60	35	-	Check allocation with mid=35	[10 ... 60]
3	Check allocation possible with mid=35	10	60	35	Yes	Update high=mid-1=34	[10 ... 34]
4	Calculate mid=(10+34)//2=22	10	34	22	-	Check allocation with mid=22	[10 ... 34]
5	Check allocation possible with mid=22	10	34	22	No	Update low=mid+1=23	[23 ... 34]
6	Calculate mid=(23+34)//2=28	23	34	28	-	Check allocation with mid=28	[23 ... 34]
7	Check allocation possible with mid=28	23	34	28	Yes	Update high=mid-1=27	[23 ... 27]
8	Calculate mid=(23+27)//2=25	23	27	25	-	Check allocation with mid=25	[23 ... 27]
9	Check allocation possible with mid=25	23	27	25	No	Update low=mid+1=26	[26 ... 27]
10	Calculate mid=(26+27)//2=26	26	27	26	-	Check allocation with mid=26	[26 ... 27]
11	Check allocation possible with mid=26	26	27	26	Yes	Update high=mid-1=25	[26 ... 25]
12	low=26 > high=25	26	25	-	-	Exit loop	Final minimum max pages = 26

💡 Loop ends when low > high; low is the minimum maximum pages allocation possible.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 9	After Step 11	Final
low	10	10	23	23	26	26	26
high	60	34	34	27	27	25	25
mid	-	35	22	28	25	26	-
Allocation Possible?	-	Yes	No	Yes	No	Yes	-

Key Moments - 3 Insights

Why do we set low to max(pages) at the start?

Because no student can get fewer pages than the largest single book, so max(pages) is the minimum possible max pages. See Step 1 in execution_table.

Why do we update high to mid-1 when allocation is possible?

Because if allocation is possible with mid, we try to find a smaller max pages to minimize it. This is shown in Steps 3, 7, and 11.

What does it mean when allocation is not possible with mid?

It means mid is too small to allocate all books to students without exceeding mid pages. So we increase low to mid+1 as in Steps 5 and 9.

Visual Quiz - 3 Questions

Look at the execution_table at Step 6, what is the value of mid?

A 28

B 22

C 35

D 26

💡 Hint

Check the 'mid' column in execution_table row for Step 6.

At which step does the condition low > high become true, ending the loop?

A Step 10

B Step 12

C Step 9

D Step 11

💡 Hint

Look at the exit_note and Step 12 in execution_table.

If the initial low was set to 5 instead of max(pages)=10, what would happen?

A The algorithm would still work correctly.

B The algorithm would run faster.

C The algorithm might incorrectly say allocation is possible with too small mid.

D The algorithm would never terminate.

💡 Hint

Refer to key_moments about why low starts at max(pages).

Concept Snapshot

Allocate Minimum Pages using Binary Search:
- Set low = max(pages), high = sum(pages)
- While low <= high:
  - mid = (low + high)//2
  - Check if allocation possible with mid
  - If yes, high = mid - 1
  - Else, low = mid + 1
- Return low as minimum max pages

Full Transcript

This concept uses binary search to find the minimum maximum pages that can be allocated to students. We start low at the largest single book pages and high at the total pages. We repeatedly check if allocation is possible with mid value. If yes, we try smaller max pages by moving high down. If no, we increase low. The process continues until low exceeds high, and low is the answer.

Practice

1. What is the main goal of the Allocate Minimum Pages problem when using binary search on the answer?

easy

A. To find the minimum maximum number of pages assigned to a student

B. To maximize the total pages assigned to all students

C. To find the maximum number of students that can be assigned books

D. To minimize the total number of books assigned

Solution

1. Step 1: Understand the problem goal

   The problem aims to allocate books to students such that the maximum pages assigned to any student is minimized.

2. Step 2: Clarify the binary search target

   Binary search is used on the range of possible maximum pages to find the smallest feasible maximum.

3. Final Answer:

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

4. Quick Check:

   Minimum max pages = goal [OK]

Hint: Focus on minimizing the largest pages any student gets [OK]

Common Mistakes:

• Confusing total pages with maximum pages per student
• Trying to maximize pages instead of minimizing
• Ignoring the binary search on answer approach

2. Which of the following is the correct way to initialize the binary search boundaries for the Allocate Minimum Pages problem given an array pages of book pages?

easy

A. low = sum(pages), high = max(pages)

B. low = max(pages), high = sum(pages)

C. low = 0, high = max(pages)

D. low = 0, high = sum(pages) * 2

Solution

1. Step 1: Determine minimum and maximum possible max pages

   The minimum max pages is at least the largest single book (max of pages), and the maximum is the sum of all pages if one student gets all books.

2. Step 2: Set binary search boundaries accordingly

   Set low = max(pages) and high = sum(pages) to cover all feasible max page values.

3. Final Answer:

   low = max(pages), high = sum(pages) -> Option B

4. Quick Check:

   Boundaries = max and sum [OK]

Hint: Low starts at max single book, high at total pages sum [OK]

Common Mistakes:

• Setting low to 0 which is invalid
• Swapping low and high values
• Using arbitrary high value instead of sum

3. What will be the output of the following code snippet?

def is_possible(pages, students, mid):
    count = 1
    total = 0
    for p in pages:
        if total + p > mid:
            count += 1
            total = p
            if count > students:
                return False
        else:
            total += p
    return True

pages = [10, 20, 30, 40]
students = 2
mid = 50
print(is_possible(pages, students, mid))

medium

A. True

B. None

C. SyntaxError

D. False

Solution

1. Step 1: Trace the allocation with mid = 50

   pages=[10,20,30,40], students=2: count=1, total=0 - p=10: total=10 <=50 - p=20: total=30 <=50 - p=30: 30+30=60 >50 -> count=2, total=30 - p=40: 30+40=70 >50 -> count=3 >2 -> return False

2. Step 2: Check if count exceeds students

   Count reaches 3 > 2, so return False.

3. Final Answer:

   False -> Option D

4. Quick Check:

   Allocation not possible with mid=50 = False [OK]

Hint: Check if books fit in mid pages per student [OK]

Common Mistakes:

• Miscounting students needed
• Adding pages incorrectly
• Confusing > and >= in condition

4. Identify the bug in the following code snippet for the Allocate Minimum Pages problem:

def allocate_pages(pages, students):
    low = max(pages)
    high = sum(pages)
    result = 0
    while low <= high:
        mid = (low + high) // 2
        if is_possible(pages, students, mid):
            result = mid
            low = mid + 1
        else:
            high = mid - 1
    return result

medium

A. The high pointer should move to mid - 1 when allocation is possible

B. The low pointer should move to mid + 1 when allocation is possible

C. The low pointer should move to mid when allocation is possible

D. The low pointer should move to mid - 1 when allocation is not possible

Solution

1. Step 1: Understand binary search update rules

   If allocation is possible at mid, we try to find smaller max pages by moving high to mid - 1. If not possible, we increase low to mid + 1.

2. Step 2: Identify incorrect pointer update

   The code incorrectly moves low to mid + 1 when allocation is possible instead of high to mid - 1, which searches larger values instead of smaller.

3. Final Answer:

   The high pointer should move to mid - 1 when allocation is possible -> Option A

4. Quick Check:

   Move high down when allocation succeeds [OK]

Hint: Move low up only when allocation fails, else move high down [OK]

Common Mistakes:

• Swapping low and high updates
• Not updating pointers correctly causing infinite loop
• Returning wrong result due to pointer errors

5. Given the books with pages [12, 34, 67, 90] and 3 students, what is the minimum possible maximum number of pages assigned to a student using binary search on answer?

hard

A. 67

B. 113

C. 90

D. 34

Solution

1. Step 1: Set binary search boundaries

   low = max(pages) = 90, high = sum(pages) = 203.

2. Step 2: Check feasibility at mid values

   Try mid = (90+203)//2 = 146: count=1, total=0 - p=12: total=12 <=146 - p=34: total=46 <=146 - p=67: total=113 <=146 - p=90: 113+90=203 >146 -> count=2, total=90 count=2 <=3, possible -> try smaller max Try mid = (90+145)//2 = 117: count=1, total=0 - p=12: total=12 <=117 - p=34: total=46 <=117 - p=67: total=113 <=117 - p=90: 113+90=203 >117 -> count=2, total=90 count=2 <=3, possible -> try smaller max Try mid = (90+116)//2 = 103: count=1, total=0 - p=12: total=12 <=103 - p=34: total=46 <=103 - p=67: 46+67=113 >103 -> count=2, total=67 - p=90: 67+90=157 >103 -> count=3, total=90 count=3 <=3, possible -> try smaller max Try mid = (90+102)//2 = 96: count=1, total=0 - p=12: total=12 <=96 - p=34: total=46 <=96 - p=67: 46+67=113 >96 -> count=2, total=67 - p=90: 67+90=157 >96 -> count=3, total=90 count=3 <=3, possible -> try smaller max Try mid = (90+95)//2 = 92: count=1, total=0 - p=12: total=12 <=92 - p=34: total=46 <=92 - p=67: 46+67=113 >92 -> count=2, total=67 - p=90: 67+90=157 >92 -> count=3, total=90 count=3 <=3, possible -> try smaller max Try mid = (90+91)//2 = 90: count=1, total=0 - p=12: total=12 <=90 - p=34: total=46 <=90 - p=67: 46+67=113 >90 -> count=2, total=67 - p=90: 67+90=157 >90 -> count=3, total=90 count=3 <=3, possible Try mid = (90+89)//2 = 89 (low > high, stop) Minimum max pages = 90

3. Final Answer:

   90 -> Option C

4. Quick Check:

   Minimum max pages = 90 [OK]

Hint: Binary search between max single book and total pages sum [OK]

Common Mistakes:

• Choosing max single book as answer without checking feasibility
• Not updating low and high correctly during binary search
• Misallocating books to students during checks