Allocate Minimum Pages Binary Search on Answer in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if allocation is possible with given max pages.

DSA Python

def is_possible(arr, n, m, [1]):
    students = 1
    pages_sum = 0
    for i in range(n):
        if arr[i] > max_pages:
            return False
        if pages_sum + arr[i] > max_pages:
            students += 1
            pages_sum = arr[i]
            if students > m:
                return False
        else:
            pages_sum += arr[i]
    return True

Drag options to blanks, or click blank then click option'

Amax_sum

Bmin_pages

Cmid

Dmax_pages

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not defined in the function parameters.

Confusing max_pages with mid or min_pages.

The function checks if allocation is possible with the given maximum pages allowed per student, so the parameter should be max_pages.

2fill in blank

medium

Complete the code to calculate mid value for binary search.

DSA Python

def find_pages(arr, n, m):
    start = max(arr)
    end = sum(arr)
    result = end
    while start <= end:
        [1] = (start + end) // 2
        if is_possible(arr, n, m, mid):
            result = mid
            end = mid - 1
        else:
            start = mid + 1
    return result

Drag options to blanks, or click blank then click option'

Amid

Bmidpoint

Cmiddle

Dmedian

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names not matching the rest of the code.

Using float division instead of integer division.

The variable mid is commonly used to represent the middle value in binary search calculations.

3fill in blank

hard

Fix the error in the loop condition to correctly perform binary search.

DSA Python

def find_pages(arr, n, m):
    start = max(arr)
    end = sum(arr)
    result = end
    while [1]:
        mid = (start + end) // 2
        if is_possible(arr, n, m, mid):
            result = mid
            end = mid - 1
        else:
            start = mid + 1
    return result

Drag options to blanks, or click blank then click option'

Astart < end

Bstart == end

Cstart <= end

Dstart >= end

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= causing missing the last check.

Using >= or == which breaks the loop logic.

The binary search loop should continue while start is less than or equal to end to cover all possible values.

4fill in blank

hard

Fill both blanks to correctly update the binary search range after checking feasibility.

DSA Python

def find_pages(arr, n, m):
    start = max(arr)
    end = sum(arr)
    result = end
    while start <= end:
        mid = (start + end) // 2
        if is_possible(arr, n, m, mid):
            result = mid
            [1] = mid - 1
        else:
            [2] = mid + 1
    return result

Drag options to blanks, or click blank then click option'

Aend

Bstart

Cresult

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping start and end updates.

Updating result instead of start or end.

When allocation is possible, we try to reduce the upper bound by setting end = mid - 1. Otherwise, we increase the lower bound by setting start = mid + 1.

5fill in blank

hard

Fill all three blanks to complete the function that returns minimum pages allocation.

DSA Python

def allocate_min_pages(arr, n, m):
    if m > n:
        return -1
    start = max(arr)
    end = sum(arr)
    result = end
    while start <= end:
        mid = (start + end) // 2
        if is_possible(arr, n, m, mid):
            result = [1]
            end = [2]
        else:
            start = [3]
    return result

Drag options to blanks, or click blank then click option'

Amid

Bmid - 1

Cmid + 1

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variables for start or end updates.

Not updating result correctly.

When allocation is possible, update result to mid and move end to mid - 1. Otherwise, move start to mid + 1.

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