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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum number of pages in the books array.

DSA Go

maxPages := 0
for _, pages := range books {
    if pages > [1] {
        maxPages = pages
    }
}

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

A0

Bpages

CmaxPages

Dlen(books)

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing pages with 0 instead of maxPages.

Using the length of books instead of pages.

We compare each book's pages with the current maxPages to find the maximum pages.

2fill in blank

medium

Complete the code to check if allocation of books is possible with given maxPages.

DSA Go

func isPossible(books []int, students int, maxPages int) bool {
    requiredStudents := 1
    currentSum := 0
    for _, pages := range books {
        if currentSum + pages > [1] {
            requiredStudents++
            currentSum = pages
        } else {
            currentSum += pages
        }
    }
    return requiredStudents <= students
}

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

Astudents

BmaxPages

Clen(books)

DcurrentSum

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with students instead of maxPages.

Using currentSum incorrectly in the condition.

We check if adding pages exceeds maxPages to allocate books to next student.

3fill in blank

hard

Fix the error in the binary search loop condition to find minimum pages.

DSA Go

low := max(books)
high := sum(books)
result := high
for low <= [1] {
    mid := (low + high) / 2
    if isPossible(books, students, mid) {
        result = mid
        high = mid - 1
    } else {
        low = mid + 1
    }
}
return result

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Amid

Blow

Cresult

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using low <= low which causes infinite loop.

Using mid in loop condition which is incorrect.

The loop should continue while low is less than or equal to high.

4fill in blank

hard

Fill both blanks to correctly calculate the sum of pages and maximum pages in the books slice.

DSA Go

sumPages := 0
maxPages := 0
for _, pages := range books {
    sumPages += [1]
    if pages > [2] {
        maxPages = pages
    }
}

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

Apages

BsumPages

CmaxPages

Dlen(books)

Attempts:

3 left

💡 Hint

Common Mistakes

Adding maxPages instead of pages to sumPages.

Comparing pages with sumPages instead of maxPages.

Add pages to sumPages and compare pages with maxPages to find maximum.

5fill in blank

hard

Fill all three blanks to implement the binary search for minimum pages allocation.

DSA Go

low := [1]
high := [2]
result := high
for low <= high {
    mid := (low + high) / 2
    if [3](books, students, mid) {
        result = mid
        high = mid - 1
    } else {
        low = mid + 1
    }
}
return result

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

AmaxPages

BsumPages

CisPossible

Dlen(books)

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to 0 instead of maxPages.

Using wrong function name instead of isPossible.

Setting high to maxPages instead of sumPages.

Set low to maxPages, high to sumPages, and use isPossible to check feasibility.

Practice

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

easy

A. To maximize the total number of pages assigned

B. To minimize the maximum number of pages assigned to a student

C. To find the student with the least pages

D. To count the total number of pages in all books

Solution

1. Step 1: Understand the problem objective

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

2. Step 2: Identify the binary search approach

   We use binary search on the range of possible maximum pages to find the minimum feasible maximum.

3. Final Answer:

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

4. Quick Check:

   Goal = minimize max pages [OK]

Hint: Minimize max pages per student using binary search [OK]

Common Mistakes:

• Confusing minimizing max pages with maximizing total pages
• Thinking the problem counts students instead of pages
• 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 in Go?

easy

A. low := 0; high := sum of all pages

B. low := 0; high := max pages in a single book

C. low := max pages in a single book; high := sum of all pages

D. low := number of students; high := number of books

Solution

1. Step 1: Determine minimum and maximum boundaries

   The minimum possible max pages is the largest single book pages, because no book can be split.

2. Step 2: Set high boundary as total pages

   The maximum possible max pages is the sum of all pages if one student reads all books.

3. Final Answer:

   low := max pages in a single book; high := sum of all pages -> Option C

4. Quick Check:

   Boundaries = max single book to total pages [OK]

Hint: Low = max single book pages, High = sum of all pages [OK]

Common Mistakes:

• Setting low to 0 which is invalid
• Setting high less than sum of pages
• Confusing number of students with page boundaries

3. Given books with pages [10, 20, 30, 40] and 2 students, what is the minimum maximum pages allocated per student?

medium

A. 40

B. 70

C. 50

D. 60

Solution

1. Step 1: Check possible max pages allocations

   Try max pages = 50: allocation could be [10,20,30] and [40], but first student has 60 pages, which is >50, so invalid.

2. Step 2: Try max pages = 60

   Allocation: [10,20,30] = 60 pages, [40] = 40 pages. Both <= 60, valid allocation.

3. Final Answer:

   60 -> Option D

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Try mid values to check feasibility [OK]

Common Mistakes:

• Choosing max single book pages as answer without checking allocation
• Assuming equal split always works
• Not verifying allocation feasibility

4. Identify the bug in this Go code snippet for checking allocation feasibility:

func isPossible(books []int, students int, mid int) bool {
    count := 1
    sum := 0
    for _, pages := range books {
        if pages > mid {
            return false
        }
        sum += pages
        if sum > mid {
            count++
            sum = pages
        }
    }
    return count <= students
}

medium

A. No bug, code is correct

B. count should increment before sum addition

C. sum should reset to 0 instead of pages after count++

D. Function should return true if count >= students

Solution

1. Step 1: Analyze sum reset logic

   When sum exceeds mid, count increments and sum resets to current pages, which is correct to start new allocation.

2. Step 2: Check return condition

   Returns true if count of students needed is less than or equal to available students, which is correct.

3. Final Answer:

   No bug, code is correct -> Option A

4. Quick Check:

   Sum reset and count logic correct [OK]

Hint: Reset sum to current pages after exceeding mid [OK]

Common Mistakes:

• Resetting sum to 0 instead of current pages
• Returning true when count exceeds students
• Incrementing count too early

5. You have books with pages [12, 34, 67, 90] and 2 students. Using binary search on answer, what is the minimum maximum pages allocated? Also, explain why allocating [12, 34] to one student and [67, 90] to another is not optimal.

hard

A. 113; Because max pages is 157 which is more than minimum possible

B. 67; Because 67 is the median of pages

C. 90; Because 90 is the largest single book pages

D. 113; Because max pages is 90 which is less than 113

Solution

1. Step 1: Calculate sum and max single book pages

   Sum = 12+34+67+90 = 203; max single book = 90; low=90, high=203.

2. Step 2: Check allocation with mid = (90+203)/2 = 146

   Try allocation: [12,34,67] = 113 pages, [90] = 90 pages; max = 113 <= 146, feasible.

3. Step 3: Try smaller mid to find minimum max pages

   Try mid=113: allocation same as above, feasible. Try mid=112: allocation fails.

4. Step 4: Explain why [12,34] and [67,90] is not optimal

   Allocation sums: 46 and 157; max is 157 which is greater than minimum max pages 113.

5. Final Answer:

   113; Because max pages is 157 which is more than minimum possible -> Option A

6. Quick Check:

   Minimum max pages = 113; given allocation max = 157 [OK]

Hint: Check allocations sums to find minimal max pages [OK]

Common Mistakes:

• Assuming largest single book pages is answer
• Not verifying allocation sums
• Confusing median with allocation max pages