Allocate Minimum Pages Binary Search on Answer in DSA Go - Time & Space Complexity

Time Complexity: Allocate Minimum Pages Binary Search on Answer

O(n * log m)

Understanding Time Complexity

We want to find how fast the binary search approach solves the problem of allocating minimum pages.

Specifically, how the number of steps grows as the number of books or pages increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

func allocateMinimumPages(books []int, students int) int {
    start, end := 0, 0
    for _, pages := range books {
        if pages > start {
            start = pages
        }
        end += pages
    }
    result := end
    for start <= end {
        mid := (start + end) / 2
        if isPossible(books, students, mid) {
            result = mid
            end = mid - 1
        } else {
            start = mid + 1
        }
    }
    return result
}

This code tries to find the smallest maximum number of pages assigned to any student by using binary search on the answer range.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The binary search loop that runs while narrowing the page range.
• How many times: The binary search runs about log of the sum of pages times.
• Inside each binary search step: The isPossible function loops through all books once (n times).
• Dominant operation: The repeated full traversal of the books array inside each binary search step.

How Execution Grows With Input

As the number of books (n) grows, each check scans all books once.

The binary search cuts the search space of possible page sums roughly in half each time.

Input Size (n)	Approx. Operations
10 books	About 10 * log(sum of pages) checks
100 books	About 100 * log(sum of pages) checks
1000 books	About 1000 * log(sum of pages) checks

Pattern observation: The total steps grow linearly with the number of books and logarithmically with the total pages.

Final Time Complexity

Time Complexity: O(n * log m)

This means the time grows linearly with the number of books and logarithmically with the sum of all pages.

Common Mistake

[X] Wrong: "The binary search alone makes this O(log n) time."

[OK] Correct: Each binary search step requires scanning all books to check feasibility, so the linear scan dominates.

Interview Connect

Understanding this time complexity helps you explain how binary search can be applied beyond simple arrays, showing problem-solving skills.

Self-Check

"What if we used a different feasibility check that skips some books? How would that affect the time complexity?"

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