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

Concept Flow - Allocate Minimum Pages Binary Search on Answer

Set low = max(pages array)

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Set high = sum(pages array)

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While low <= high

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Calculate mid = (low + high) / 2

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Check if allocation possible with mid

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Update high = mid - 1

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Repeat loop

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Return low as minimum pages

We use binary search on the range of possible maximum pages to allocate, checking feasibility each time, narrowing down to the minimum possible maximum.

Execution Sample

DSA Go

pages := []int{10, 20, 30, 40}
students := 2
low := max(pages)
high := sum(pages)
for low <= high {
  mid := (low + high) / 2
  if canAllocate(pages, students, mid) {
    high = mid - 1
  } else {
    low = mid + 1
  }
}
return low

This code finds the minimum maximum pages to allocate to students using binary search on the answer.

Execution Table

Step	Operation	mid (max pages)	Allocation Possible?	Update low/high	Visual State (pages allocated)
1	Initialize low and high	N/A	N/A	low=40, high=100	[ ]
2	Calculate mid	70	Yes	high=69	[10,20,30] | [40]
3	Calculate mid	54	No	low=55	[10,20] | [30,40]
4	Calculate mid	62	Yes	high=61	[10,20,30] | [40]
5	Calculate mid	58	No	low=59	[10,20] | [30,40]
6	Calculate mid	60	Yes	high=59	[10,20,30] | [40]
7	Calculate mid	59	No	low=60	[10,20] | [30,40]
8	Loop ends	N/A	N/A	low=60, high=59	[10,20,30] | [40]
9	Return result	60	Minimum max pages found	N/A	[10,20,30] | [40]

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

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
low	40	40	55	55	59	59	60	60
high	100	69	69	61	61	59	59	59
mid	N/A	70	54	62	58	60	59	N/A
allocation_possible	N/A	Yes	No	Yes	No	Yes	No	N/A

Key Moments - 3 Insights

Why do we set low to the maximum pages in the array at the start?

Because no student can be allocated less than the largest single book's pages, so low starts at max(pages). See execution_table Step 1 where low=40 (max of 10,20,30,40).

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

Because if allocation is possible with mid as max pages, we try to find a smaller max by reducing high. See execution_table Steps 2,4,6 where high decreases.

What does it mean when low becomes greater than high?

It means we have found the smallest max pages allocation possible. The loop ends and low is returned. See execution_table Step 8 where low=60 > high=59.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the value of mid and was allocation possible?

A mid=54, allocation not possible

B mid=54, allocation possible

C mid=70, allocation possible

D mid=70, allocation not possible

💡 Hint

Check the 'mid (max pages)' and 'Allocation Possible?' columns in Step 3.

At which step does the loop end because low becomes greater than high?

A Step 7

B Step 9

C Step 8

D Step 6

💡 Hint

Look at the 'Update low/high' column and see when low > high.

If the initial pages array had a larger maximum page count, how would the initial low value change?

A low would decrease

B low would increase

C low would stay the same

D low would become zero

💡 Hint

Refer to variable_tracker 'low' at Start and the explanation in key_moments about low initialization.

Concept Snapshot

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

Full Transcript

This concept uses binary search on the answer space 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 sum of all pages. We repeatedly check if allocation is possible with mid as the max pages. If yes, we try smaller max by moving high down. If no, we increase low. When low exceeds high, low is the answer. 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_possible change. Key moments clarify why low starts at max pages, why high moves down when allocation is possible, and what ending condition means. The quiz tests understanding of mid values, loop ending, and initial low value changes.

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