Allocate Minimum Pages Binary Search on Answer in DSA Javascript - 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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mid = Math.floor((low+high)/2)

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Check if mid can allocate

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high = mid - 1

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low = mid + 1

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Return low as minimum pages allocation

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

Execution Sample

DSA Javascript

function isPossible(pages, students, mid) {
  let required = 1, currentSum = 0;
  for (let page of pages) {
    if (page > mid) return false;
    if (currentSum + page > mid) {
      required++;
      currentSum = page;
    } else {
      currentSum += page;
    }
  }
  return required <= students;
}

This function checks if it is possible to allocate books so that no student reads more than mid pages.

Execution Table

Step	Operation	low	high	mid	Check Feasibility	Result	Action	Visual State
1	Initialize low and high	max pages=40	sum pages=100	-	-	-	-	[Pages: 10,20,30,40]
2	Calculate mid	40	100	70	Check if max pages per student <= 70	Yes	high = mid - 1 = 69	[Allocations possible with max 70]
3	Calculate mid	40	69	54	Check if max pages per student <= 54	No	low = mid + 1 = 55	[Allocations not possible with max 54]
4	Calculate mid	55	69	62	Check if max pages per student <= 62	Yes	high = mid - 1 = 61	[Allocations possible with max 62]
5	Calculate mid	55	61	58	Check if max pages per student <= 58	No	low = mid + 1 = 59	[Allocations not possible with max 58]
6	Calculate mid	59	61	60	Check if max pages per student <= 60	Yes	high = mid - 1 = 59	[Allocations possible with max 60]
7	Calculate mid	59	59	59	Check if max pages per student <= 59	No	low = mid + 1 = 60	[Allocations not possible with max 59]
8	Exit loop	60	59	-	-	-	-	[Minimum max pages = 60]

💡 Loop ends when low (60) > high (59), minimum max pages allocation found.

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	-	70	54	62	58	60	59	-
Feasibility	-	Yes	No	Yes	No	Yes	No	-

Key Moments - 3 Insights

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

Because no student can be assigned less than the largest single book, as books cannot be split. See execution_table row 1 where low is set to max pages 40.

Why do we move low up when allocation is not possible with mid?

If mid is too small to allocate all books within student limits, we increase low to mid + 1 to try a larger max allocation. See execution_table rows 3 and 5 where feasibility is No and low increases.

Why do we move high down when allocation is possible with mid?

If allocation is possible with mid, we try to find a smaller max by reducing high to mid - 1. See execution_table rows 2, 4, 6 where feasibility is Yes and high decreases.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the value of low after this step?

A 40

B 55

C 46

D 53

💡 Hint

Check the 'low' column in execution_table row 4 after step 4.

At which step does the feasibility check first return 'No'?

A Step 4

B Step 2

C Step 3

D Step 5

💡 Hint

Look at the 'Feasibility' column in execution_table to find the first 'No'.

If the maximum pages in the array were 50 instead of 40, what would be the initial value of low?

A 50

B 40

C 100

D 70

💡 Hint

Low is set to the maximum pages in the array at start, see variable_tracker Start column.

Concept Snapshot

Allocate Minimum Pages using Binary Search:
- Set low = max pages in array
- Set high = sum of all pages
- While low <= high:
  - mid = Math.floor((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 allocated to students. We start low at the largest single book pages and high at the sum of all pages. Each iteration calculates mid and checks if allocation is possible without exceeding mid pages per student. If possible, we try smaller max by moving high down; if not, we increase low. The process repeats until low passes high, and low is the answer. The execution table shows step-by-step how low, high, mid, and feasibility change. The variable tracker records these values after each step. Key moments clarify why low starts at max pages, and how low and high adjust based on feasibility. The visual quiz tests understanding of these steps and values.

Practice

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

easy

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

B. To maximize the total number of pages assigned

C. To find the student with the least pages

D. To sort the pages in ascending order

Solution

1. Step 1: Understand the problem goal

   The problem aims to divide books among students so that the student with the most pages has as few pages as possible.

2. Step 2: Identify the binary search target

   Binary search is used on the maximum pages assigned to any student to find the minimum possible maximum.

3. Final Answer:

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

4. Quick Check:

   Goal = minimize max pages [OK]

Hint: Focus on minimizing the largest pages assigned [OK]

Common Mistakes:

• Thinking the goal is to maximize pages
• Confusing sorting with allocation
• Assuming equal pages per student

2. Which of the following is the correct way to initialize the binary search boundaries for the Allocate Minimum Pages problem in JavaScript?

easy

A. let low = 0, high = sum of all pages;

B. let low = max pages in a single book, high = sum of all pages;

C. let low = 0, high = max pages in a single book;

D. let low = sum of all pages, high = max pages in a single book;

Solution

1. Step 1: Determine low boundary

   Low should be at least the maximum pages in a single book because no student can get less than that.

2. Step 2: Determine high boundary

   High is the sum of all pages, representing the case where one student reads all books.

3. Final Answer:

   let low = max pages in a single book, high = sum of all pages; -> Option B

4. Quick Check:

   low = max book pages, high = total pages [OK]

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

Common Mistakes:

• Setting low to 0 which is too low
• Swapping low and high values
• Using arbitrary boundaries

3. Given books with pages [10, 20, 30, 40] and 2 students, what is the minimum maximum pages assigned to a student using binary search on answer?

medium

A. 40

B. 50

C. 60

D. 70

Solution

1. Step 1: Calculate boundaries

   Max single book pages = 40, sum = 100. So low=40, high=100.

2. Step 2: Apply binary search and check feasibility

   Mid = 70: Can we allocate so max pages ≤ 70? Yes: [10,20,30] and [40]. Try lower mid.

3. Step 3: Check mid = 55

   Allocate: [10,20,30] sum=60 > 55, so no. Increase low to 56.

4. Step 4: Check mid = 60

   Allocate: [10,20,30] sum=60 ≤ 60, [40] next. Possible, try lower.

5. Step 5: Check mid = 50

   Allocate: [10,20] sum=30 ≤ 50, next [30,40] sum=70 > 50 no. So 50 not possible.

6. Final Answer:

   60 -> Option C

7. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Check mid feasibility by simulating allocation [OK]

Common Mistakes:

• Choosing sum/2 without checking feasibility
• Ignoring max single book pages
• Misallocating books to students

4. Identify the bug in this JavaScript snippet for checking feasibility in Allocate Minimum Pages problem:

function isPossible(books, students, maxPages) {
  let count = 1, sum = 0;
  for (let i = 0; i < books.length; i++) {
    sum += books[i];
    if (sum > maxPages) {
      count++;
      sum = books[i];
    }
  }
  return count <= students;
}

medium

A. It does not check if a single book has pages more than maxPages

B. It does not reset sum correctly after exceeding maxPages

C. It increments count even when sum equals maxPages

D. It uses a for loop instead of while loop

Solution

1. Step 1: Analyze sum accumulation

   The code adds book pages to sum and resets sum when sum > maxPages, increasing count.

2. Step 2: Check for single book pages exceeding maxPages

   There is no check if any single book's pages exceed maxPages, which makes allocation impossible.

3. Final Answer:

   It does not check if a single book has pages more than maxPages -> Option A

4. Quick Check:

   Missing single book page check [OK]

Hint: Always check single book pages ≤ maxPages first [OK]

Common Mistakes:

• Ignoring single book page size check
• Resetting sum incorrectly
• Confusing loop types

5. You have books with pages [12, 34, 67, 90] and 2 students. Using binary search on answer, what is the minimum maximum pages a student can get? Also, explain why allocating [12, 34, 67] to one student and [90] to another is optimal.

hard

A. Minimum max pages is 101; allocation is not optimal because 67+34=101 < 113

B. Minimum max pages is 90; allocation is optimal

C. Minimum max pages is 113; allocation is not optimal because 12+34+67=113 is too high

D. Minimum max pages is 113; allocation is optimal

Solution

1. Step 1: Calculate boundaries and try allocations

   Max single book = 90, sum = 203. Low=90, High=203.

2. Step 2: Check mid = 146

   Allocate: [12,34,67] sum=113 ≤146, [90] sum=90 ≤146. Possible, try lower.

3. Step 3: Check mid = 113

   Allocate: [12,34,67] sum=113 ≤113, [90] sum=90 ≤113. Possible, try lower.

4. Step 4: Check mid = 101

   Allocate: [12,34,67] sum=113 >101 no. Try [12,34] sum=46 ≤101, next [67,90] sum=157 >101 no. Not possible.

5. Step 5: Conclusion

   Minimum max pages = 113. Allocation [12,34,67] and [90] achieves max(113,90)=113, which is the minimum possible maximum, so it is optimal.

6. Final Answer:

   Minimum max pages is 113; allocation is optimal -> Option D

7. Quick Check:

   Minimum max pages = 113, allocation optimal [OK]

Hint: Try mid values and simulate allocation carefully [OK]

Common Mistakes:

• Assuming 90 is minimum max pages
• Ignoring sum of allocated books
• Not testing all allocation splits