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

Concept Flow - Allocate Minimum Pages Binary Search on Answer

Set low = max(pages array)

↓

Set high = sum(pages array)

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While low <= high

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Calculate mid = Math.floor((low + high) / 2)

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Check if allocation possible with mid

↓

Update high = mid-1

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Repeat loop

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Return low as minimum max pages

We use binary search on the range of possible max pages to find the minimum max pages per student that allows allocation.

Execution Sample

DSA Typescript

function isPossible(pages: number[], students: number, mid: number): boolean {
  let allocated = 1;
  let currentSum = 0;
  for (const page of pages) {
    if (page > mid) return false;
    if (currentSum + page > mid) {
      allocated++;
      currentSum = page;
      if (allocated > students) return false;
    } else {
      currentSum += page;
    }
  }
  return true;
}

This function checks if pages can be allocated to students such that no student gets more than mid pages.

Execution Table

Step	Operation	low	high	mid	Allocation Possible?	Action	Visual State (Allocation)
1	Initialize low and high	max pages = 40	sum pages = 110	-	-	-	Pages: [10,20,30,40,10]
2	Calculate mid	40	110	75	-	-	-
3	Check allocation with mid=75	-	-	75	Yes	high = mid - 1 = 74	Allocations: [10+20+30=60], [40], [10] (3 students)
4	Calculate mid	40	74	57	-	-	-
5	Check allocation with mid=57	-	-	57	Yes	high = mid - 1 = 56	Allocations: [10+20=30], [30], [40+10=50] (3 students)
6	Calculate mid	40	56	48	-	-	-
7	Check allocation with mid=48	-	-	48	No	low = mid + 1 = 49	Allocations fail: [10+20=30], [30], [40+10=50 > 48]
8	Calculate mid	49	56	52	-	-	-
9	Check allocation with mid=52	-	-	52	Yes	high = mid - 1 = 51	Allocations: [10+20=30], [30], [40+10=50]
10	Calculate mid	49	51	50	-	-	-
11	Check allocation with mid=50	-	-	50	Yes	high = mid - 1 = 49	Allocations: [10+20=30], [30], [40+10=50]
12	Calculate mid	49	49	49	-	-	-
13	Check allocation with mid=49	-	-	49	No	low = mid + 1 = 50	Allocations fail: [10+20=30], [30], [40+10=50 > 49]
14	Loop ends	50	49	-	-	-	low > high, stop
15	Return result	-	-	-	-	Return low = 50	-

💡 Loop ends when low (50) > high (49), minimum max pages per student is 50.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 9	After Step 11	After Step 13	Final
low	40	40	40	49	49	49	50	50
high	110	74	56	56	51	49	49	49
mid	-	75	57	48	52	50	49	-
Allocation Possible?	-	Yes	Yes	No	Yes	Yes	No	-

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 (Step 1 in execution_table).

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

Because if allocation is possible with mid, we try to find a smaller max by reducing high (see Steps 3, 5, 9, 11).

Why do we update low to mid + 1 when allocation is not possible?

Because mid is too small to allocate pages properly, so we increase low to search higher (see Steps 7, 13).

Visual Quiz - 3 Questions

Look at the execution_table at Step 7, what is the value of low after this step?

A 49

B 48

C 50

D 47

💡 Hint

Check the 'Action' column at Step 7 where low is updated to mid + 1 = 49.

At which step does the allocation become impossible for the first time?

A Step 3

B Step 7

C Step 5

D Step 9

💡 Hint

Look at the 'Allocation Possible?' column to find the first 'No' value.

If the number of students increased, how would the final minimum max pages change?

A It would stay the same

B It would increase

C It would decrease

D It would become zero

💡 Hint

More students means pages can be split more, so max pages per student decreases.

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 mid
  - If yes, high = mid - 1
  - Else, low = mid + 1
- Return low as minimum max pages

Full Transcript

This concept uses binary search on the answer space to allocate minimum pages to students. We start low at the largest single book pages and high at the total pages. We calculate mid and check if allocation is possible without exceeding mid pages per student. If possible, we try smaller mid by moving high down. If not, we increase low. We repeat until low passes high. The final low is the minimum maximum pages per student. The execution table shows each step's low, high, mid, allocation check, and actions. Variable tracker shows how low, high, mid, and allocation results change. Key moments clarify why low starts at max pages, why high moves down on success, and low moves up on failure. The quiz tests understanding of these steps and effects of changing students.

Practice

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

easy

A. To assign pages randomly among students

B. To maximize the total number of pages assigned

C. To find the average number of pages per student

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

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 D

4. Quick Check:

   Goal = minimize max pages [OK]

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

Common Mistakes:

• Trying to maximize pages instead of minimizing
• Ignoring the maximum pages constraint
• Confusing total pages with max pages per student

2. Which of the following is the correct TypeScript syntax to declare the binary search boundaries for the Allocate Minimum Pages problem given an array pages?

easy

A. let low = pages.length; let high = Math.max(...pages);

B. let low = Math.min(...pages); let high = pages.length;

C. let low = 0; let high = pages.reduce((a, b) => a + b, 0);

D. let low = 0; let high = Math.min(...pages);

Solution

1. Step 1: Determine low and high boundaries

   Low is minimum possible max pages (0 or max single book pages), high is sum of all pages (max total pages).

2. Step 2: Check correct syntax for sum

   Using reduce sums all pages correctly for high boundary.

3. Final Answer:

   let low = 0; let high = pages.reduce((a, b) => a + b, 0); -> Option C

4. Quick Check:

   low = 0, high = sum pages [OK]

Hint: Sum array for high, zero or max page for low [OK]

Common Mistakes:

• Setting low as pages.length instead of page count
• Using min pages as high boundary
• Incorrect sum calculation syntax

3. Consider the following TypeScript code snippet for checking feasibility in Allocate Minimum Pages problem:

function isPossible(pages: number[], students: number, maxPages: number): boolean {
  let requiredStudents = 1;
  let currentSum = 0;
  for (let page of pages) {
    if (page > maxPages) return false;
    if (currentSum + page > maxPages) {
      requiredStudents++;
      currentSum = page;
    } else {
      currentSum += page;
    }
  }
  return requiredStudents <= students;
}

What is the output of isPossible([10, 20, 30, 40], 2, 50)?

medium

A. false

B. SyntaxError

C. true

D. RuntimeError

Solution

1. Step 1: Trace allocation with maxPages = 50

   Assign pages: 10+20=30 (still <=50), add 30 would exceed 50, so new student starts with 30, then 40 exceeds 50 if added, so new student starts with 40.

2. Step 2: Count required students

   We used 3 students but only 2 allowed, so requiredStudents=3 > 2 means return false. But check carefully: 10+20=30, next 30 alone fits in second student, 40 alone needs third student.

3. Final Answer:

   false -> Option A

4. Quick Check:

   Required students = 3 > 2, so false [OK]

Hint: Count students needed, compare with allowed [OK]

Common Mistakes:

• Ignoring that a single book can exceed maxPages
• Miscounting students when sum exceeds maxPages
• Assuming sum of all pages fits in one student

4. The following TypeScript function attempts to find the minimum maximum pages but has a bug. Identify the bug.

function findMinPages(pages: number[], students: number): number {
  let low = Math.min(...pages);
  let high = pages.reduce((a, b) => a + b, 0);
  let result = high;
  while (low <= high) {
    let mid = Math.floor((low + high) / 2);
    if (isPossible(pages, students, mid)) {
      result = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  return result;
}

medium

A. High should be min page, not sum of pages

B. Low should start from max page, not min page

C. Mid calculation causes infinite loop

D. Result variable is not updated

Solution

1. Step 1: Analyze low boundary initialization

   Low should be the maximum single book pages, because no student can get less than the largest book.

2. Step 2: Identify impact of wrong low

   Starting low at minimum page can cause incorrect feasibility checks and longer search.

3. Final Answer:

   Low should start from max page, not min page -> Option B

4. Quick Check:

   Low = max page [OK]

Hint: Low boundary = max single book pages [OK]

Common Mistakes:

• Setting low to min page instead of max page
• Confusing high boundary with low
• Not updating result inside loop

5. Given books with pages [12, 34, 67, 90] and 3 students, what is the minimum maximum pages allocated to any student using binary search on answer?

hard

A. 90

B. 113

C. 67

D. 34

Solution

1. Step 1: Set search boundaries

   Low = max pages = 90, High = sum pages = 12+34+67+90=203.

2. Step 2: Binary search feasibility checks

   Mid = (90+203)//2=146. Check if possible with maxPages=146: allocate [12+34+67=113] to student1, 90 to student2, only 2 students needed <=3, possible. So high=145.

3. Step 3: Narrow down to minimum max pages

   Continuing: mid=103 possible [12+34=46,67,90] 3 students. mid=96 possible. mid=92 possible. mid=90: [12+34=46], [67], [90] all <=90, 3 students <=3 feasible, high=89. Loop ends with result=90.

4. Step 4: Re-evaluate minimum max pages

   Check if 113 is feasible with 3 students: [12+34+67=113], [90] 2 students needed <=3, so 113 is feasible but not minimum.

5. Step 5: Check if 90 is feasible

   Allocations: [12+34=46], [67], [90] 3 students needed <=3, feasible.

6. Final Answer:

   90 -> Option A

7. Quick Check:

   Minimum max pages = 90 [OK]

Hint: Binary search between max page and sum pages [OK]

Common Mistakes:

• Misallocating at max=90 thinking 4 students needed
• Miscounting students during allocation
• Confusing with 2 students (113)