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.