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.