Binary Search on Answer Technique in DSA Go - Execution Trace

Concept Flow - Binary Search on Answer Technique

Set low = min possible answer

↓

Set high = max possible answer

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While low <= high

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mid = (low + high) / 2

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Check if mid is a valid answer

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Update low = mid + 1

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Loop until low > high

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Return best valid answer

This flow shows how we guess an answer, check if it works, and adjust our search range until we find the best answer.

Execution Sample

DSA Go

low, high := 1, 10
for low <= high {
  mid := (low + high) / 2
  if can(mid) {
    low = mid + 1
  } else {
    high = mid - 1
  }
}
return high

This code tries to find the maximum value for which can(mid) is true by binary searching between low and high.

Execution Table

Step	Operation	low	high	mid	can(mid)?	Action	Search Range After
1	Initialize	1	10				[1..10]
2	Calculate mid	1	10	5			[1..10]
3	Check can(5)	1	10	5	Yes	low = mid + 1 = 6	[6..10]
4	Calculate mid	6	10	8			[6..10]
5	Check can(8)	6	10	8	No	high = mid - 1 = 7	[6..7]
6	Calculate mid	6	7	6			[6..7]
7	Check can(6)	6	7	6	Yes	low = mid + 1 = 7	[7..7]
8	Calculate mid	7	7	7			[7..7]
9	Check can(7)	7	7	7	No	high = mid - 1 = 6	[7..6]
10	Exit loop	7	6				low > high, stop
11	Return					Return high = 6

💡 Loop stops when low (7) becomes greater than high (6), best valid answer is high = 6

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 9	Final
low	1	6	6	7	7	7
high	10	10	7	7	6	6
mid		5	8	6	7

Key Moments - 3 Insights

Why do we update low to mid + 1 when can(mid) is true?

Because mid is a valid answer, we try to find a bigger valid answer by moving low above mid (see step 3 and 7 in execution_table).

Why do we return high at the end, not low?

When the loop ends, low is just above the best valid answer, so high holds the largest valid answer (see step 11 in execution_table).

What happens if can(mid) is false?

We reduce the search space by moving high below mid to exclude invalid answers (see step 5 and 9 in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of mid at step 6?

A 6

B 7

C 8

D 5

💡 Hint

Check the 'mid' column in execution_table row with Step '6'

At which step does the condition low <= high become false?

A Step 9

B Step 10

C Step 11

D Step 8

💡 Hint

Look at the exit_note and Step 10 in execution_table

If can(mid) was always true, what would happen to the variable high?

A It would decrease each step

B It would stay the same

C It would increase each step

D It would never be updated

💡 Hint

See that high only changes when can(mid) is false in execution_table

Concept Snapshot

Binary Search on Answer Technique:
- Set low and high as min and max possible answers
- While low <= high:
  - mid = (low + high) / 2
  - If mid is valid (can(mid) == true), move low to mid + 1
  - Else, move high to mid - 1
- Return high as the best valid answer
This finds the maximum valid answer efficiently.

Full Transcript

Binary Search on Answer Technique helps find the best answer by guessing a middle value and checking if it works. We start with low and high as the smallest and largest possible answers. We calculate mid as the average of low and high. If mid is valid, we try bigger answers by moving low above mid. If mid is not valid, we try smaller answers by moving high below mid. We repeat until low passes high. The best valid answer is then high. This method efficiently narrows down the answer range by half each time.