Binary Search on Answer Technique in DSA C++ - Execution Trace

Concept Flow - Binary Search on Answer Technique

Set low and high bounds

↓

Calculate mid = (low + high) / 2

↓

Check if mid is a valid answer

↓

Update high = mid

↓

Repeat until low >= high

↓

Return low as answer

Start with a range of possible answers, check the middle, adjust range based on validity, repeat until range narrows to the answer.

Execution Sample

DSA C++

int low = 1, high = 10;
while (low < high) {
  int mid = (low + high) / 2;
  if (isValid(mid)) high = mid;
  else low = mid + 1;
}
return low;

This code finds the smallest valid number between 1 and 10 using binary search on the answer.

Execution Table

Step	low	high	mid	isValid(mid)	Action	Range After Action
1	1	10	5	No	low = mid + 1	[6, 10]
2	6	10	8	Yes	high = mid	[6, 8]
3	6	8	7	No	low = mid + 1	[8, 8]
Exit	8	8	-	-	low >= high, stop	Answer = 8

💡 low equals high, so the smallest valid answer is found.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
low	1	6	6	8	8
high	10	10	8	8	8
mid	-	5	8	7	-

Key Moments - 3 Insights

Why do we update 'low' to 'mid + 1' when isValid(mid) is false?

Because mid is not a valid answer, so the answer must be greater than mid. See Step 1 and Step 3 in execution_table where low moves past mid.

Why do we update 'high' to 'mid' when isValid(mid) is true?

Because mid might be the smallest valid answer, so we keep it by setting high = mid. See Step 2 where high moves down to mid.

Why does the loop stop when low equals high?

Because the search range has narrowed to a single number, which is the smallest valid answer. See Exit row in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'mid' at Step 3?

A 8

B 7

C 5

D 6

💡 Hint

Check the 'mid' column in the row labeled Step 3 in execution_table.

At which step does the 'low' variable first change from its initial value?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'low' column in execution_table rows to see when it changes from 1.

If isValid(mid) returned true at Step 1, what would be the new value of 'high' after Step 1?

A 10

B 6

C 5

D 1

💡 Hint

When isValid(mid) is true, high is set to mid. Check Step 2 for example.

Concept Snapshot

Binary Search on Answer Technique:
- Define low and high bounds for answer
- Calculate mid = (low + high) / 2
- Check if mid is valid
- If valid, set high = mid
- Else, set low = mid + 1
- Repeat until low >= high
- Return low as smallest valid answer

Full Transcript

Binary Search on Answer Technique starts by setting a low and high range for possible answers. We calculate the middle value mid and check if it is valid. If mid is valid, we move the high bound down to mid to find a smaller valid answer. If mid is not valid, we move the low bound up to mid plus one to search higher values. This repeats until low equals high, which is the smallest valid answer. The example code shows this process with low starting at 1 and high at 10. The execution table traces each step, showing how low, high, and mid change, and what action is taken based on validity. Key moments explain why we update low or high in each case and why the loop stops when low equals high. The visual quiz tests understanding of mid values, variable changes, and hypothetical changes in validity. This technique helps solve problems where the answer lies within a range and can be checked for validity efficiently.