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

↓

While low <= high

↓

mid = (low + high) / 2

↓

Check if mid is a valid answer

↓

Update low = mid + 1

↓

Loop until low > high

↓

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.

Practice

1. What is the main idea behind the Binary Search on Answer technique?

easy

A. We guess an answer and check if it satisfies the problem, then adjust the guess using binary search.

B. We sort the input array before applying binary search.

C. We use binary search to find an element in a sorted array.

D. We use recursion to divide the problem into smaller subproblems.

Solution

1. Step 1: Understand the technique concept

   Binary Search on Answer guesses a possible answer and checks if it meets the problem's conditions.

2. Step 2: Adjust guess based on check

   If the guess is too high or too low, we adjust it using binary search until the best answer is found.

3. Final Answer:

   We guess an answer and check if it satisfies the problem, then adjust the guess using binary search. -> Option A

4. Quick Check:

   Guess and check with binary search = D [OK]

Hint: Think guess-check-adjust using binary search [OK]

Common Mistakes:

• Confusing with normal binary search on arrays
• Thinking it always requires sorting input
• Assuming recursion is mandatory

2. Which of the following is the correct Go syntax to start a binary search on answer between low = 1 and high = 100?

easy

A. while low < high { mid = (low + high) / 2 }

B. for i := low; i < high; i++ { mid := (low + high) / 2 }

C. for low <= high { mid := (low + high) / 2 }

D. if low < high { mid := (low + high) / 2 }

Solution

1. Step 1: Identify Go loop syntax for binary search

   Go uses for loops for iteration; for low <= high { ... } is correct for binary search.

2. Step 2: Check other options for syntax errors

   while is not valid in Go; for i := low; i < high; i++ is a counting loop, not binary search; if is conditional, not loop.

3. Final Answer:

   for low <= high { mid := (low + high) / 2 } -> Option C

4. Quick Check:

   Go binary search loop uses for low <= high [OK]

Hint: Use for loop with condition low <= high in Go [OK]

Common Mistakes:

• Using while loop (not valid in Go)
• Using for with increment instead of condition
• Using if instead of loop

3. Consider the following Go code snippet using binary search on answer to find the minimum x such that x*x >= 10:

low, high := 1, 10
var ans int
for low <= high {
  mid := (low + high) / 2
  if mid*mid >= 10 {
    ans = mid
    high = mid - 1
  } else {
    low = mid + 1
  }
}
fmt.Println(ans)

What will be printed?

medium

A. 3

B. 4

C. 10

D. 1

Solution

1. Step 1: Trace binary search for x where x*x >= 10

   Check mid values: - mid=5: 25 >=10 true, ans=5, high=4 - mid=2: 4 >=10 false, low=3 - mid=3: 9 >=10 false, low=4 - mid=4: 16 >=10 true, ans=4, high=3 Loop ends, ans=4

2. Step 2: Find minimum x satisfying condition

   Minimum x with x*x >=10 is 4, so ans=4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Minimum x with x²≥10 is 4 [OK]

Hint: Check mid² against target, adjust low/high accordingly [OK]

Common Mistakes:

• Choosing mid incorrectly (integer division errors)
• Not updating ans when condition is true
• Confusing >= with >

4. The following Go code tries to find the maximum length of a rope that can be cut into at least k pieces using binary search on answer. Identify the bug:

func maxRopeLength(ropes []int, k int) int {
  low, high := 1, 0
  for _, r := range ropes {
    if r > high {
      high = r
    }
  }
  ans := 0
  for low < high {
    mid := (low + high) / 2
    count := 0
    for _, r := range ropes {
      count += r / mid
    }
    if count >= k {
      ans = mid
      low = mid + 1
    } else {
      high = mid - 1
    }
  }
  return ans
}

medium

A. The loop condition should be low <= high, not low < high.

B. The initial high value should be 1, not 0.

C. The count calculation should use float division, not integer division.

D. The ans variable should be initialized to high, not 0.

Solution

1. Step 1: Analyze binary search loop condition

   The loop uses for low < high, which misses the case when low == high, possibly skipping the final answer.

2. Step 2: Correct loop condition for binary search

   Changing to for low <= high ensures all candidates are checked, including when low equals high.

3. Final Answer:

   The loop condition should be low <= high, not low < high. -> Option A

4. Quick Check:

   Binary search loops use low <= high [OK]

Hint: Use low <= high to include all candidates in binary search [OK]

Common Mistakes:

• Using low < high and missing final check
• Confusing integer and float division here
• Initializing ans incorrectly

5. You want to find the minimum maximum workload to assign n tasks to m workers so that no worker has workload exceeding that value. You have an array of task times. Which approach correctly applies binary search on answer to solve this problem?

hard

A. Use binary search on the number of workers instead of workload.

B. Set low as 0, high as max task time; binary search on this range checking if sum of tasks is less than high.

C. Sort tasks and binary search for the task with median time as answer.

D. Set low as max task time, high as sum of all tasks; binary search on this range checking feasibility of workload limit.

Solution

1. Step 1: Define search space for answer

   The minimum maximum workload must be at least the largest single task time (low) and at most the sum of all tasks (high).

2. Step 2: Check feasibility in binary search

   For each mid in [low, high], check if tasks can be assigned to m workers without exceeding mid workload per worker.

3. Step 3: Adjust search based on feasibility

   If feasible, try smaller max workload (high = mid - 1), else increase (low = mid + 1).

4. Final Answer:

   Set low as max task time, high as sum of all tasks; binary search on this range checking feasibility of workload limit. -> Option D

5. Quick Check:

   Search between max task and sum, check feasibility [OK]

Hint: Search between max single task and total sum [OK]

Common Mistakes:

• Setting low to 0 ignoring max task time
• Searching on wrong variable like number of workers
• Not checking feasibility properly