Binary Search on Answer Technique in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the binary search boundaries for the answer.

DSA Go

low, high := 0, [1]

Drag options to blanks, or click blank then click option'

Alen(arr)

Bmath.MaxInt64

C1000000

Darr[len(arr)-1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using array length as upper bound which may be too small

Using last element of array which may not represent max answer

We set high to a large number as the upper bound for the answer search space.

2fill in blank

medium

Complete the code to calculate the middle value in binary search safely.

DSA Go

mid := low + (high - [1]) / 2

Drag options to blanks, or click blank then click option'

Amid

Blow

Chigh

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using (low + high) / 2 directly which can overflow

Using wrong variable in subtraction

We subtract low from high to avoid overflow, then add low back.

3fill in blank

hard

Fix the error in the condition to check if mid is a valid answer.

DSA Go

if canAchieve(arr, [1]) {

Drag options to blanks, or click blank then click option'

Amid

Bhigh

Clow

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Passing high or low instead of mid

Passing mid + 1 which is not the current candidate

We check if the current middle value mid can achieve the goal.

4fill in blank

hard

Fill both blanks to update the binary search boundaries correctly.

DSA Go

if canAchieve(arr, mid) {
    high = [1]
} else {
    low = [2]
}

Drag options to blanks, or click blank then click option'

Amid

Bmid + 1

Cmid - 1

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to mid instead of mid + 1 causing infinite loop

Setting high to mid - 1 which may skip valid answers

If mid works, we try smaller or equal values by setting high = mid. Otherwise, we try larger values by setting low = mid + 1.

5fill in blank

hard

Fill all three blanks to return the final answer after binary search.

DSA Go

for low < high {
    mid := low + (high - low) / 2
    if canAchieve(arr, mid) {
        high = [1]
    } else {
        low = [2]
    }
}
return [3]

Drag options to blanks, or click blank then click option'

Amid

Blow

Cmid + 1

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Returning mid instead of low after loop

Incorrectly updating boundaries causing infinite loop

Inside the loop, update high to mid if condition is true, else update low to mid + 1. After the loop, low holds the smallest valid answer.

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