Binary Search on Answer Technique in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the Binary Search on Answer Technique?

It is a method to find an unknown value by guessing an answer range and using binary search to narrow down the correct answer based on a condition check.

beginner

In Binary Search on Answer, what does the 'answer' represent?

The 'answer' is the value we want to find, often a number like minimum or maximum that satisfies a problem's condition.

intermediate

Why do we use Binary Search on Answer instead of normal binary search on arrays?

Because sometimes the answer is not directly in a list but lies in a range of values, and we can check if a guess is valid to narrow down the range.

beginner

What is the role of the 'check' function in Binary Search on Answer?

It tests if a guessed answer meets the problem's condition, helping decide if we should search higher or lower.

intermediate

Give a simple example where Binary Search on Answer is useful.

Finding the minimum maximum load to split an array into subarrays with limited sum, by guessing load values and checking feasibility.

What does Binary Search on Answer typically require?

A A range of possible answers and a way to check if a guess is valid

B A sorted array only

C A linked list

D Random guessing

Binary Search on Answer needs a range and a check function to narrow down the answer.

If the check function returns true for a guess, what should you do next?

A Stop immediately

B Search in the upper half

C Search in the lower half to find a better answer

D Ignore and guess randomly

If guess is valid, try to find a smaller valid answer by searching lower.

Binary Search on Answer is best used when:

A The array is unsorted

B You want to find duplicates

C You want to sort an array

D The answer lies in a continuous or discrete range and can be checked

It works when answer is in a range and can be validated by a check.

What is the time complexity of Binary Search on Answer if the check function runs in O(n)?

A O(n log m), where m is the answer range size

B O(n^2)

C O(log n)

D O(n)

Binary search does log m steps, each with O(n) check.

Which of these problems is a good candidate for Binary Search on Answer?

A Reversing a linked list

B Finding minimum maximum subarray sum for splitting

C Finding duplicates in an array

D Sorting an array

Minimum maximum subarray sum can be found by guessing sums and checking feasibility.

Explain the steps to solve a problem using Binary Search on Answer Technique.

Think about guessing and validating answers.

Describe a real-life scenario where Binary Search on Answer can be applied.

Imagine dividing tasks or resources evenly.

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