Home/Blog/Interview Prep/10 common binary search interview questions

10 common binary search interview questions

Adeel Qayyum

Nov 14, 2024

9 min read

content

Curated list of problems Binary search Code template Binary search interview questions list Problem 1: Search insert position Problem 2: First and last position of an element in a sorted array Problem 3: Peak element in an array Further reading Conclusion

The importance of the binary search algorithm in technical interviews cannot be overstated. It tests a candidate’s understanding of algorithmic thinking and their ability to optimize solutions. Top tech companies, including Google, Amazon, and Microsoft, frequently include binary search problems in their interviews, making it important for candidates to master this algorithm.

In this blog, we'll explore ten common binary search interview questions. We’ll break down each problem, explain how binary search applies, and provide step-by-step solutions to help you understand the approach. By the end, you'll have a deeper understanding of how to tackle these problems with confidence, making binary search a valuable tool in your coding interview.

Curated list of problems#

The table below lists the top 10 binary search interview questions commonly featured in coding interviews at top tech companies like Google, Facebook, and Netflix. The problems will be selected and arranged based on their difficulty and the extent to which they require modifications to the basic binary search algorithm.

Problem #	Name	Selection Criteria	Frequency	Difficulty
1	Search insert position	This is the most basic binary search problem. Selected because it extends the basic binary search problem by handling cases where the element is not found.	Commonly asked in interviews by companies like Amazon and Facebook.	Easy
2	First and last position of an element in a sorted array	Selected because it requires finding both the first and last occurrence of an element, which is a common variation of binary search problems.	Frequently asked in interviews by companies like Google and Microsoft.	Medium
3	Peak element in an array	Selected because it demonstrates the use of binary search in non-trivial array conditions, showcasing its efficiency in finding an element that meets specific criteria beyond simple searching.	Companies like Google frequently ask it in interviews, making it important for interview preparation.	Medium
4	Single element in a sorted array	This problem tests the understanding of binary search in identifying unique elements within a sorted array.	Common, frequently asked by companies like Amazon and Google.	Medium
5	Finding minimum in a rotated sorted array	This problem introduces the concept of rotated arrays, adding complexity to binary search.	Common, particularly in interviews with Amazon and Facebook.	Medium
6	Search in rotated sorted array	This problem builds on the previous problem, adding an extra layer of difficulty.	Very common, often asked by Google, Amazon, and Microsoft.	Medium
7	Find the smallest missing positive number	This problem tests the application of binary search in an unconventional context.	Common, frequently asked by companies like Facebook.	Medium
8	Find the duplicate number	This problem demonstrates the use of binary search in solution space rather than the input array.	Frequently asked by companies like Amazon and Google.	Medium
9	Median of two sorted arrays	This complex problem tests the understanding of binary search on multiple arrays.	Very common, especially in interviews with top tech companies.	Hard
10	Kth smallest element in a sorted matrix	This problem tests binary search in a 2D context, adding complexity.	Common, frequently asked by companies like Google and Microsoft.	Hard

Let’s first look at the basic binary search algorithm. After that, we will explore some of the above problems in detail with algorithm explanation, code solution, and illustrations to explain how the algorithm works.

Binary search#

Binary search is an algorithm to find an element in a sorted array or list. The process involves repeatedly dividing the search interval in half. If the value of the search key is less than the item in the middle of the interval, the algorithm narrows the interval to the lower half. Otherwise, it narrows it to the upper half. The steps are as follows:

Initialize: Set two pointers, one at the beginning (left) and one at the end (right) of the array.
Calculate the midpoint: Compute the middle index of the current interval as mid = left + (right - left) // 2.
Compare: Compare the target value with the value at the middle index.
1. If they are equal, return the middle index.
2. If the target value is less than the middle value, adjust the right pointer to mid - 1.
3. If the target value is greater than the middle value, adjust the left pointer to mid + 1.
Repeat: Continue the process until the left pointer exceeds the right pointer, indicating that the target is not in the array.

Binary Search for Coding Interviews

Binary Search (BS) is, arguably, the most popular question in coding interviews. Even when you see that a problem can be solved with BS, it can be tricky to handle all of the indexes and base cases in the solution. What if there was an approach that could work for all BS problems? This course was designed from scratch with this one goal in mind. We'll explore one technique that can handle any BS problem. The practice problems in this course were carefully chosen to cover the most frequently asked BS questions in coding interviews. At the end of this course, you will feel confident to handle BS questions during phone or on-site interviews, as well as with all of the follow-ups about its time and space complexity.

3hrs

Beginner

16 Challenges

3 Quizzes

This algorithm is efficient with a time complexity of $O(\log n)$ , making it suitable for large datasets.

Binary search interview questions list#

This section will cover ten common interview problems that can be efficiently solved using the binary search algorithm. Each problem is selected based on its relevance and frequency in coding interviews, ranging from fundamental to advanced levels.

Problem 1: Search insert position#

This is one of the most basic binary search example problems. It was selected because it extends the basic binary search problem by handling cases where the element is not found. Companies like Amazon and Facebook commonly ask this problem in interviews and consider it easy in terms of difficulty.

Statement: Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if inserted in order. You must write an algorithm with $O(\log n)$ runtime complexity.

Solution: Binary search is an algorithm for finding the position of a target value within a sorted array. Here, we will modify the standard binary search to handle cases where the element is not found by returning the index where the target value should be inserted to maintain the order.

Here is the step-by-step explanation of the algorithm:

Initialization: Start with two pointers, left and right, initialized to the start and end of the array, respectively.
Iteration:
- Calculate the middle index mid = left + (right - left) // 2.
- Compare the target value with the middle element of the array:
  - If the target value equals the middle element, return mid as the index.
  - If the target value is less than the middle element, adjust the right pointer to mid - 1 to continue searching in the left half.
  - If the target value is greater than the middle element, adjust the left pointer to mid + 1 to continue searching in the right half.
Completion: When the left pointer exceeds the right pointer, the search interval is exhausted. The left pointer will be where the target value should be inserted. Return left as the insert position.

Let’s visualize the above mentioned steps:

def search_insert_position(arr, target):
    # Initialize left and right pointers
    left, right = 0, len(arr) - 1
    while left <= right:
        # Calculate the mid-ndex
        mid = left + (right - left) // 2
        # Check if the target is at the mid-index
        if arr[mid] == target:
            return mid
        # If the target is greater, ignore the left half
        elif arr[mid] < target:
            left = mid + 1
        # If the target is smaller, ignore the right half
        else:
            right = mid - 1
    # Return the position where the target should be inserted
    return left
# Driver code with test cases
test_cases = [
    ([1, 3, 5, 6], 5),  # Target found
    ([1, 3, 5, 6], 2),  # Target not found, insert at index 1
    ([1, 3, 5, 6], 7)   # Target not found, insert at index 4
]
for arr, target in test_cases:
    print(f"Array: {arr}, Target: {target}, Insert Position: {search_insert_position(arr, target)}")

Search insert position

The time complexity of the algorithm is $O(\log n)$ , and the space complexity of the algorithm is $O(1)$ .

Problem 2: First and last position of an element in a sorted array#

This problem is selected because it requires finding an element’s first and the last occurrence, a common variation of binary search problems. It is frequently asked in interviews by companies like Google and Microsoft and is considered medium in terms of difficulty.

Statement: Given a sorted array of integers nums and a target value target, return the starting and ending position of target in the array. If target is not found in the array, return [-1, -1]. You must write an algorithm with $O(\log n)$ runtime complexity.

Solution: To solve this problem, we can perform two binary searches: one to find the first occurrence of the target and another to find the last occurrence. By modifying the standard binary search, we can locate these positions.

Here is the step-by-step explanation of the algorithm:

Finding the first occurrence:
- Initialize left and right pointers to the start and end of the array, respectively.
- Use a binary search to find the first occurrence:
  - Calculate the middle index mid = left + (right - left) // 2.
  - If the middle element is greater than or equal to the target, adjust the right pointer to mid - 1.
  - If the middle element is less than the target, adjust the left pointer to mid + 1.
  - If the middle element equals the target, record the index and continue searching in the left half by adjusting right = mid - 1.
Finding the last occurrence:
- Reinitialize left and right pointers to the start and end of the array, respectively.
- Use a binary search to find the last occurrence:
  - Calculate the middle index mid = left + (right - left) // 2.
  - If the middle element is less than or equal to the target, adjust the left pointer to mid + 1.
  - If the middle element is greater than the target, adjust the right pointer to mid - 1.
  - If the middle element equals the target, record the index and continue searching in the right half by adjusting left = mid + 1.
Completion:
- After both searches, return the recorded indexes of the first and last occurrences. If either search does not find the target, return [-1, -1].

Let’s visualize the above mentioned steps with an example:

def search_range(nums, target):
    # Function to find the first occurrence of the target
    def find_first(nums, target):
        left, right = 0, len(nums) - 1
        first_pos = -1
        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] >= target:
                right = mid - 1
            else:
                left = mid + 1
            if nums[mid] == target:
                first_pos = mid
        return first_pos
    # Function to find the last occurrence of the target
    def find_last(nums, target):
        left, right = 0, len(nums) - 1
        last_pos = -1
        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] <= target:
                left = mid + 1
            else:
                right = mid - 1
            if nums[mid] == target:
                last_pos = mid
        return last_pos
    # Return the positions of the first and last occurrences
    return [find_first(nums, target), find_last(nums, target)]
# Driver code with test cases
if __name__ == "__main__":
    test_cases = [
        ([5, 7, 7, 8, 8, 10], 8),
        ([5, 7, 7, 8, 8, 10], 6),
        ([], 0)
    ]
    
    for nums, target in test_cases:
        print(f"nums: {nums}, target: {target}, result: {search_range(nums, target)}")

First and last position of an element in a sorted array

The time complexity of the algorithm is $O(\log n)$ for both find_first and find_last functions, where $n$ is the number of elements in the array. Because both functions are called separately, the total time complexity is $O(\log n) + O(\log n) = O(\log n)$ . The space complexity of the algorithm is $O(1)$ because no additional space proportional to the input size is used; only a few variables are maintained.

Problem 3: Peak element in an array#

This problem is selected because it demonstrates the use of binary search in non-trivial array conditions. That showcases its efficiency in finding an element that meets specific criteria beyond simple searching. Companies like Google frequently ask it in interviews, making it important for interview preparation. It is considered medium in terms of difficulty.

Statement: Given an array of integers nums, find a peak element, and return its index. A peak element is an element that is strictly greater than its neighbors. You may imagine that nums[-1] = -∞ and nums[n] = -∞ (for the boundary elements). You must write an algorithm with O(log n) runtime complexity.

Solution: To solve this problem, we use a binary search approach to find a peak element. We continuously narrow down the search space by comparing the middle element with its neighbors and adjusting the search boundaries accordingly.

Here is the step-by-step explanation of the algorithm:

Initialization: Start with two pointers, left and right, initialized to the start and end of the array, respectively.
Iteration:
1. Calculate the middle index mid = left + (right - left) // 2.
2. Compare the middle element with its neighbors:
  1. If the middle element is greater than its right neighbor, it means the peak is on the left side, so adjust the right pointer to mid.
  2. Otherwise, the peak is on the right side, so adjust the left pointer to mid + 1.
Completion: When the left pointer meets the right pointer, it points to the peak element. Return left as the index of the peak element.

Let’s visualize the above mentioned steps with an example:

def find_peak_element(nums):
    # Initialize left and right pointers
    left, right = 0, len(nums) - 1
    while left < right:
        mid = left + (right - left) // 2
        # Check if the middle element is greater than its right neighbor
        if nums[mid] > nums[mid + 1]:
            right = mid  # Peak is on the left side
        else:
            left = mid + 1  # Peak is on the right side
    return left  # Left pointer points to the peak element
# Driver code with test cases
if __name__ == "__main__":
    test_cases = [
        [1, 2, 3, 1],
        [1, 2, 1, 3, 5, 6, 4],
        [1]
    ]
    
    for nums in test_cases:
        print(f"nums: {nums}, peak index: {find_peak_element(nums)}")

Peak element in an array

The algorithm’s time complexity is $O(\log n)$ , where $n$ is the number of elements in the array. This is because we are using a binary search approach. The algorithm’s space complexity is $O(1)$ because no additional space proportional to the input size is used; only a few variables are maintained.

Conclusion#

Mastering binary search is essential for excelling in coding interviews. As we've seen through these ten problems, understanding when and how to apply binary search can significantly reduce the time complexity of your solutions. This efficiency not only impresses interviewers but also demonstrates your ability to think critically and solve complex problems. By mastering binary search, you gain a valuable skill that sets you apart from other candidates, giving you a competitive edge in your job search.

Here’s where our mock interviews can become a game changer. We offer a range of mock interviews, including data structures interviews, allowing you to practice with various problem types. During these mock interviews, pay close attention to the interviewer’s feedback. Use the feedback provided to identify areas for improvement and refine your skills.

Frequently Asked Questions

What is a real-world example of binary search?

A real-world example of binary search is looking up a word in a dictionary. Instead of starting from the first page, you open the dictionary around the middle, check whether the word you’re searching for is before or after the current page, and then repeatedly narrow down your search range by halving it until you find the word. This efficient method significantly reduces the number of pages you need to check, much like how binary search operates on a sorted list.

Is binary search important for interviews?

Yes, binary search is important, especially in technical interviews for software engineering roles. It’s a fundamental algorithm demonstrating a candidate’s understanding of search techniques, algorithmic thinking, and efficiency. Many problems, particularly those involving sorted data or search optimizations, can be effectively solved using binary search, making it a common topic in coding interviews at companies like FAANG. Mastery of binary search also helps solve more complex problems requiring modifications or variations of the basic algorithm.

What is the best-case scenario for binary search?

The best-case scenario for binary search occurs when the target element is found at the very middle of the sorted array on the first attempt. In this case, the algorithm immediately identifies the desired element without needing to perform any further comparisons or divide the search range, resulting in a time complexity of O(1).

What is the main advantage of binary search?

The main advantage of binary search is its efficient search through large, sorted datasets. By repeatedly dividing the search range in half, binary search significantly reduces the number of comparisons needed to find a target element, achieving a time complexity of O(log n). This makes it much faster than linear search, especially when dealing with large data sets, where the speed and performance gains are substantial.

What are the different cases of binary search?

Binary search has three main cases based on its performance:

Best Case: This occurs when the target element is located in the middle of the array on the first comparison, resulting in a time complexity of O(1).

Average Case: This happens when the target element is found after a few divisions of the search range. On average, the time complexity is O(log n), as the algorithm cuts the search space in half with each step.

Worst Case: The worst case occurs when the target element is either not in the array or located at the beginning or end of the search range. In this scenario, the algorithm needs to perform the maximum number of comparisons, which results in a time complexity of O(log n).

Free Resources

Learn in-demand tech skills in half the time

PRODUCTS

Mock Interview

New

Courses

Skill Paths

Projects

Assessments

10 common binary search interview questions

Curated list of problems#

Binary search#

Code template#

Binary search interview questions list#

Problem 1: Search insert position#

Problem 2: First and last position of an element in a sorted array#

Problem 3: Peak element in an array#

Further reading#

Conclusion#

Frequently Asked Questions

What is a real-world example of binary search?

Is binary search important for interviews?

What is the best-case scenario for binary search?

What is the main advantage of binary search?

What are the different cases of binary search?