Interview

Ace Your Next Interview

Get ready to nail your tech interviews with expert tips and advice.

Interview preparation is key to landing your dream job in tech. Whether you’re preparing for a technical coding challenge or a behavioral interview, being well-prepared can make all the difference. This page offers everything you need to excel in your interview—from common coding questions and system design scenarios to essential soft skills tips. Our guide provides actionable advice and insider tips to help you build confidence and stand out to recruiters.

We’ve got you covered with practice questions, real-world examples, and proven strategies to help you master both technical and non-technical interviews.

Common Technical Interview Questions

Prepare for the most asked coding challenges and technical queries

1. Explain how a binary search algorithm works. Can you implement it?

How to Answer the Binary Search Question:

When asked to explain the binary search algorithm, the interviewer wants to assess your understanding of:

  1. Algorithmic efficiency (why binary search is faster than linear search).
  2. How you handle sorted data.
  3. Your ability to explain edge cases and implementation logic.

Key Points to Include:

  • Concept: Binary search works on a sorted array by repeatedly dividing the search interval in half. Start by checking the middle element, and if the target value is smaller, eliminate the right half; if larger, eliminate the left half. Repeat until the target is found or the search interval is empty.

  • Efficiency: Explain that binary search has a time complexity of O(log n), which is much faster than O(n) of a linear search for large datasets.

  • Edge cases: Discuss what happens when the array is empty, when the element is at the start/end, or when the element is not found.

Sample Answer:

Explanation:

“Binary search is an efficient algorithm used to find an element in a sorted array by repeatedly dividing the search interval in half. It starts by comparing the target value to the middle element of the array. If the target is equal to the middle element, the search is complete. If the target is smaller, the algorithm discards the right half and continues searching the left half. If the target is larger, the left half is discarded, and the search continues on the right half. This process repeats until the element is found or the search interval becomes empty.”

“Binary search works only on sorted arrays and has a time complexity of O(log n) because it halves the search space with every step.”

Python Code Snippet:

def binary_search(arr, target):
   # Initialize pointers to the start and end of the array
    left, right = 0, len(arr) – 1

    while left <= right:
        # Find the middle index
        mid = (left + right) // 2

        # Check if the middle element is the target
        if arr[mid] == target:
            return mid # Target found at index mid

        # If the target is smaller, search the left half
        elif arr[mid] > target:
            right = mid – 1

        # If the target is larger, search the right half
        else:
            left = mid + 1

    # If the loop ends, the target is not in the array
    return -1

# Example usage:
arr = [1, 3, 5, 7, 9, 11]
target = 7
result = binary_search(arr, target)

if result != -1:
print(f”Target found at index {result}”)
else:
print(“Target not found”)

2. How would you reverse a linked list

How to Answer the Question:

When asked to reverse a linked list, the interviewer wants to:

  1. Assess your understanding of linked list operations.
  2. See how you handle pointers or references in a data structure.
  3. Evaluate your ability to walk through the problem step-by-step.

Key Points to Include:

  • Concept: In a singly linked list, each node points to the next node. To reverse it, you need to reverse the direction of these pointers so that the head becomes the tail and vice versa.
  • Approach: You’ll need to traverse the list and at each step, reverse the pointer of the current node to point to the previous node.
  • Efficiency: Reversing a linked list has a time complexity of O(n) since you only need one pass through the list to reverse the links.

Sample Answer:

Explanation:

“To reverse a singly linked list, we need to iterate through the list and reverse the direction of the links between the nodes. This can be done by keeping track of three pointers: previous, current, and next. As we traverse the list, we change the next pointer of the current node to point to the previous node. This continues until the list is fully reversed, and the original head becomes the new tail, and we update the head pointer to the new start of the list.”

“The time complexity is O(n) because we visit each node exactly once.”

Python Code Snippet:

# Definition for singly-linked list node
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next

def reverse_linked_list(head):
# Initialize pointers: previous, current, and next
prev = None
current = head

while current is not None:
# Store the next node temporarily
next_node = current.next

# Reverse the current node’s pointer
current.next = prev

# Move the previous and current pointers one step forward
prev = current
current = next_node

# Return the new head of the reversed list (prev points to the last node)
return prev

# Example usage:
def print_list(head):
while head:
print(head.val, end=” -> “)
head = head.next
print(“None”)

# Creating a linked list: 1 -> 2 -> 3 -> 4 -> None
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)

print(“Original List:”)
print_list(head)

# Reversing the linked list
reversed_head = reverse_linked_list(head)

print(“Reversed List:”)
print_list(reversed_head)

Explanation of the Code:

  1. prev is initialized to None because the new tail will point to None.
  2. current starts at the head of the list, and next_node temporarily stores the next node to maintain traversal.
  3. The current.next pointer is reversed to point to prev.
  4. The pointers are moved forward: prev becomes current, and current becomes next_node.
  5. The loop continues until all nodes are reversed, and prev will now point to the new head of the list.
  6. The new head is returned, and the linked list is printed to confirm the reversal.

This approach runs in O(n) time and uses O(1) space, making it an efficient and common solution to reverse a linked list.

3. How would you find the longest substring without repeating characters?

How to Answer the Question:

When asked this question, the interviewer wants to:

  1. Test your understanding of strings and substrings.
  2. Assess your knowledge of sliding window techniques.
  3. Evaluate how efficiently you can solve problems related to time complexity.

Key Points to Include:

  • Concept: The problem is about finding the longest substring in a given string that does not have any repeating characters.
  • Approach: A sliding window technique works well here. You can use two pointers (or indices) to represent the current substring, and a set (or dictionary) to track characters and their positions.
  • Efficiency: The sliding window approach works in O(n) time because each character is processed at most twice (once when added to the window and once when removed).

Sample Answer:

Explanation:

“To solve this problem, we can use the sliding window technique. We’ll maintain a window with two pointers, left and right, that represent the current substring without repeating characters. As we move the right pointer through the string, we’ll check if the character is already in our window (using a set or dictionary). If it is, we’ll move the left pointer to shrink the window until there are no duplicates. The size of the window gives us the length of the substring, and we update the maximum length as we go.”

“The time complexity of this approach is O(n), where n is the length of the string, since we only traverse the string once.”

Python Code Snippet:

def longest_substring_without_repeating(s):

    # Dictionary to store the last positions of characters

    char_index = {}

    # Initialize pointers and max length

    left = 0

    max_len = 0

    for right in range(len(s)):

        # If character is already in the window, move the left pointer

        if s[right] in char_index and char_index[s[right]] >= left:

            left = char_index[s[right]] + 1

        # Update the last seen index of the current character

        char_index[s[right]] = right

        # Calculate the length of the current window

        max_len = max(max_len, right – left + 1)

    return max_len

# Example usage:

s = “abcabcbb”

result = longest_substring_without_repeating(s)

print(f”The length of the longest substring without repeating characters is: {result}”)

Explanation of the Code:

  1. char_index is a dictionary that stores the most recent index of each character as we traverse the string.
  2. left and right pointers represent the start and end of the current substring.
  3. If the character at right is already in the substring (i.e., it exists in char_index and its value is within the current window), we update left to shrink the window.
  4. right pointer moves through the string, and at each step, we calculate the length of the current window and update max_len to store the maximum length seen so far.
  5. The final result is the length of the longest substring without repeating characters.

Example Output:

The length of the longest substring without repeating characters is: 3

The longest substring without repeating characters in “abcabcbb” is “abc”, which has a length of 3.

Key Edge Cases:

  • Empty string: Should return 0.
  • String with all unique characters: The whole string is the longest substring.
  • String with all the same character: Longest substring is of length 1.

This approach runs in O(n) time and uses O(min(n, m)) space, where n is the length of the string and m is the number of distinct characters.