Google Coding Interview Questions and Answers
Did you know that Google ranks third on the 2024 Global 500 brand value list, with a brand worth of $333.4 billion? Landing a job at Google is a goal for many developers, but it requires thorough preparation, especially for coding interviews. Google’s coding interviews are recognized for their high evaluation process, where candidates are evaluated on their problem-solving abilities, coding skills, and ability to create efficient algorithms. Knowing the types of coding questions can also significantly aid your preparation. In this guide, we’ll delve into Google coding interview questions and answers for various experience levels and offer tips on how to effectively prepare irrespective of your professional level.
Google Interview Coding Questions and Answers for Freshers
For recent graduates or those with limited work experience, Google tends to focus on fundamental programming concepts and data structures. Here are some basic Google coding interview questions and answers to help you prepare:
Q1. Write a function to reverse a singly linked list.
Answer: To reverse a linked list, we can iterate through the list and change the next pointer of each node to its previous node. The previous node is initialized as None, and as we traverse the list, we update the pointers, eventually returning the new head of the reversed list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def reverse_linked_list(head):
prev = None
current = head
while current:
next_node = current.next
current.next = prev
prev = current
current = next_node
return prev
Q2. Check if a given string is a palindrome (reads the same forwards and backwards).
Answer: simple approach to check if a string is a palindrome is by comparing the string with its reversed version. First, we remove all non-alphanumeric characters and convert the string to lowercase to ensure a case-insensitive check.
def is_palindrome(s):
s = ''.join(char.lower() for char in s if char.isalnum())
return s == s[::-1]
# Example
print(is_palindrome("A man, a plan, a canal: Panama")) # Output: True
Q3. Given two sorted arrays, merge them into a single sorted array.
Answer: The most efficient way to merge two sorted arrays is to use two pointers. We compare the elements from both arrays and insert the smaller one into the merged result. This can be done in O(n) time, where n is the total number of elements in both arrays.
def merge_sorted_arrays(arr1, arr2):
merged = []
i, j = 0, 0
while i < len(arr1) and j < len(arr2):
if arr1[i] < arr2[j]:
merged.append(arr1[i])
i += 1
else:
merged.append(arr2[j])
j += 1
merged.extend(arr1[i:])
merged.extend(arr2[j:])
return merged
# Example
print(merge_sorted_arrays([1, 3, 5], [2, 4, 6])) # Output: [1, 2, 3, 4, 5, 6]
Q4. Given an array containing n distinct numbers in the range [0, n], find the missing number.
Answer: To find the missing number in an array of distinct integers ranging from 0 to n, we can use the formula for the sum of the first n integers: n * (n + 1) / 2. By subtracting the sum of the given array from this expected sum, we can find the missing number in O(n) time.
def missing_number(nums):
n = len(nums)
total = n * (n + 1) // 2
return total - sum(nums)
# Example
print(missing_number([3, 0, 1])) # Output: 2
Q5. Given an array of integers, find two numbers such that they add up to a specific target. Return their indices.
Answer: To solve this Google coding interview question efficiently, we can use a hash map (dictionary) to store the numbers we’ve encountered so far. For each number in the array, we check if the complement (i.e., target – num) already exists in the hash map. This method allows us to find a solution in O(n) time complexity instead of using the O(n²) approach.
def two_sum(nums, target):
num_map = {}
for i, num in enumerate(nums):
complement = target - num
if complement in num_map:
return [num_map[complement], i]
num_map[num] = i
return []
# Example
print(two_sum([2, 7, 11, 15], 9)) # Output: [0, 1]
Q6. Given a string containing only the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[‘ and ‘]’, determine if the input string is valid (properly balanced).
Answer: A stack is an ideal data structure for this problem. As we encounter opening brackets ((, {, [), we push them onto the stack. When we encounter a closing bracket (), }, ]), we pop from the stack and check if the popped element matches the expected opening bracket. If our stack is empty after processing all characters, then the string is considered balanced.
def is_valid(s):
stack = []
mapping = {')': '(', '}': '{', ']': '['}
for char in s:
if char in mapping:
top_element = stack.pop() if stack else '#'
if mapping[char] != top_element:
return False
else:
stack.append(char)
return not stack
# Example
print(is_valid("()[]{}")) # Output: True
Q7. Write a function to return the nth Fibonacci number using both iterative and recursive methods.
Answer: The iterative solution calculates the Fibonacci number in O(n) time using a bottom-up approach. The recursive solution, while simple, has exponential time complexity and is less efficient, but it’s often used for educational purposes.
Iterative Solution:
def fibonacci_iterative(n):
a, b = 0, 1
for _ in range(n):
a, b = b, a + b
return a
# Example
print(fibonacci_iterative(5)) # Output: 5
Recursive Solution:
def fibonacci_recursive(n):
if n <= 1:
return n
return fibonacci_recursive(n - 1) + fibonacci_recursive(n - 2)
# Example
print(fibonacci_recursive(5)) # Output: 5
Q8. Implement binary search to find the index of a given target element in a sorted array.
Answer: Binary search works only on sorted arrays. It repeatedly divides the search space in half by comparing the target with the middle element. This reduces the time complexity to O(log n), making it much faster than linear search methods.
def binary_search(arr, target):
left, right = 0, len(arr) - 1
while left <= right:
mid = (left + right) // 2
if arr[mid] == target:
return mid
elif arr[mid] < target:
left = mid + 1
else:
right = mid - 1
return -1
# Example
print(binary_search([1, 2, 3, 4, 5], 4)) # Output: 3
Q9. Given a sorted array, remove the duplicates in-place such that each element appears only once and return the new length.
Answer: Since the array is sorted, duplicates will always appear next to each other. By using a two-pointer technique, we can efficiently remove duplicates in-place, maintaining the time complexity of O(n).
def remove_duplicates(arr):
if not arr:
return 0
i = 0
for j in range(1, len(arr)):
if arr[j] != arr[i]:
i += 1
arr[i] = arr[j]
return i + 1 # Returns the new length
# Example
print(remove_duplicates([1, 1, 2, 2, 3])) # Output: 3
Q10. Given two integer arrays, return their intersection (elements that appear in both arrays).
Answer: To find common elements between two integer arrays, we can convert both arrays into sets and use the set intersection operation. This approach allows us to find the common elements in O(min(n, m)) time, where n and m are the lengths of the two arrays.
def intersection(arr1, arr2):
return list(set(arr1) & set(arr2))
# Example
print(intersection([1, 2, 2, 1], [2, 2])) # Output: [2]
Google Coding Interview Questions with Answers for Mid-Level Candidates
Mid-level candidates are generally expected to have a few years of industry experience, and the interview questions tend to be more complex. The focus shifts towards solving problems that require a deeper understanding of algorithms and system design. Here are some Google coding interview questions and answers for mid-level candidates:
Q11. Given a string s, find the length of the longest substring without repeating characters.
Answer: This problem can be solved using a sliding window approach. We use a hash map to store the index of each character and maintain two pointers (left and right) to represent the current window. As we move the right pointer, if we encounter a repeating character (already in the hash map), we move the left pointer to skip past it. We update the hash map with the new index of the character and track the maximum length of the window.
def length_of_longest_substring(s):
char_map = {}
left, max_length = 0, 0
for right in range(len(s)):
if s[right] in char_map and char_map[s[right]] >= left:
left = char_map[s[right]] + 1
char_map[s[right]] = right
max_length = max(max_length, right - left + 1)
return max_length
# Example
print(length_of_longest_substring("abcabcbb")) # Output: 3 ("abc")
Q12. Given an integer array nums and an integer k, return the k most frequent elements.
Answer: To find the k most frequent elements in an array, we can use a hash map to count the frequency of each element. Then, we use a min-heap to store the top k elements based on their frequencies. This allows us to efficiently retrieve the most frequent elements.
import heapq
from collections import Counter
def top_k_frequent(nums, k):
count = Counter(nums)
return heapq.nlargest(k, count.keys(), key=count.get)
# Example
print(top_k_frequent([1, 1, 1, 2, 2, 3], 2)) # Output: [1, 2]
Q13. Given an array nums, return an array output where output[i] is the product of all elements of nums except nums[i].
Answer: We can solve this problem in O(n) time without using division by performing two passes through the array. The first pass calculates the prefix product, and the second pass calculates the suffix product. We multiply these to get the result.
def product_except_self(nums):
n = len(nums)
result = [1] * n
prefix, suffix = 1, 1
for i in range(n):
result[i] *= prefix
prefix *= nums[i]
for i in range(n - 1, -1, -1):
result[i] *= suffix
suffix *= nums[i]
return result
# Example
print(product_except_self([1, 2, 3, 4])) # Output: [24, 12, 8, 6]
Q14. Find the kth largest element in an unsorted array.
Answer: To identify the kth largest element in an unsorted array, we can use a min-heap that maintains only k elements at any given time. By doing so, the root of this heap will represent the kth largest element once all elements have been processed.
import heapq
def find_kth_largest(nums, k):
return heapq.nlargest(k, nums)[-1]
# Example
print(find_kth_largest([3, 2, 1, 5, 6, 4], 2)) # Output: 5
Q15. You are given an array of integers height where height[i] represents the height of a vertical line at index i. The two lines form a container, and you need to find the maximum amount of water it can hold.
Answer: This Google coding interview question can be solved using a two-pointer approach. We initialize two pointers, one at the beginning and one at the end of the array. We calculate the area using the minimum of the two heights and the distance between them. We then move the pointer pointing to the shorter height inward, as this gives a chance to find a taller line that might form a larger area. We continue this process until the pointers meet.
def max_area(height):
left, right = 0, len(height) - 1
max_area = 0
while left < right:
width = right - left
current_area = min(height[left], height[right]) * width
max_area = max(max_area, current_area)
if height[left] < height[right]:
left += 1
else:
right -= 1
return max_area
# Example
print(max_area([1, 8, 6, 2, 5, 4, 8, 3, 7])) # Output: 49
Q16. Given an array of strings, group the anagrams together.
Answer: We can use a hash map where the key is a sorted tuple of characters. Anagrams will have the same sorted tuple, allowing us to group them.
from collections import defaultdict
def group_anagrams(strs):
anagrams = defaultdict(list)
for s in strs:
key = tuple(sorted(s))
anagrams[key].append(s)
return list(anagrams.values())
# Example
print(group_anagrams(["eat", "tea", "tan", "ate", "nat", "bat"]))
# Output: [['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]
Q17. Given an n x n matrix, rotate it 90 degrees clockwise.
Answer: First, transpose the matrix by swapping elements across the diagonal. Then, reverse each row to complete the rotation. This method ensures that all elements are repositioned correctly in just two passes over the matrix.
def rotate(matrix):
n = len(matrix)
for i in range(n):
for j in range(i, n):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
for row in matrix:
row.reverse()
# Example
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
rotate(matrix)
print(matrix) # Output: [[7, 4, 1], [8, 5, 2], [9, 6, 3]]
Q18. Given a 2D board and a word, find if the word exists in the grid.
Answer: We can solve this problem using backtracking to explore each possible path starting from every cell. If the current character matches, we move in four possible directions (up, down, left, right).
def exist(board, word):
rows, cols = len(board), len(board[0])
def backtrack(r, c, index):
if index == len(word):
return True
if r < 0 or r >= rows or c < 0 or c >= cols or board[r][c] != word[index]:
return False
temp, board[r][c] = board[r][c], '#'
found = any(backtrack(r + dr, c + dc, index + 1) for dr, dc in [(0, 1), (1, 0), (0, -1), (-1, 0)])
board[r][c] = temp
return found
for i in range(rows):
for j in range(cols):
if backtrack(i, j, 0):
return True
return False
# Example
print(exist([["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]], "ABCCED"))
# Output: True
Q19. Given a list of coin denominations and a target amount, find the minimum number of coins needed to make up that amount.
Answer: We use dynamic programming to build a table where each entry represents the minimum number of coins needed for that amount.
def coin_change(coins, amount):
dp = [float('inf')] * (amount + 1)
dp[0] = 0
for coin in coins:
for x in range(coin, amount + 1):
dp[x] = min(dp[x], dp[x - coin] + 1)
return dp[amount] if dp[amount] != float('inf') else -1
# Example
print(coin_change([1, 2, 5], 11)) # Output: 3
Q20. Given a string s containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[‘, and ‘]’, determine if the input string is valid. An input string is valid if:
- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
Answer: We can solve this problem using a stack. As we traverse the string, whenever we encounter an opening bracket, we push it onto the stack. When we encounter a closing bracket, we check if it matches the top element of the stack. If it does not match or if the stack is empty, the string is invalid. At the end, the stack should be empty if the input string is valid.
def is_valid(s):
stack = []
mapping = {')': '(', '}': '{', ']': '['}
for char in s:
if char in mapping:
top_element = stack.pop() if stack else '#'
if mapping[char] != top_element:
return False
else:
stack.append(char)
return not stack
# Example
print(is_valid("()[]{}")) # Output: True
print(is_valid("(]")) # Output: False
Google Coding Interview Questions and Answers for Experienced Professionals
For senior-level or highly experienced professionals, Google expects candidates to demonstrate strong technical expertise, problem-solving skills, and an understanding of large-scale system design. The questions may include complex coding challenges and design problems. Here are some Google coding interview questions for practice:
Q21. Given two sorted arrays, find the median of the combined array. The overall run time complexity should be O(log (m+n)).
Answer: We can solve this problem using binary search. The idea is to partition both arrays so that the left half contains the smaller elements and the right half contains the larger elements. We adjust the partition using binary search on the smaller array.
def find_median_sorted_arrays(nums1, nums2):
if len(nums1) > len(nums2):
nums1, nums2 = nums2, nums1
m, n = len(nums1), len(nums2)
left, right = 0, m
while left <= right:
partition1 = (left + right) // 2
partition2 = (m + n + 1) // 2 - partition1
maxLeft1 = float('-inf') if partition1 == 0 else nums1[partition1 - 1]
minRight1 = float('inf') if partition1 == m else nums1[partition1]
maxLeft2 = float('-inf') if partition2 == 0 else nums2[partition2 - 1]
minRight2 = float('inf') if partition2 == n else nums2[partition2]
if maxLeft1 <= minRight2 and maxLeft2 <= minRight1:
if (m + n) % 2 == 0:
return (max(maxLeft1, maxLeft2) + min(minRight1, minRight2)) / 2
else:
return max(maxLeft1, maxLeft2)
elif maxLeft1 > minRight2:
right = partition1 - 1
else:
left = partition1 + 1
# Example
print(find_median_sorted_arrays([1, 3], [2])) # Output: 2.0
Q22. Given a binary matrix filled with 0s and 1s, find the largest rectangle containing only 1s and return its area.
Answer: This problem can be reduced to the “Largest Rectangle in Histogram” problem. We maintain a heights array where each element represents the height of columns of 1s. For each row, update the heights and find the largest rectangle.
def maximal_rectangle(matrix):
if not matrix:
return 0
max_area = 0
heights = [0] * len(matrix[0])
for row in matrix:
for i in range(len(row)):
heights[i] = heights[i] + 1 if row[i] == '1' else 0
max_area = max(max_area, largest_rectangle_area(heights))
return max_area
def largest_rectangle_area(heights):
stack = []
max_area = 0
heights.append(0)
for i, h in enumerate(heights):
while stack and heights[stack[-1]] > h:
height = heights[stack.pop()]
width = i if not stack else i - stack[-1] - 1
max_area = max(max_area, height * width)
stack.append(i)
heights.pop()
return max_area
# Example
matrix = [
["1", "0", "1", "0", "0"],
["1", "0", "1", "1", "1"],
["1", "1", "1", "1", "1"],
["1", "0", "0", "1", "0"]
]
print(maximal_rectangle(matrix)) # Output: 6
Q23. Design an algorithm to serialize and deserialize a binary tree. Serialization is converting the tree to a string, and deserialization is reconstructing the tree from the string.
Answer: We can use pre-order traversal for serialization and then reconstruct the tree during deserialization using the same order. We use ‘None’ for null nodes.
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Codec:
def serialize(self, root):
def dfs(node):
if not node:
return ['None']
return [str(node.val)] + dfs(node.left) + dfs(node.right)
return ','.join(dfs(root))
def deserialize(self, data):
def dfs(nodes):
val = nodes.pop(0)
if val == 'None':
return None
node = TreeNode(int(val))
node.left = dfs(nodes)
node.right = dfs(nodes)
return node
nodes = data.split(',')
return dfs(nodes)
# Example
codec = Codec()
root = TreeNode(1, TreeNode(2), TreeNode(3, TreeNode(4), TreeNode(5)))
serialized = codec.serialize(root)
print(serialized) # Output: "1,2,None,None,3,4,None,None,5,None,None"
deserialized = codec.deserialize(serialized)
Q24. Given two words (beginWord and endWord), and a dictionary’s word list, find all the shortest transformation sequences from beginWord to endWord, such that only one letter can be changed at a time.
Answer: We use a breadth-first search (BFS) to explore the transformation sequences, while using a dictionary to backtrack and find all paths.
from collections import defaultdict, deque
def find_ladders(beginWord, endWord, wordList):
wordList = set(wordList)
if endWord not in wordList:
return []
layer = {}
layer[beginWord] = [[beginWord]]
while layer:
new_layer = defaultdict(list)
for word in layer:
if word == endWord:
return layer[word]
for i in range(len(word)):
for c in 'abcdefghijklmnopqrstuvwxyz':
new_word = word[:i] + c + word[i+1:]
if new_word in wordList:
new_layer[new_word] += [j + [new_word] for j in layer[word]]
wordList -= set(new_layer.keys())
layer = new_layer
return []
# Example
print(find_ladders("hit", "cog", ["hot", "dot", "dog", "lot", "log", "cog"]))
# Output: [["hit", "hot", "dot", "dog", "cog"], ["hit", "hot", "lot", "log", "cog"]]
Q25. Given an array of integers representing the histogram’s bar height where the width of each bar is 1, find the area of the largest rectangle in the histogram.
Answer: To solve this Google coding interview question efficiently, we use a stack-based approach. The idea is to maintain a stack of indices of the histogram bars. We push indices onto the stack until we encounter a bar that is shorter than the bar at the stack’s top. When that happens, we pop from the stack and calculate the area of the rectangle formed using the popped height.
def largest_rectangle_area(heights):
stack = []
max_area = 0
heights.append(0) # Add a sentinel value to flush out the remaining elements
for i, h in enumerate(heights):
while stack and heights[stack[-1]] > h:
height = heights[stack.pop()]
width = i if not stack else i - stack[-1] - 1
max_area = max(max_area, height * width)
stack.append(i)
return max_area
# Example
print(largest_rectangle_area([2, 1, 5, 6, 2, 3])) # Output: 10
Q26. Given an integer n, return all distinct solutions to the n-queens puzzle. Each solution contains an n x n chessboard where ‘Q’ and ‘.’ represent a queen and an empty space, respectively.
Answer: We use backtracking to place queens on the chessboard. We ensure that no two queens attack each other by checking columns, diagonals, and anti-diagonals.
def solve_n_queens(n):
def backtrack(row, cols, diagonals, anti_diagonals, board):
if row == n:
result.append([''.join(r) for r in board])
return
for col in range(n):
diag = row - col
anti_diag = row + col
if col in cols or diag in diagonals or anti_diag in anti_diagonals:
continue
cols.add(col)
diagonals.add(diag)
anti_diagonals.add(anti_diag)
board[row][col] = 'Q'
backtrack(row + 1, cols, diagonals, anti_diagonals, board)
cols.remove(col)
diagonals.remove(diag)
anti_diagonals.remove(anti_diag)
board[row][col] = '.'
result = []
board = [['.' for _ in range(n)] for _ in range(n)]
backtrack(0, set(), set(), set(), board)
return result
# Example
print(solve_n_queens(4))
# Output: [['.Q..', '...Q', 'Q...', '..Q.'], ['..Q.', 'Q...', '...Q', '.Q..']]
Q27. Given a string containing only digits, restore it by returning all possible valid IP address combinations. A valid IP address consists of four integers (0 to 255), separated by periods.
Answer: We use backtracking to generate valid IP addresses. We split the string into four parts, ensuring each part is a valid integer between 0 and 255.
def restore_ip_addresses(s):
def backtrack(start, path):
if len(path) == 4 and start == len(s):
result.append('.'.join(path))
return
if len(path) == 4 or start == len(s):
return
for i in range(1, 4):
if start + i > len(s):
break
part = s[start:start + i]
if (part[0] == '0' and len(part) > 1) or int(part) > 255:
continue
backtrack(start + i, path + [part])
result = []
backtrack(0, [])
return result
# Example
print(restore_ip_addresses("25525511135"))
# Output: ['255.255.11.135', '255.255.111.35']
Time Complexity: O(1) (fixed number of combinations) Space Complexity: O(1)
Q28. Given an n x n matrix where each row and column is sorted in ascending order, find the kth smallest element in the matrix.
Answer: We use a min-heap to efficiently extract the smallest elements. We push elements from the first row and pop from the heap until we reach the kth smallest element.
import heapq
def kth_smallest(matrix, k):
n = len(matrix)
min_heap = [(matrix[i][0], i, 0) for i in range(n)]
heapq.heapify(min_heap)
for _ in range(k - 1):
val, r, c = heapq.heappop(min_heap)
if c + 1 < n:
heapq.heappush(min_heap, (matrix[r][c + 1], r, c + 1))
return heapq.heappop(min_heap)[0]
# Example
matrix = [[1, 5, 9], [10, 11, 13], [12, 13, 15]]
print(kth_smallest(matrix, 8)) # Output: 13
Q29. Given a string and a list of words, find all starting indices of substring(s) in the given string that is a concatenation of each word in the list exactly once without any intervening characters.
Answer: We use a sliding window with a hash map to count word frequencies. The window size is fixed to the total length of all words combined.
from collections import Counter
def find_substring(s, words):
if not s or not words:
return []
word_length = len(words[0])
word_count = len(words)
total_length = word_length * word_count
word_map = Counter(words)
result = []
for i in range(word_length):
left = i
current_map = Counter()
for j in range(i, len(s) - word_length + 1, word_length):
word = s[j:j + word_length]
if word in word_map:
current_map[word] += 1
while current_map[word] > word_map[word]:
current_map[s[left:left + word_length]] -= 1
left += word_length
if j + word_length - left == total_length:
result.append(left)
else:
current_map.clear()
left = j + word_length
return result
# Example
print(find_substring("barfoothefoobarman", ["foo", "bar"])) # Output: [0, 9]
Q30. Given two strings s and t, find the minimum window in s which contains all the characters of t. If there is no such window, return an empty string.
Answer: We use a sliding window and a hash map to track the required and current counts of characters.
from collections import Counter
def min_window(s, t):
if not s or not t:
return ""
t_count = Counter(t)
current_count = Counter()
left, right = 0, 0
min_length = float('inf')
start = 0
required = len(t_count)
formed = 0
while right < len(s):
char = s[right]
current_count[char] += 1
if char in t_count and current_count[char] == t_count[char]:
formed += 1
while left <= right and formed == required:
char = s[left]
if right - left + 1 < min_length:
min_length = right - left + 1
start = left
current_count[char] -= 1
if char in t_count and current_count[char] < t_count[char]:
formed -= 1
left += 1
right += 1
return s[start:start + min_length] if min_length != float('inf') else ""
# Example
print(min_window("ADOBECODEBANC", "ABC")) # Output: "BANC"
How to Prepare for Coding Interview Questions at Google?
Getting ready for a coding interview at Google requires a thoughtful strategy that balances essential skills with advanced problem-solving skills. Here are a few tips to help you prepare for your next coding interview at Google:
- Master Data Structures and Algorithms: Google emphasizes a solid understanding of data structures such as arrays, linked lists, trees, graphs, stacks, and queues. To help you prepare, practice solving problems on platforms like LeetCode, HackerRank, and Codeforces.
- Practice Mock Interviews: Engage in mock interviews with friends or use platforms like Pramp and Interviewing.io to replicate the Google interview experience. Pay attention to articulating your thought process and refining your solutions.
- Learn System Design: For those with more experience, system design questions are vital. Get acquainted with scalable architecture, load balancing, and caching strategies.
- Review Previous Interview Experiences: Look through the experiences shared by past Google interview candidates on sites like Glassdoor. This can provide valuable insights into the types of questions you might encounter and the overall interview format.
- Build a Strong Portfolio: Highlight your coding abilities through personal projects, contributions to open-source initiatives, or by participating in competitive coding events. A robust GitHub profile can leave a positive impression on interviewers.
Pro Tip: To prepare better, you can also enroll in our online course on how to ace the coding interviews to land your dream job at Google. Other than that, you can also check out our placement-guaranteed full-stack developer training course to get started as a coder.
Job Roles Available for Developers at Google
Getting a developer job at Google is a significant achievement for most developers. This is due to the company’s its reputation as a leader in technology and innovation. Google offers roles suited to different experience levels.
Here is a list of job roles available for developers at Google:
| Job Roles | Experience |
| Software Engineering Intern | 0 years (Student/Graduate) |
| Associate Software Engineer | 0-1 years |
| Software Engineer | 2-5 years |
| Frontend Engineer | 2-5 years |
| Backend Engineer | 2-5 years |
| Senior Software Engineer | 5-10 years |
| Site Reliability Engineer (SRE) | 5-10 years |
| Data Engineer | 3-8 years |
| Machine Learning Engineer | 3-8 years |
| Staff Software Engineer | 10+ years |
Conclusion
Craking a Google coding job interview questions is easy with the right preparation and attitude. By familiarizing yourself with the types of questions asked at various experience levels and sharpening your problem-solving abilities, you can boost your chances of securing a developer position at one of the most prestigious tech companies in the world. Keep in mind that regular practice, a solid understanding of the basics, and a well-organized study plan are essential for excelling in the Google coding interview. To explore more career opportunities you can also read our blog on Accenture associate software engineer interview questions and enhance your career prospects.
