Coding Projects
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Kanal postlari
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| 2 | What will this code print?
text = "I like Java"
new_text = text.replace("Java", "Python") print(new_text) | 1 141 |
| 3 | Which method is used to split a string into smaller parts? | 1 057 |
| 4 | What will be the output?
text = "Programming" print(text[0:4]) | 1 043 |
| 5 | What is the index of the first character in a Python string? | 1 005 |
| 6 |  𝗗𝗮𝘁𝗮 𝗦𝗰𝗶𝗲𝗻𝗰𝗲 & 𝗠𝗮𝗰𝗵𝗶𝗻𝗲 𝗟𝗲𝗮𝗿𝗻𝗶𝗻𝗴 𝗙𝗥𝗘𝗘 𝗢𝗻𝗹𝗶𝗻𝗲 𝗠𝗮𝘀𝘁𝗲𝗿𝗰𝗹𝗮𝘀𝘀 📊
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📢 Share this opportunity with your friends and classmates. | 1 590 |
| 7 |  𝗠𝗜𝗖𝗥𝗢𝗦𝗢𝗙𝗧 𝗙𝗥𝗘𝗘 𝗗𝗔𝗧𝗔 𝗦𝗖𝗜𝗘𝗡𝗖𝗘 𝗥𝗘𝗦𝗢𝗨𝗥𝗖𝗘𝗦 𝟮𝟬𝟮𝟲! 🎓
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| 8 |  𝗠𝗜𝗖𝗥𝗢𝗦𝗢𝗙𝗧 𝗙𝗥𝗘𝗘 𝗗𝗔𝗧𝗔 𝗦𝗖𝗜𝗘𝗡𝗖𝗘 𝗥𝗘𝗦𝗢𝗨𝗥𝗖𝗘𝗦 𝟮𝟬𝟮𝟲! 🎓
Microsoft offers valuable FREE learning resources to help you develop in-demand technical skills! 💻
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✓ Show your resume, projects, and links
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Frontend: React
Backend: Node.js + Express
Database: MongoDB
✓ Users can write, edit, and delete posts
3️⃣ Task Manager
Frontend: Vue.js
Backend: Django REST
Database: PostgreSQL
✓ Add, update, mark complete/incomplete tasks
4️⃣ E-commerce Store
Frontend: Next.js
Backend: Express.js
Database: MongoDB
✓ Product listing, cart, payment (Stripe API)
5️⃣ Chat App (Real-time)
Frontend: React
Backend: Node.js + Socket.io
✓ Users can send/receive messages live
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Frontend: HTML + Bootstrap
Backend: Flask
✓ Admin can post jobs, users can apply
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Frontend: Vanilla JS
Backend: Express + JWT
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Frontend: React
Backend: Node + MongoDB
✓ Create, edit, and preview markdown notes
💬 Tap ❤️ for more! | 1 888 |
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| 12 | Time complexity:
O(n)
This is a very important array interview problem because it tests whether you can optimize a seemingly expensive problem.
🔹 14. What is the difference between a subarray and a subsequence?
Answer:
A subarray must contain contiguous elements.
Example:
[2, 3, 4]
is a subarray of:
[1, 2, 3, 4, 5]
A subsequence does not need to be contiguous, but the original order must be maintained.
Example:
[1, 3, 5]
is a subsequence of:
[1, 2, 3, 4, 5]
but it is not a contiguous subarray.
This distinction becomes extremely important in Dynamic Programming problems such as Longest Common Subsequence.
🔹 15. How do you rotate an array?
Example:
Input:
[1, 2, 3, 4, 5]
Rotate right by 2:
[4, 5, 1, 2, 3]
One efficient approach uses the reversal technique.
Conceptually:
Reverse entire array
Reverse first k elements
Reverse remaining elements
For:
[1, 2, 3, 4, 5]
rotate right by "2":
Reverse all:
[5, 4, 3, 2, 1]
Reverse first 2:
[4, 5, 3, 2, 1]
Reverse remaining:
[4, 5, 1, 2, 3]
This technique can achieve:
Time: O(n)
Space: O(1)
Learn to recognize patterns:
✔ Two Pointers
✔ Sliding Window
✔ Prefix Sum
✔ HashMap / HashSet
✔ Binary Search
✔ Sorting
✔ Kadane's Algorithm
✔ In-place modification
Once you recognize the pattern, many seemingly different array problems become much easier to solve.
Double Tap ❤️ For More
-----
2.86 ₽ · /balance_help | 1 652 |
| 13 | Average time complexity:
O(n)
Additional space:
O(n)
This is usually faster than comparing every pair, which can take O(n²).
🔹 8. What is the Two Sum problem?
Answer:
You are given an array and a target value. Find two elements whose sum equals the target.
Example:
Array: [2, 7, 11, 15]
Target: 9
The answer is:
2 + 7 = 9
A HashMap can solve this efficiently.
arr = [2, 7, 11, 15]
target = 9
seen = {}
for i, num in enumerate(arr):
complement = target - num
if complement in seen:
print(seen[complement], i)
break
seen[num] = i
Output:
0 1
Average time complexity:
O(n)
Space complexity:
O(n)
This is one of the most important patterns to recognize in array interviews.
🔹 9. How do you move all zeroes to the end?
Example:
Input:
[0, 1, 0, 3, 12]
Output:
[1, 3, 12, 0, 0]
A Two Pointer approach can solve this in-place.
arr = [0, 1, 0, 3, 12]
position = 0
for i in range(len(arr)):
if arr[i] != 0:
arr[position], arr[i] = arr[i], arr[position]
position += 1
print(arr)
Output:
[1, 3, 12, 0, 0]
Time complexity:
O(n)
Space complexity:
O(1)
🔹 10. How do you find the missing number?
Suppose the array contains numbers from "0" to "n", but one number is missing.
Example:
[3, 0, 1]
The missing number is:
2
One efficient approach uses the expected sum:
arr = [3, 0, 1]
n = len(arr)
expected_sum = n * (n + 1) // 2
actual_sum = sum(arr)
missing = expected_sum - actual_sum
print(missing)
Output:
2
Time complexity:
O(n)
Space complexity:
O(1)
🔹 11. What is the difference between Linear Search and Binary Search?
Answer:
Linear Search checks elements one by one.
Time: O(n)
It can work on unsorted data.
Binary Search repeatedly divides the search space approximately in half.
Time: O(log n)
It requires suitable sorted data.
For example:
Linear Search:
10 → 20 → 30 → 40 → 50
Binary Search:
30
/ \
10 40...
The key interview point is not simply memorizing the complexities. You should understand when Binary Search can actually be applied.
🔹 12. What is an in-place algorithm?
Answer:
An in-place algorithm modifies the input data structure while using only a small amount of additional memory.
For example, reversing an array using two pointers:
arr = [1, 2, 3, 4]
left = 0
right = len(arr) - 1
while left < right:
arr[left], arr[right] = arr[right], arr[left]
left += 1
right -= 1
No second array is created.
Therefore, the extra space used is:
O(1)
🔹 13. What is the Maximum Subarray problem?
Answer:
You are given an array of integers and need to find the contiguous subarray with the maximum sum.
Example:
[-2, 1, -3, 4, -1, 2, 1, -5, 4]
The maximum-sum contiguous subarray is:
[4, -1, 2, 1]
Its sum is:
6
A famous solution uses Kadane's Algorithm.
arr = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
current = arr[0]
maximum = arr[0]
for i in range(1, len(arr)):
current = max(arr[i], current + arr[i])
maximum = max(maximum, current)
print(maximum)
Output:
6 | 1 162 |
| 14 | 📦 Common Array Interview Questions with Answers
After understanding the fundamentals of Arrays, let's look at the questions frequently asked in coding and technical interviews.
🔹 1. What is an Array?
Answer:
An array is a data structure used to store multiple elements in an ordered sequence.
Each element can generally be accessed using an index.
Example:
arr = [10, 20, 30, 40]
print(arr[2])
Output:
30
The index of "30" is "2".
The main advantage of an array is efficient index-based access.
🔹 2. What is the time complexity of accessing an array element?
Answer:
Accessing an element by its index is typically:
O(1)
Example:
arr = [10, 20, 30, 40, 50]
print(arr[3])
The system can directly access the element at index "3" rather than searching through all previous elements.
This is one of the biggest advantages of arrays.
🔹 3. What is the difference between an Array and a Linked List?
Answer:
An array stores elements in a contiguous structure in traditional array implementations, while a linked list stores elements as separate nodes connected using references/pointers.
Arrays provide fast index-based access:
arr[i] → O(1)
Linked lists generally require traversal to reach a particular position:
Access → O(n)
However, linked lists can be advantageous when inserting or deleting nodes at a known position because elements do not need to be shifted like they may in an array.
The choice depends on the problem.
🔹 4. How do you find the largest element in an array?
Answer:
Traverse the array while maintaining the largest value found so far.
Example:
arr = [10, 45, 23, 67, 12]
largest = arr[0]
for num in arr:
if num > largest:
largest = num
print(largest)
Output:
67
Time complexity:
O(n)
Space complexity:
O(1)
🔹 5. How do you find the second-largest element?
Answer:
We can maintain two variables:
largest
second_largest
Example:
arr = [10, 45, 23, 67, 12]
largest = float('-inf')
second_largest = float('-inf')
for num in arr:
if num > largest:
second_largest = largest
largest = num
elif num > second_largest and num != largest:
second_largest = num
print(second_largest)
Output:
45
This can be solved in:
Time: O(n)
Space: O(1)
An important interview follow-up is:
"What if all elements are equal?"
A good solution should explicitly handle the case where a distinct second-largest element does not exist.
🔹 6. How do you reverse an array?
Answer:
One approach is to use two pointers.
Example:
arr = [1, 2, 3, 4, 5]
left = 0
right = len(arr) - 1
while left < right:
arr[left], arr[right] = arr[right], arr[left]
left += 1
right -= 1
print(arr)
Output:
[5, 4, 3, 2, 1]
Time complexity:
O(n)
Space complexity:
O(1)
This is an example of an in-place algorithm because we modify the original array instead of creating another array.
🔹 7. How do you find duplicate elements in an array?
Answer:
A HashSet can be used to keep track of elements already encountered.
Example:
arr = [1, 2, 3, 2, 4, 1]
seen = set()
duplicates = set()
for num in arr:
if num in seen:
duplicates.add(num)
else:
seen.add(num)
print(duplicates)
Output:
{1, 2} | 1 171 |
| 15 |  𝗜𝗕𝗠 𝗙𝗥𝗘𝗘 𝗧𝗘𝗖𝗛 𝗖𝗢𝗨𝗥𝗦𝗘𝗦 𝟮𝟬𝟮𝟲! 🎓
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