Coding Projects
Channel specialized for advanced concepts and projects to master: * Python programming * Web development * Java programming * Artificial Intelligence * Machine Learning Managed by: @love_data
إظهار المزيد📈 نظرة تحليلية على قناة تيليجرام Coding Projects
تُعد قناة Coding Projects (@programming_experts) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 67 949 مشتركاً، محتلاً المرتبة 1 910 في فئة التكنولوجيات والتطبيقات والمرتبة 4 739 في منطقة الهند.
📊 مؤشرات الجمهور والحراك
منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 67 949 مشتركاً.
بحسب آخر البيانات بتاريخ 25 أكتوبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار 172، وفي آخر 24 ساعة بمقدار 6، مع بقاء الوصول العام مرتفعاً.
- حالة التحقق: غير موثّقة
- معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 3.69%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.06% من ردود الفعل نسبةً إلى إجمالي المشتركين.
- وصول المنشورات: يحصل كل منشور على متوسط 2 510 مشاهدة. وخلال اليوم الأول يجمع عادةً 717 مشاهدة.
- التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 4.
- الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل |--, algorithm, array, framework, javascript.
📝 الوصف وسياسة المحتوى
يصف المؤلف القناة بأنها مساحة للتعبير عن الآراء الذاتية:
“Channel specialized for advanced concepts and projects to master:
* Python programming
* Web development
* Java programming
* Artificial Intelligence
* Machine Learning
Managed by: @love_data”
بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 26 أكتوبر, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التكنولوجيات والتطبيقات.
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:
6arr = [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}