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
Mostrar más📈 Análisis del canal de Telegram Coding Projects
El canal Coding Projects (@programming_experts) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 67 949 suscriptores, ocupando la posición 1 910 en la categoría Tecnologías y Aplicaciones y el puesto 4 739 en la región India.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 67 949 suscriptores.
Según los últimos datos del 25 octubre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 172, y en las últimas 24 horas de 6, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 3.69%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.06% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 2 510 visualizaciones. En el primer día suele acumular 717 visualizaciones.
- Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 4.
- Intereses temáticos: El contenido se centra en temas clave como |--, algorithm, array, framework, javascript.
📝 Descripción y política de contenido
El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“Channel specialized for advanced concepts and projects to master:
* Python programming
* Web development
* Java programming
* Artificial Intelligence
* Machine Learning
Managed by: @love_data”
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 26 octubre, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Tecnologías y Aplicaciones.
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}