Binary search exercise solution could be further optimized

[Kimonili]

Instead of linearly searching for all the occurrences we could do that faster by searching them all in a binary way.

Here:

while i >=0:
    if numbers[i] == number_to_find:
        indices.append(i)
    else:
        break
    i = i - 1

And here:

while i<len(numbers):
    if numbers[i] == number_to_find:
        indices.append(i)
    else:
        break
    i = i + 1

I would be happy to make a pull request 😄

PS: Your tutorial playlist on DSA is amazing!

[Muhammad-ismail786]

thanks bhai jan