Binary search exercise solution could be further optimized #13
Description
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!