File Transfer Protocol

Here is a project I worked on with python sockets for transferring information to and from a server. I will continue to work on functionalities such as transferring different file types and other improvements. 

Project Description:

This project gives an example of a file transfer protocol client-server model.

Contributors (github)

Austin Suarez: austinmsuarez
Blake Molina:  molinab297
Paul Miller:     
Hassan Hamod:
Curtis Laudenslayer: ibcurly

Language Used

Python 3.6.4 was used for the implementation of this project

How to Execute Program

Open two terminal windows, one for client and one for server.
In one window, type python3 <hostname> <port number>
In the other window, type python3 <port number>


ftp> get <file name> (downloads file <file name> from the server)
ftp> put <filename> (uploads file <file name> to the server)
ftp> ls (lists files on the server)
ftp> lls (lists files on the client)
ftp> quit (disconnects from the server and exits)

Reverse Words in a String

Problem 🤔

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

Input: “Let’s take LeetCode contest”

Output: “s’teL ekat edoCteeL tsetnoc”

Note: In the string, each word is separated by single space and there will not be any extra space in the string.


def reverseWords(self, s):
    s = s.split(" ")

    for words in range(0,len(s)):
        s[words] = s[words][::-1]

    s = ' '.join(s)
    return s

Developer’s Notes 👨‍💻

Here I took advantage of what python had to offer such as the split and join methods.

I first split the string where there was a ” ” and put them into a list.

I then cycled through the list reassigning the index with the reverse of the word.

    I reversed the word using a reverse step method s[begin:end: step] with step being -1

After I had reversed all the words I joined them back into a string with a space between each index.

Array Partition

Problem 🤔

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

Input: [1,4,3,2]

Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).


def quickSort(self,nums, low, high):
    if low < high:
        part = self.partition(nums, low,high)

def partition(self, nums, low, high):
    i = (low - 1)
    pivot = nums[high]
    for j in range(low, high):
        if nums[j] <= pivot:
            i += 1
            nums[i],nums[j] = nums[j],nums[i]
    nums[i+1],nums[high] = nums[high],nums[i+1]

def arrayPairSum(self, nums):
    temp = 0
    for index in range(0,len(nums)-1,2):
        temp += min(nums[index],nums[index+1])
    return temp

Developer’s Notes 👨‍💻

For this I ordered the list using a quicksort, then I cycled through the array took the pairs of numbers and summed the mins. I ordered the array since it would give the largest possible minimum for each pair.




Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123

Output:  321

Example 2:

Input: -123

Output: -321

Example 3:

Input: 120

Output: 21


def reverse(self, x):

    y = str(x)

    #checks the number if negative it removes the - sign

    if y[0] == '-':

        negative = True

        y = y[1:]


            negative = False

    #reverses the string [begin:end:step]

    x = y[::-1]

    #if it is negative reverse the sign

    if negative:

        x =  '-' + x

    #checks to make sure it doesnt overflow the 32bit

    if(abs(int(x)) > (2 ** 31 - 1)):

        x = 0


Developer’s  Notes

There are a couple of ways to tackle this problem. Here I decide to take advantage of Python’s ability of string manipulation.

I converted the number to a string, checked for a negative, then reversed the string adding the negative if necessary.  At the end of the script, I made sure it was less than the 32bit requirement by a size comparison and returned the result.

Hamming Distance


The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

1   (0 0 0 1)

4   (0 1 0 0)



def distanceFinder(self,x,y,count):

    if x%2 != y%2:



    if x/2 == 0 and y/2 == 0:

        return count

    return self.distanceFinder(int(x/2),int(y/2),count)

def hammingDistance(self, x, y):

    return self.distanceFinder(x,y,0)

Developer’s Notes

For this problem, I took a basic recursion approach and tested the distance by finding if then modulus of 2 is different. If it is different then I increment a counter. Once the numbers are 0 then it returns the counter. Since the algorithm uses recursion and depths by half each time the BigO looks to be O(log x). It is due to the fact that x and y are the same lengths.

Back to Basics

It has been a while since I have posted and for somewhat good reasons. This past couple of weeks I have been bombarded with interviews, projects, and an assortment of random things. I have also taken some time to reflect on what I want this blog to be about. At the basic level, I want it to be a place where I can work out my thoughts and post interesting problems and that’s what I’m gonna do. Hope anyone reading this enjoys or can follow my randomness! 🙃