jsbonso

# Get the latest file that's added in the folder
LATEST_FILE=`ls -tp | grep -v /$ | head -1`
echo $LATEST_FILE

# Upload the new file to your Amazon S3 Bucket. 
# In this example, the "jonbonso-media" S3 bucket has an associated CloudFront distribution
# with a custom domain of "media.jonbonso.com"
aws s3 cp $LATEST_FILE s3://jonbonso-media

jsbonso

#!/bin/bash
for i in {1..250}
do 
    curl -o /dev/null \
    -w "%{time_starttransfer}\n" \
    https://google.com >> prod.txt
done

jsbonso

MY_IP=`curl checkip.amazonaws.com`
echo $MY_IP

aws ec2 authorize-security-group-ingress \
    --group-id sg-01298cde13c80c75 \
    --protocol tcp \
    --port 22 \
    --cidr $MY_IP/32 \
    --region us-east-1

jsbonso

class FindTheLengthOfLongestSubstringUsingWhileLoop {
    public int lengthOfLongestSubstring(String s) {
        
        /**
            Examples: 
          
              #1 - 
                eabcdabcbb = 4
                "abcd" is the longest substring
                4 is the length of "abcd"

jsbonso

class FindTheLengthOfLongestSubstring {
    public int lengthOfLongestSubstring(String s) {
        
        /**
            Examples: 
          
              #1 - 
                eabcdabcbb = 4
                "abcd" is the longest substring
                4 is the length of "abcd"

jsbonso

public class Palindrome
{
    public static void main(String[] args) {
        
        String input = "ababa";
        String input1 = "mom";
        String input2 = "nono";
        String input3 = "A man, a plan, a canal: Panama!";

jsbonso

/**
 This tracks the Elapsed Time for the 2 Solutions for the Classic Two Sum Problem
 
 You will discover how immensely better an (O)n solution is (twoSumOptimized) 
 than a brute force (O)n^2 solution (twoSumBruteForce) by checking the start and end 
 times of both methods
*/


import java.util.Map;

jsbonso

/**
*
Two Sum Solution #2 - Using a HashMap and iterates the array in a single pass approach

Author: Jon Bonso
  
Goal: Get the two indices from the array that sums up to the value of the given target

MATH THEORY:

jsbonso

/**
*

Two Sum Solution #1 - Brute Force Approach
  
  Approach
    - Brute force approach by using a nested loop
    - Create the two loops to process the array, wherein the second loop is one step ahead of the previous loop and traversing the array elements in a quadratic manner
    - The first loop uses the variable i to hold the first index while the second loop uses the variable j to hold the latter one.
    - Compare the values of the i and j indices. If they are the same, return the indices as output

jsbonso

/*

Simple Odd or Even Coding Example

Author: Jon Bonso 
Demo Link: https://www.online-java.com/BYJCIjwhpg

Mathematical Theory:
   - Parity is the property of an integer which determines if it is even or odd.
   - An even number is an integer that can be divided exactly by 2 is an even number.