Jon Bonso jsbonso

## FindTheLengthOfLongestSubstringUsingWhileLoop.java
class FindTheLengthOfLongestSubstringUsingWhileLoop {
    public int lengthOfLongestSubstring(String s) {

        /**
            Examples:

              #1 -
                eabcdabcbb = 4
                "abcd" is the longest substring
                4 is the length of "abcd"

## FindTheLengthOfLongestSubstring.java
class FindTheLengthOfLongestSubstring {
    public int lengthOfLongestSubstring(String s) {

        /**
            Examples:

              #1 -
                eabcdabcbb = 4
                "abcd" is the longest substring
                4 is the length of "abcd"

## TwoSumBruteForce.java
/**
*

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

## Palindrome.java


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!";

## TwoSumMethodsPerformanceChecks.java
/**
 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;

## TwoSumUsingHashMapSinglePass.java
/**
*
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:

## OddOrEven.java
/*

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.

## index.html
<!DOCTYPE html>
<html>
<body>

<h1>Simple Checkout with Price Rules</h1>

<div style="padding-bottom: 25px;">
    <h3>Test Case 1: 3 CARI Products</h3>
    <p>SKUs Scanned: <code>cari, cari, cari, peti </code></p>
    <p>Total expected: $249.00</p>

## How to connect Java to MySQL.java
package com.tutorialsdojo;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import java.sql.ResultSetMetaData;

/**

## VoteCount.java
import java.util.*;
import java.util.concurrent.*;
import java.util.Map.Entry;
import java.util.stream.Collectors;

public class VoteCount{
    public static void main(String[] args) {

       	// Populate Data
		String[] votes = {"Jon", "Bonso", "Stacey", "Jon", "Bonso", "Stacey", "Jon", "Bonso", "Brie", "Jon", "Bonso", "Jon"};
	class FindTheLengthOfLongestSubstringUsingWhileLoop {
	public int lengthOfLongestSubstring(String s) {

	/**
	Examples:

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

	/**
	Examples:

	#1 -
	eabcdabcbb = 4
	"abcd" is the longest substring
	4 is the length of "abcd"
	/**
	*

	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


	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!";
	/**
	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;
	/**
	*
	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:
	/*

	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.
	<!DOCTYPE html>
	<html>
	<body>

	<h1>Simple Checkout with Price Rules</h1>

	<div style="padding-bottom: 25px;">
	<h3>Test Case 1: 3 CARI Products</h3>
	<p>SKUs Scanned: <code>cari, cari, cari, peti </code></p>
	<p>Total expected: $249.00</p>
	package com.tutorialsdojo;

	import java.sql.Connection;
	import java.sql.DriverManager;
	import java.sql.ResultSet;
	import java.sql.SQLException;
	import java.sql.Statement;
	import java.sql.ResultSetMetaData;

	/**
	import java.util.*;
	import java.util.concurrent.*;
	import java.util.Map.Entry;
	import java.util.stream.Collectors;

	public class VoteCount{
	public static void main(String[] args) {

	// Populate Data
	String[] votes = {"Jon", "Bonso", "Stacey", "Jon", "Bonso", "Stacey", "Jon", "Bonso", "Brie", "Jon", "Bonso", "Jon"};