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+/*Iteration 1: Names and Input
+1.1 Create a variable hacker1 with the driver's name.*/
+let hacker1 = "Juergen";
+
+
+//1.2 Print "The driver's name is XXXX".//
+
+console.log(`The driver´s name is ${hacker1}`);
+
+//1.3 Create a variable hacker2 with the navigator's name.//
+
+let hacker2 = "AlinaSergej";
+/*Iteration 1: Names and Input
+1.1 Create a variable hacker1 with the driver's name.*/
+let hacker1 = "Juergen";
+
+
+//1.2 Print "The driver's name is XXXX".//
+
+console.log(`The driver´s name is ${hacker1}`);
+
+//1.3 Create a variable hacker2 with the navigator's name.//
+
+let hacker2 = "AlinaSergej";
+
+//1.4 Print "The navigator's name is YYYY".//
+
+console.log(`The navigator´s name is ${hacker2}`);
+
+/*Iteration 2: Conditionals
+2.1. Depending on which name is longer, print: - The Driver has the longest name, it has XX characters or - Yo, navigator got the longest name, it has XX characters or - Wow, you both got equally long names, XX characters!*/
+
+if (hacker1.length >=hacker2.length){
+  if (hacker1.length == hacker2.length){
+    console.log(`Wow, you both got equally long names, ${hacker1.length} characters!`)
+  } else {
+    console.log(`The Driver has the longest name, it has ${hacker1.length} characters`)
+  } 
+} else {
+  console.log(`Yo, navigator got the longest name, it has ${hacker2.length} characters`)
+}
+
+/*Iteration 3: Loops
+3.1 Print all the characters of the driver's name, separated by a space and in capitals i.e. "J O H N"*/
+
+let uppercasedDriver=hacker1.toUpperCase();
+/*let split="";
+. Solution:
+for (let i = 0; i<hacker1.length; i++){
+   split += `${uppercasedDriver[i]}  `;
+  
+}
+console.log(split);*/
+
+let splittedName = uppercasedDriver.split("");
+let nameFinal = splittedName.join(" ")
+console.log(nameFinal)
+
+//3.2 Print all the characters of the navigator's name, in reverse order. i.e. "nhoJ"//
+
+
+let reversedName = hacker2.split("").reverse().join("")
+console.log(reversedName)
+
+//3.3 Depending on the lexicographic order of the strings, print: - The driver's name goes first. - Yo, the navigator goes first definitely. - What?! You both got the same name?//
+
+let abc = "abcdefghijklmnopqrstuvwxyz";
+let firstLetterHacker1 = hacker1[0];
+let firstLetterHacker2 = hacker2[0];
+  if (abc.indexOf(firstLetterHacker1) < abc.indexOf(firstLetterHacker2)){
+    if(abc.indexOf(firstLetterHacker1) == abc.indexOf(firstLetterHacker2)){
+      console.log("Both have the same first letter");
+    } else {
+    console.log(`${hacker1} goes first`);
+    }
+  }
+    else{
+    console.log(`${hacker2} goes first`);
+  }
+
+/*Bonus Time!
+Go to lorem ipsum generator and:
+
+Generate 3 paragraphs. Store the text in a variable type of string.
+Make your program count the number of words in the string.
+Make your program count the number of times the Latin word et appears.*/
+//1.4 Print "The navigator's name is YYYY".//
+
+console.log(`The navigator´s name is ${hacker2}`);
+
+/*Iteration 2: Conditionals
+2.1. Depending on which name is longer, print: - The Driver has the longest name, it has XX characters or - Yo, navigator got the longest name, it has XX characters or - Wow, you both got equally long names, XX characters!*/
+
+if (hacker1.length >=hacker2.length){
+  if (hacker1.length == hacker2.length){
+    console.log(`Wow, you both got equally long names, ${hacker1.length} characters!`)
+  } else {
+    console.log(`The Driver has the longest name, it has ${hacker1.length} characters`)
+  } 
+} else {
+  console.log(`Yo, navigator got the longest name, it has ${hacker2.length} characters`)
+}
+
+/*Iteration 3: Loops
+3.1 Print all the characters of the driver's name, separated by a space and in capitals i.e. "J O H N"*/
+
+//3.2 Print all the characters of the navigator's name, in reverse order. i.e. "nhoJ"//
+
+//3.3 Depending on the lexicographic order of the strings, print: - The driver's name goes first. - Yo, the navigator goes first definitely. - What?! You both got the same name?//
+
+/*Bonus Time!
+Go to lorem ipsum generator and:
+
+Generate 3 paragraphs. Store the text in a variable type of string.
+Make your program count the number of words in the string.
+Make your program count the number of times the Latin word et appears.*/
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