Mastering String Manipulation and Nested Loops in Kotlin
Introduction
Hello, and welcome to today's lesson! We are going to unravel a compelling challenge that will refine our skills in string manipulation. This lesson will place particular emphasis on nested loops. Prepare yourself for an intriguing adventure as we explore how to extract odd-indexed characters from each word in a sentence, but only if the word has an even number of characters. Does that sound exciting? Let's dive in!
Task Statement
The task we'll be demonstrating is as follows: We will work with a string representing a sentence in which words are separated by spaces. Your challenge involves creating a Kotlin function that identifies the odd-indexed characters of words that have an even number of characters, and then combines these characters into a single string, maintaining the order in which they appeared in the sentence.
Consider this example: "Kotlin programming language is versatile." The word Kotlin has 6 characters (an even number), and we'll select the odd-indexed characters from this word: specifically, o, l, and n. Similarly, we'll select a, g, a, and e from language, s from is, and e, s, t, l, and . from versatile.. We'll skip the word programming because it has an odd length.
If our function is working correctly, it should return "olnageesestl.". Isn't it fascinating to see what we can extract from a simple sentence?
Solution Building: Step 1
We will start by splitting the sentence into words using the split method. This will help us create a list containing all the words in the sentence.
Solution Building: Step 2
We now introduce nested loops: an outer loop that iterates over every single word, and an inner loop that checks every character within each word. First, we'll use an if condition to verify whether a word has an even length. We can determine this by checking whether the length of the word mod 2 equals zero. If it does, the word has an even length!
