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# Solution: The verse with the maximum number of appearances

Problem: Given a poem, with each verse on a new line, Find the verse (or verses) that have the maximum number of appearances and how many times it appears in the text.

Solution:

``````public static List<Pair> max2(String fullText) throws Exception {
// If there is no text provided, throw an exception
if (fullText == null || fullText.isBlank()) throw new Exception("No text provided");

// We split it by line separator so that we get all the verses. We also do a 'toLowerCase' so that
// we analyze it case-insensitive
String[] split = fullText.toLowerCase(Locale.ROOT).split("\n");

// Create a List from the verses (so that we can use Streams
// And a HashSet, so that we have each verse only once
List<String> verses = Arrays.asList(split);
Set<String> uniqueVerses = new HashSet<>(verses);

// Now, let's process it
return uniqueVerses.stream()
// We map each line to a Pair that has the line and the frequency it appears in the text
.map(line -> new Pair(line, Collections.frequency(verses, line)))

// We collect it and group them by the number of appearances, now, we will have groups (Map) with the
// Number of appearances as the key and the list of verses as the value
.collect(Collectors.groupingBy(p -> p.getValue()))

// We make the result a Stream so that we can find the MAX
.entrySet().stream()

// Get the group that has the maximum number of appearances
.max(Comparator.comparingInt(p -> p.getKey().intValue()))

// Retrieve the result and get the List. It will be a List of Pair<String, Integer>,
// where the Key is the verse and the value is the number of appearances
.get().getValue();
}
``````

Conclusions:
Using the power of streams, we can easily manipulate our data set to do groupings, counting, and finding the maximum. No need for complex for-loops or external libraries.