From 2390cb3375065c3ecf8ac1d3a8321337979144a1 Mon Sep 17 00:00:00 2001
From: Packy Anderson <packy@cpan.org>
Date: Wed, 13 Sep 2023 00:11:32 -0400
Subject: Add challenge 234 solutions in Java

---
 challenge-234/packy-anderson/java/Ch1.java | 81 ++++++++++++++++++++++++++++++
 1 file changed, 81 insertions(+)
 create mode 100644 challenge-234/packy-anderson/java/Ch1.java

(limited to 'challenge-234/packy-anderson/java/Ch1.java')

diff --git a/challenge-234/packy-anderson/java/Ch1.java b/challenge-234/packy-anderson/java/Ch1.java
new file mode 100644
index 0000000000..70bb98fa43
--- /dev/null
+++ b/challenge-234/packy-anderson/java/Ch1.java
@@ -0,0 +1,81 @@
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.HashSet;
+import java.util.Set;
+
+// Press Shift twice to open the Search Everywhere dialog and type `show whitespaces`,
+// then press Enter. You can now see whitespace characters in your code.
+public class Ch1 {
+    public static HashMap<Character, Integer> charFrequency(String word) {
+        HashMap<Character, Integer> freq = new HashMap<Character, Integer>();
+        for (int i=0; i < word.length(); i++) {
+            char c = word.charAt(i);
+            freq.put(c, freq.getOrDefault(c,0)+1);
+        }
+        return freq;
+    }
+
+    public static ArrayList<String> commonCharacters(String[] words) {
+        // get the character frequencies for each word in words
+        ArrayList<HashMap<Character, Integer>> freq =
+                new ArrayList<HashMap<Character, Integer>>();
+        for (String word : words) {
+            freq.add(charFrequency(word));
+        }
+        // grab the character frequency map for the first word
+        HashMap<Character, Integer> first = freq.remove(0);
+        // now check the characters in the first word against
+        // the characters in all the subsequent words
+        Set<Character> keys = new HashSet<>(first.keySet());
+        for (Character c : keys) {
+            for (HashMap<Character, Integer> subsequent : freq) {
+                if (! subsequent.containsKey(c)) {
+                    // this character isn't in subsequent words,
+                    // so let's remove it from the frequency map
+                    // of the first word
+                    first.remove(c);
+                }
+                else {
+                    // the character IS in subsequent words,
+                    // so let's set the frequency count to be
+                    // the minimum count found in those words
+                    first.put(c, Math.min(first.get(c), subsequent.get(c)));
+                }
+            }
+        }
+        // now we generate a list of characters in the order they
+        // appear in the first word
+        ArrayList<String> output = new ArrayList<String>();
+        // once again, loop over the characters in the first word
+        for (int i=0; i < words[0].length(); i++) {
+            Character c = words[0].charAt(i);
+            if (first.containsKey(c)) {
+                if (first.get(c) > 1) {
+                    first.put(c, first.get(c) - 1);
+                }
+                else {
+                    first.remove(c);
+                }
+                output.add("" + c);
+            }
+        }
+        return output;
+    }
+
+    public static void solution(String[] words) {
+        System.out.println("Input: @words = (\"" + String.join("\", \"", words) + "\")");
+        ArrayList<String> output = commonCharacters(words);
+        System.out.println("Output: (\"" + String.join("\", \"", output) + "\")");
+    }
+
+    public static void main(String[] args) {
+        System.out.println("Example 1:");
+        solution(new String[] {"java", "javascript", "julia"});
+
+        System.out.println("\nExample 2:");
+        solution(new String[] {"bella", "label", "roller"});
+
+        System.out.println("\nExample 3:");
+        solution(new String[] {"cool", "lock", "cook"});
+    }
+}
\ No newline at end of file
-- 
cgit