Chap 22, Ex: 02

Author: HarryDulaney

ch_22/Exercise22_01.java

@@ -13,7 +13,9 @@
  * Enter a string: abcabcdgabmnsxy
  * abmnsxy
  * <p>
- * _____________________________________________
+ * __________________________________________________________________________________________
+ * ----------------------------- Time Complexity Analysis -----------------------------------
+ * __________________________________________________________________________________________
  * If input string is length 20:
  * ---- T(n) = 20 * (input string loop) * (substring loop) = O(n^2) -----
  * Time complexity is O(n^2).
@@ -39,7 +41,6 @@ public static void main(String[] args) {
             }
         }
         System.out.println(maxSub);
-
         input.close();
     }
 

ch_22/Exercise22_02.java

@@ -0,0 +1,43 @@
+package ch_22;
+
+import java.util.Scanner;
+
+/**
+ * *22.2 (Maximum increasingly ordered subsequence) Write a program that prompts
+ * the user to enter a string and displays the maximum increasingly ordered subsequence of characters.
+ * Analyze the time complexity of your program. Here is a sample run:
+ * <p>
+ * Enter a string: Welcome
+ * Welo
+ * __________________________________________________________________________________________
+ * ----------------------------- Time Complexity Analysis -----------------------------------
+ * __________________________________________________________________________________________
+ * Linear Growth Rate (Linear Algorithm)
+ * For example:
+ * If input string is length 20:
+ * ---- T(n) = 20 * (input string loop) = O(n) -----
+ * So, Time complexity is O(n) because
+ * as the length of the input grows, the time will increase at a linear rate
+ * based on the size of the input.
+ * __________________________________________________________________________________________
+ */
+public class Exercise22_02 {
+    public static void main(String[] args) {
+        Scanner in = new Scanner(System.in);
+        System.out.print("Enter a string: ");
+        String input = in.next().trim();
+        String result = "";
+        char lastSeqChar = input.charAt(0);
+        /* Loop always executes one time, so growth rate of time is constant as the input string gets larger */
+        for (int j = 1; j < input.length(); j++) {
+            if (lastSeqChar < input.charAt(j)) {
+                result += lastSeqChar + "";
+                lastSeqChar = input.charAt(j);
+            }
+        }
+        result += lastSeqChar;
+        System.out.println(result);
+
+        in.close();
+    }
+}