ch 22 04

Author: Harry Dulaney

ch_22/Exercise22_03.java

@@ -28,8 +28,7 @@
  * 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 s1.
- * The size of s2 is not relevant since we are only check that the string have the same starting char
- * and then if they do checking for equality of s1 substring and s2;
+ * The size of s2 is not relevant to time complexity because we iterate based on the length of s1.
  *
  * --------------------------------- Additional efficiency ---------------------------------
  * This could be made more efficient by utilizing the constraint of neighboring characters must be

ch_22/Exercise22_04.java

@@ -0,0 +1,65 @@
+package ch_22;
+
+
+import java.util.Scanner;
+
+/**
+ * (Pattern matching) Write a program that prompts the user to enter two strings
+ * and tests whether the second string is a substring of the first string. (Don’t use
+ * the indexOf method in the String class.) Analyze the time complexity of
+ * your algorithm. Here is a sample run of the program:
+ * <p>
+ * Enter a string s1: Mississippi
+ * Enter a string s2: sip
+ * matched at index 6
+ * <p>
+ * __________________________________________________________________________________________
+ * ----------------------------- Time Complexity Analysis -----------------------------------
+ * __________________________________________________________________________________________
+ * Time complexity is: O(n) (Linear Time)
+ * <p>
+ * If s1.length is 100 and s2.length is 4:
+ * Worst case is that the loop will execute over all values of s1
+ * <p>
+ * T(n) = 100 * (s1 String loop) = O(n)
+ * <p>
+ * 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 s1.
+ * The size of s2 is not relevant to time complexity because we iterate based on the length of s1.
+ */
+public class Exercise22_04 {
+    public static void main(String[] args) {
+        Scanner scanner = new Scanner(System.in);
+        System.out.print("Enter a string s1: ");
+        String s1 = scanner.nextLine();
+
+        System.out.print("Enter a string s2: ");
+        String s2 = scanner.nextLine();
+        int result = new Exercise22_04().indexOfSubstring(s1, s2);
+        if (result == -1) {
+            System.out.println("s2 is not a substring of s1...");
+            return;
+        }
+
+        System.out.println("matched at index " + result);
+        scanner.close();
+    }
+
+    private int indexOfSubstring(String s1, String s2) {
+        int index = 0;
+        int matchLength = s2.length();
+        char s2Start = s2.charAt(0);
+        while (index < (s1.length() - matchLength)) {
+            /* Try to match first char in s2 with char at next index */
+            if (s1.charAt(index) == s2Start) {
+                String s1Sub = s1.substring(index, index + matchLength);
+                if (s1Sub.equals(s2)) {
+                    return index;
+                }
+            }
+            index++;
+        }
+        return -1;
+    }
+}