Make a code using Recursion The countSubstring function will take two strings as parameters and will return an integer that is the count of how many times the substring (the second parameter) appears in the first string without overlapping with itself. This method will be case insensitive.   For example: countSubstring(“catwoman loves cats”, “cat”) would return 2 countSubstring(“aaa nice”, “aa”) would return 1 because “aa” only appears once without overlapping itself.     public static int countSubstring(String s, String x) { if (s.length() == 0 || x.length() == 0) return 1;   if (s.length() == 1 || x.length() == 1){ if (s.substring(0,1).equals(x.substring(0,1))){ s.replaceFirst((x), " "); return 1 + countSubstring(s.substring(1), x); } else { return 0 + countSubstring(s.substring(1), x); } } return countSubstring(s.substring(0,1), x) + countSubstring(s.substring(1), x); }   public class Main { public static void main(String[] args) { System.out.println(Recursion.countSubstring("catwoman loves cats","cat")); System.out.println(Recursion.countSubstring("aaa nice","aa")); } }

C++ Programming: From Problem Analysis to Program Design
8th Edition
ISBN:9781337102087
Author:D. S. Malik
Publisher:D. S. Malik
Chapter15: Recursion
Section: Chapter Questions
Problem 8SA
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Make a code using Recursion

The countSubstring function will take two strings as parameters and will return an integer that is the count of how many times the substring (the second parameter) appears in the first string without overlapping with itself. This method will be case insensitive.

 

For example:

countSubstring(“catwoman loves cats”, “cat”) would return 2

countSubstring(“aaa nice”, “aa”) would return 1 because “aa” only appears once without overlapping itself.

 

 

public static int countSubstring(String s, String x) {

if (s.length() == 0 || x.length() == 0)

return 1;

 

if (s.length() == 1 || x.length() == 1){

if (s.substring(0,1).equals(x.substring(0,1))){

s.replaceFirst((x), " ");

return 1 + countSubstring(s.substring(1), x);

}

else {

return 0 + countSubstring(s.substring(1), x);

}

}

return countSubstring(s.substring(0,1), x) + countSubstring(s.substring(1), x);

}

 

public class Main {

public static void main(String[] args) {

System.out.println(Recursion.countSubstring("catwoman loves cats","cat"));

System.out.println(Recursion.countSubstring("aaa nice","aa"));

}

}

Expert Solution
Step 1

The source code of the program

#Main.java

class Main 
{
    public static void main(String[] args) 
    {
        String testString3 = "catwoman loves cats";
        String testString2 = "aaa nice";
        System.out.println(Recursion.countSubstring(testString3, "cat"));
        System.out.println(Recursion.countSubstring(testString2, "aa"));
    }
}

 

#Recursion.java

class Recursion 
{
    public static int countSubstring(String s, String sub) 
    {
        int subLength = sub.length();
        int sLength = s.length();
        if (sLength < subLength) 
        {
            return 0;
        } 
        else 
        {
            String sLower = s.toLowerCase();
            String subLower = sub.toLowerCase();
            if (sLower.substring(0, subLength).equals(subLower)) 
            {
                return 1 + countSubstring(s.substring(subLength), sub);
            } 
            else 
            {
                return countSubstring(s.substring(1), sub);
            }
        }
    }
}

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