How to find common suffix using recursion in java

Question

So I have seen on Stack Overflow that common suffix can be derived using methods which do not use recursion. I was curious if i could achieve the same with recursion. Tried with below code but getting java.lang.StackOverflowError. Can someone help?

public static String commonSuffixWithRecursion(String s1, String s2) {
    if(s1.length()==0 || s2.length()==0) {
        return "Strings are empty";
    }
    if(s1.charAt(s1.length()-1)==s2.charAt(s2.length()-1)) {
        return s1.length()-1 + commonSuffixWithRecursion(s1,s2);
    }
    return "No common Suffix";
}

Answer 1

• It should return empty string when there is not common suffix.

• You can use subString to get the input for next round

• Current char should be appended after the result of next rount

Sample code:

public static String commonSuffixWithRecursion(String s1, String s2) {
    if (s1.length() == 0 || s2.length() == 0) {
        return "";
    }
    if (s1.charAt(s1.length() - 1) == s2.charAt(s2.length() - 1)) {
        String nextS1 = s1.substring(0, s1.length() - 1);
        String nextS2 = s2.substring(0, s2.length() - 1);
        return commonSuffixWithRecursion(nextS1, nextS2 ) + s1.charAt(s1.length() - 1);
    }
    return "";
}

Test case:

public static void main(String[] args) {
    System.out.println(commonSuffixWithRecursion("dbc", "abc"));    // bc
    System.out.println(commonSuffixWithRecursion("a", "a"));    // a
    System.out.println(commonSuffixWithRecursion("ab", "b"));   // b
    System.out.println(commonSuffixWithRecursion("a", "b"));    // empty
}

Answer 2

One simple way to do it recursively is

public static String commonSuffixWithRecursion(String x, String y, int m, int n) {
        if(m <= 0 || n <= 0)
            return "";
        if (x.charAt(m-1) == y.charAt(n-1)) 
            return commonSuffixWithRecursion(x, y, m-1, n-1) + x.charAt(m-1);
        else 
         return "";
    }

Here m=x.length() and n=y.length()

Answer 3

Here is another option using an incremented suffix size:

public static String commonSuffixWithRecursion(String s1, String s2, int suffixSize) {
    int i1 = s1.length() - suffixSize;
    int i2 = s2.length() - suffixSize;
    if (i1 > -1 && i2 > -1 && s1.charAt(i1) == s2.charAt(i2)) {
        return commonSuffixWithRecursion(s1, s2, suffixSize + 1);
    }
    return s1.substring(i1 + 1);
}

Here are some test cases:

public static void main(String[] args) {
    System.out.println(commonSuffixWithRecursion("dbc", "abc", 1));    // bc
    System.out.println(commonSuffixWithRecursion("a", "a", 1));    // a
    System.out.println(commonSuffixWithRecursion("ab", "b", 1));   // b
    System.out.println(commonSuffixWithRecursion("a", "b", 1));    // empty
    System.out.println(commonSuffixWithRecursion("a", "bc", 1));    // empty
    System.out.println(commonSuffixWithRecursion("ac", "bc", 1));    // c
    System.out.println(commonSuffixWithRecursion("ac", "bcd", 1));    // empty
    System.out.println(commonSuffixWithRecursion("", "bcd", 1));    // empty
    System.out.println(commonSuffixWithRecursion("bcd", "bcd", 1));    // bcd
}