Finding the intersection between two list of string candidates

Question

I wrote the following Java code, to find the intersection between the prefix and the suffix of a String in Java.

// you can also use imports, for example:
// import java.math.*;
import java.util.*;
class Solution {
    public int max_prefix_suffix(String S) {
        if (S.length() == 0) {
            return 1;
        }
        // prefix candidates
        Vector<String> prefix = new Vector<String>();
        // suffix candidates
        Vector<String> suffix = new Vector<String>();
        // will tell me the difference
        Set<String> set = new HashSet<String>();

        int size = S.length();
        for (int i = 0; i < size; i++) {
            String candidate = getPrefix(S, i);
            // System.out.println( candidate );
            prefix.add(candidate);
        }

        for (int i = size; i >= 0; i--) {
            String candidate = getSuffix(S, i);
            // System.out.println( candidate );
            suffix.add(candidate);
        }

        int p = prefix.size();
        int s = suffix.size();
        for (int i = 0; i < p; i++) {
            set.add(prefix.get(i));
        }
        for (int i = 0; i < s; i++) {
            set.add(suffix.get(i));
        }

        System.out.println("set: " + set.size());
        System.out.println("P: " + p + " S: " + s);
        int max = (p + s) - set.size();
        return max;
    }

    // codility
    // y t i l i d o c
    public String getSuffix(String S, int index) {
        String suffix = "";
        int size = S.length();
        for (int i = size - 1; i >= index; i--) {
            suffix += S.charAt(i);
        }

        return suffix;
    }

    public String getPrefix(String S, int index) {
        String prefix = "";
        for (int i = 0; i <= index; i++) {
            prefix += S.charAt(i);
        }

        return prefix;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        String t1 = "";
        String t2 = "abbabba";
        String t3 = "codility";

        System.out.println(sol.max_prefix_suffix(t1));
        System.out.println(sol.max_prefix_suffix(t2));
        System.out.println(sol.max_prefix_suffix(t3));

        System.exit(0);
    }
}

Some test cases are:

String t1 = "";
String t2 = "abbabba";
String t3 = "codility";

and the expected values are:

1, 4, 0

My idea was to produce the prefix candidates and push them into a vector, then find the suffix candidates and push them into a vector, finally push both vectors into a Set and then calculate the difference. However, I'm getting 1, 7, and 0. Could someone please help me figure it out what I'm doing wrong?

Answer 1

I'd write your method as follows:

public int max_prefix_suffix(String s) {
    final int len = s.length();
    if (len == 0) {
        return 1; // there's some dispute about this in the comments to your post
    }
    int count = 0;
    for (int i = 1; i <= len; ++i) {
        final String prefix = s.substring(0, i);
        final String suffix = s.substring(len - i, len);
        if (prefix.equals(suffix)) {
            ++count;
        }
    }
    return count;
}

If you need to compare the prefix to the reverse of the suffix, I'd do it like this:

final String suffix = new StringBuilder(s.substring(len - i, len))
    .reverse().toString();

Answer 2

I see that the code by @ted Hop is good.. The question specify to return the max number of matching characters in Suffix and Prefix of a given String, which is a proper subset. Hence the entire string is not taken into consideration for this max number.

Ex. "abbabba", prefix and suffix can have abba(first 4 char) - abba (last 4 char),, hence the length 4 codility,, prefix(c, co,cod,codi,co...),, sufix (y, ty, ity, lity....), none of them are same. hence length here is 0.

By modifying the count here from

if (prefix.equals(suffix)) {
    ++count;
}

with

if (prefix.equals(suffix)) {
    count = prefix.length();// or suffix.length()
}

we get the max length. But could this be done in O(n).. The inbuilt function of string equals, i believe would take O(n), and hence overall complexity is made O(n2).....

Answer 3

i would use this code.

public static int max_prefix_suffix(String S)
{
    if (S == null)
        return -1;
    Set<String> set = new HashSet<String>();
    int len = S.length();
    StringBuilder builder = new StringBuilder();
    for (int i = 0; i < len - 1; i++)
    {
        builder.append(S.charAt(i));
        set.add(builder.toString());
    }
    int max = 0;
    for (int i = 1; i < len; i++)
    {
        String suffix = S.substring(i, len);
        if (set.contains(suffix))
        {
            int suffLen = suffix.length();
            if (suffLen > max)
                max = suffLen;
        }
    }
    return max;
}

Answer 4

@ravi.zombie If you need the length in O(n) then you just need to change Ted's code as below:

int max =0;
for (int i = 1; i <= len-1; ++i) {
    final String prefix = s.substring(0, i);
    final String suffix = s.substring(len - i, len);
    if (prefix.equals(suffix) && max < i) {
        max =i;
}
    return max;
}

I also left out the entire string comparison to get proper prefix and suffixes so this should return 4 and not 7 for an input string abbabba.