Algorithm

This algorithm is a two step process.First we create a auxiliary array lps[] and then use this array for searching the pattern.

Preprocessing :

1. We pre-process the pattern and create an auxiliary array lps[] which is used to skip characters while matching.
2. Here lps[] indicates longest proper prefix which is also suffix.A proper prefix is prefix in which whole string is not included.For example, prefixes of string ABC are “ ”, “A”, “AB” and “ABC”. Proper prefixes are “ ”, “A” and “AB”. Suffixes of the string are “ ”, “C”, “BC” and “ABC”.

Searching

1. We keep matching characters txt[i] and pat[j] and keep incrementing i and j while pat[j] and txt[i] keep matching.

2. When we see a mismatch,we know that characters pat[0..j-1] match with txt[i-j+1…i-1].We also know that lps[j-1] is count of characters of pat[0…j-1] that are both proper prefix and suffix.From this we can conclude that we do not need to match these lps[j-1] characters with txt[i-j…i-1] because we know that these characters will match anyway.

Implementaion in Java

public class KMP {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String str = "abcabdabc";
        String pattern = "abc";
        KMP obj = new KMP();
        System.out.println(obj.patternExistKMP(str.toCharArray(), pattern.toCharArray()));
    }
    
    public int[] computeLPS(char[] str){
        int lps[] = new int[str.length];
        
        lps[0] = 0;
        int j = 0;
        for(int i =1;i<str.length;i++){
            if(str[j] == str[i]){
                lps[i] = j+1;
                j++;
                i++;
            }else{
                if(j!=0){
                    j = lps[j-1];
                }else{
                    lps[i] = j+1;
                    i++;
                }
            }
            
        }
        
        return lps;
    }
    
    public boolean patternExistKMP(char[] text,char[] pat){
        int[] lps = computeLPS(pat);
        int i=0,j=0;
        while(i<text.length && j<pat.length){
            if(text[i] == pat[j]){
                i++;
                j++;
            }else{
                if(j!=0){
                    j = lps[j-1];
                }else{
                    i++;
                }
            }
        }
        
        if(j==pat.length)
            return true;
        return false;
    }

}