Rabin-Karp String matching

I am learning algorithms and have implemented the Rabin-Karp string search algorithm. I didn't implement a rolling HashCode, but it's judged correct by this online judge.

It's slower than the str.indexOf() implementation (7ms, compared to 4 ms). Can this be improved? What is the time complexity for this?

class Solution {
public int strStr(String haystack, String needle) {
if (haystack == null || needle == null)
return -1;
int needleHash = hashCode(needle);
int plen = needle.length();
boolean t = true;
int tlen = haystack.length() - needle.length() + 1;
for (int i = 0; i < tlen; i++) {
String sub = haystack.substring(i, plen + i);
int haystackHash = hashCode(sub);
if (haystackHash == needleHash) {
for (int j = 0; j < plen; j++) {
char[] charN = needle.toCharArray();
char[] charH = sub.toCharArray();
if (charN[j] != charH[j]) {
t = false;
}
}
if (t == true) {
return i;
}
}
}
return -1;
}

public int hashCode(String value) {
int h = 777;
if (value.length() > 0) {
char val[] = value.toCharArray();

for (int i = 0; i < val.length; i++) {
h = 31 * h + val[i];
}
}
return h;
}
}
• Do you have some tests that show it produces correct results? It would be useful to include them with the code, for the benefit of reviewers. Oct 19 '18 at 6:53
• Code Review is for code which works as intended, to the best of the author's knowledge. We are not a code-testing facility. You'll have to put some effort in yourself. Please take a look at the help center.
– Mast
Oct 19 '18 at 7:30
• @Mast Toby it got selected as an answer passing all the test cases in online judge. I want if anyone could tell thats the correct implementation of the algorithm Oct 19 '18 at 10:41

You're missing the essential benefit of using a rolling hash:

for (int i = 0; i < tlen; i++) {
String sub = haystack.substring(i, plen + i);
int haystackHash = hashCode(sub);

Here, we examine plen characters of haystack at every character position, which is O(mn) where m is the haystack length and n the needle length.

By using a rolling hash, we only need to examine two characters each time around the loop (one to add to the hash, and one to remove from it). That is of course O(m), which greatly improves the efficiency when n is large.

Only inspect the substring when there's a hash match, which should be infrequent.

Other issues:

• Why convert needle and sub to char arrays to compare them? You certainly don't want to do that every time around the j loop; you don't even need that loop if you write if (sub == needle) return i;.

• if (t == true) is more simply written if (t) since that's the only true value that t can hold. Again, not needed when you eliminate the j loop.

Improved code

You meant to write something more like this:

public int strStr(final String haystack, final String needle)
{
if (haystack == null || needle == null)
return -1;

final int nlen = needle.length();
final int hlen = haystack.length();
if (nlen > hlen)
return -1;

int needleHash = 0;
int hash_remove = 1;
for (char c: needle.toCharArray()) {
}

int haystackHash = 0;
for (int i = 0;  i < nlen - 1;  ++i) {
}

for (int i = 0;  i + nlen < hlen;  ++i) {
haystackHash = addHash(haystackHash, haystack.charAt(i + nlen));
if (haystackHash == needleHash && haystack.substring(i, i+nlen) == needle)
return i;
haystackHash -= haystack.charAt(i) * hash_remove;
}

return -1;
}

private int addHash(int h, char c)
{
// calculation may overflow, but that's fine
return 31 * h + c;
}

The only magic in there is the computation and use of hash_remove, to undo the effect of characters that have moved out of the match window.

• Your solution doesn't work as OP wanted, please try "hello" "ll" for the input. Apr 13 '19 at 4:07
• I just wrote "something like" - I never claimed it was correct! :-p Apr 15 '19 at 7:39
• Hahah, yeah... sorry I missed it out Apr 15 '19 at 8:45