# 28 Implement strStr() – Easy

### Problem:

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

### Thoughts:

Very straightforward. String manipulation.

### Solutions:

``````public class Solution {
public int strStr(String haystack, String needle) {
if (needle.equals("")) {
return 0;
}
for (int i = 0; i < (haystack.length() - needle.length()) + 1; i ++ ){
for (int j = 0; j < needle.length(); j ++ ) {
if (haystack.charAt(i+j) != needle.charAt(j)){
break;
}
if (j == needle.length() -1) {
return i;
}
}
}
return -1;
}
}
``````
``````public class Solution {
public int strStr(String haystack, String needle) {
if (needle.length() == 0) {
return 0;
}
int[] kmp = new int[needle.length()];
computeKMP(needle, kmp);
int i = 0, j = 0;//i is for haystack and j is for needle
while (i < haystack.length()) {
if (j == -1) {
j = 0;
i ++;
continue;
}
if (haystack.charAt(i) == needle.charAt(j)) {
if (j == needle.length() -1) {
return i - needle.length() + 1;
}
i ++;
j ++;
}
else {
j = kmp[j];
}
}
return -1;
}
private void computeKMP(String s, int[] kmp){
int prefixEnd = -1;
int suffixEnd = 0;
kmp[0] = -1;
//while loop updates kmp[suffixEnd + 1]
while (suffixEnd < s.length() - 1) {
if (prefixEnd == -1 || s.charAt(prefixEnd) == s.charAt(suffixEnd)) {
kmp[suffixEnd +1] = prefixEnd + 1;
prefixEnd ++;
suffixEnd ++;
}
else {
prefixEnd = kmp[prefixEnd];
}
}
}
}
``````