This solution is based on KMP algorithm.

**Time complexity = O(n + m)**

KMP pre-process the pattern, and saves if ith character doesn't match, what's the next position to check, to avoid duplicated comparasions.

Runtime of brute force: O(n * m), so KMP is much faster!

**JAVA Code:**

```
public int strStr(String haystack, String needle) {
//KMP algorithms
if(needle.equals("")) return 0;
if(haystack.equals("")) return -1;
char[] arr = needle.toCharArray();
int[] next = makeNext(arr);
for(int i = 0, j = 0, end = haystack.length(); i < end;){
if(j == -1 || haystack.charAt(i) == arr[j]){
j++;
i++;
if(j == arr.length) return (i - arr.length);
}
if(i < end && haystack.charAt(i) != arr[j])
j = next[j]; //<--Use prefix length
}
return -1;
}
//Pre-process the pattern string, to get the prefix length for each position i in the pattern string
private int[] makeNext(char[] arr){
int len = arr.length;
int[] next = new int[len];
next[0] = -1;
for(int i = 0, j = -1; i + 1 < len;){
if(j == -1 || arr[i] == arr[j]){
next[i+1] = j+1;
if(arr[i+1] == arr[j+1]) next[i+1] = next[j+1];
i++;
j++;
}
if(arr[i] != arr[j]) j = next[j];
}
return next;
}
```