# GeetCode Hub

Implement strStr().

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

Clarification:

What should we return when `needle` is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when `needle` is an empty string. This is consistent to C's strstr() and Java's indexOf().

Example 1:

```Input: haystack = "hello", needle = "ll"
Output: 2
```

Example 2:

```Input: haystack = "aaaaa", needle = "bba"
Output: -1
```

Example 3:

```Input: haystack = "", needle = ""
Output: 0
```

Constraints:

• `0 <= haystack.length, needle.length <= 5 * 104`
• `haystack` and `needle` consist of only lower-case English characters.

class Solution { public int strStr(String haystack, String needle) { if(haystack==null || needle== null || needle.length()>haystack.length()){ return -1; } int[] parr = KmpPreprocess(needle); int i=0; int j=0; while(i<haystack.length() && j<needle.length()){ if(haystack.charAt(i) == needle.charAt(j)){ i++; j++; } else if(j>0){ j=parr[j-1]; } else{ i++; } } if(j==needle.length()){ return i-j; } return -1; } private int[] KmpPreprocess(String pattern){ int[] parr = new int[pattern.length()]; int i=0; int j=1; while(j<pattern.length()){ if(pattern.charAt(i)==pattern.charAt(j)){ parr[j]=i+1; i++; j++; } else if(i>0){ i = parr[i-1]; } else{ j++; } } return parr; } }

We can solve this question using both the Naive Approaches as well as using KMP Algorithm.

Naive Approach: Traverse all the possible starting points of `haystack` (from `0` to `haystack.length() - needle.length()`) and see if the following characters in `haystack` match those of `needle`.

Time Complexity: O(mn)
Space Complexity: O(1)

KMP Algorithm: Finally comes the KMP algorithm. You may refer to KMP on jBoxer's blog for some explanations.

Time Complexity: O(m+n)
Space Complexity: O(n)