**Question**

Implement strstr(). Returns the index of the first occurrence of needle in haystack, or –1 if needle is not part of haystack.

**Solution:**

**O(nm) runtime, O(1) space – Brute force:**

There are known efficient algorithms such as Rabin-Karp algorithm, KMP algorithm, or

the Boyer-Moore algorithm. Since these algorithms are usually studied in an advanced

algorithms class, it is sufficient to solve it using the most direct method in an interview –

The brute force method.

The brute force method is straightforward to implement. We scan the needle with the

haystack from its first position and start matching all subsequent letters one by one. If one

of the letters does not match, we start over again with the next position in the haystack.

Assume that n = length of haystack and m = length of needle, then the runtime

complexity is O(nm).

Have you considered these scenarios ?

- needle or haystack is empty. If needle is empty, always return 0. If haystack is empty, then there will always be no match (return –1) unless needle is also empty which 0 is returned.
- needle’s length is greater than haystack’s length. Should always return –1.
- needle is located at the end of haystack. For example, “aaaba” and “ba”. Catch possible off-by-one errors.
- needle occur multiple times in haystack. For example, “mississippi” and “issi”. It should return index 2 as the first match of “issi”.
- Imagine two very long strings of equal lengths = n, haystack = “aaa…aa” and needle = “aaa…ab”. You should not do more than n character comparisons, or else your code will get Time Limit Exceeded in OJ

Below is a clean implementation – no special if statements for all the above scenarios.

```
public int strStr(String haystack, String needle) {
for (int i = 0; ; i++) {
for (int j = 0; ; j++) {
if (j == needle.length()) return i;
if (i + j == haystack.length()) return -1;
if (needle.charAt(j) != haystack.charAt(i + j)) break;
}
}
}
```