> For the complete documentation index, see [llms.txt](https://anton-veselskyi.gitbook.io/codding-problems-solutions/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://anton-veselskyi.gitbook.io/codding-problems-solutions/leetcode/easy/implement-strstr.md).

# Implement strStr()

## [Implement strStr()](https://leetcode.com/problems/implement-strstr)

Implement [strStr()](http://www.cplusplus.com/reference/cstring/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()](http://www.cplusplus.com/reference/cstring/strstr/) and Java's [indexOf()](https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf\(java.lang.String\)).

**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.

## Solutions

### 🧠 Cpp

```cpp
class Solution
{
public:
    int strStr(string haystack, string needle)
    {
        if(haystack == needle || needle.empty())
            return 0;

        for(auto iter = begin(haystack);
            iter < end(haystack);
            ++iter)
        {
            if(end(haystack) - iter < needle.size())
                return -1;

            if(*iter == needle.front())
            {
                bool are_same = false;
                //check is chunk the same
                for(auto iter_A = iter, iter_B = begin(needle);
                    iter_A < end(haystack) && iter_B < end(needle);
                    ++iter_A, ++ iter_B
                    )
                    if(*iter_A != *iter_B)
                        break;
                    else if(next(iter_B) == end(needle))
                        are_same = true;

                    if(are_same)
                        return iter - begin(haystack);
            }
        }

        return -1;
    }
};
```