## Task

Given two strings `needle`

and `haystack`

, return the index of the first occurrence of `needle`

in `haystack`

, or `-1`

if `needle`

is not part of `haystack`

.

**Example 1:**

**Input:**

haystack = "sadbutsad", needle = "sad"

**Output:**

0

**Explanation:**

"sad" occurs at index 0 and 6. The first occurrence is at index 0, so we return 0.

**Example 2:**

**Input:**

haystack = "leetcode", needle = "leeto"

**Output:**

-1

**Explanation:**

"leeto" did not occur in "leetcode", so we return -1.

**Constraints:**

`1 <= haystack.length, needle.length <= 10`

^{4}`haystack`

and`needle`

consist of only lowercase English characters.

this problem was taken from Leetcode

## Solution

/** * @param {string} haystack * @param {string} needle * @return {number} */ /** * @param {string} haystack * @param {string} needle * @return {number} */ var strStr = function(haystack, needle) { var match = 0; // Edge case: if needle is an empty string, return 0 if (needle === "") { return 0; } // Get the lengths of both strings const haystackLen = haystack.length; const needleleLen = needle.length; // Iterate through the haystack string for (let i = 0; i <= haystackLen - needleleLen; i++) { var left = 0; while(left < needleleLen && haystack[left + i] == needle[left]) { left ++; } if(left == needleleLen) { return i; } } return -1; };

### Explanation:

**Edge Case**: If`needle`

is an empty string, the function immediately returns`0`

.**Outer Loop**: The outer loop iterates over each character in`haystack`

where there is still enough remaining length to match the`needle`

(`i <= haystackLen - needleLen`

).**Inner Loop**: The inner loop checks if the substring of`haystack`

starting at`i`

matches`needle`

. It does this by comparing characters one-by-one.**Match Found**: If a match is found (`j === needleLen`

), the starting index`i`

is returned.**No Match**: If no match is found by the end of the loop, the function returns`-1`

.

This implementation mimics a basic substring search without using any built-in functions. The time complexity is O(n * m), where `n`

is the length of `haystack`

and `m`

is the length of `needle`

.