Add a Solution for Find the Index of the First Occurrence in a String (LeetCode Problem 28)

dsa-solutions/lc-solutions/0000-0099/0028-find-first-occurrence-in-string.md

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+---
+id: find-first-occurrence-in-string
+title: Find the Index of the First Occurrence in a String
+difficulty: Easy
+sidebar_label: 0028-FindFirstOccurrenceInString
+tags:
+  - String
+---
+
+## Problem Description
+
+| Problem Statement                                                                                           | Solution Link                                                                                                                               | LeetCode Profile                                   |
+| :----------------------------------------------------------------------------------------------------------- | :------------------------------------------------------------------------------------------------------------------------------------------ | :------------------------------------------------- |
+| [Remove Duplicates from Sorted Array](https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string/)                                         | [Remove Duplicates from Sorted Array Solution on LeetCode](https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string/solutions/) | [VijayShankerSharma](https://leetcode.com/u/darkknight648/) |
+
+## Problem Description
+
+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`.
+
+### Examples
+
+#### 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.
+
+### Approach
+
+To solve the problem, we can use the following approach:
+
+1. **Sliding Window Technique:**
+   - We iterate through the `haystack` string with a window of size equal to the length of the `needle`.
+   - At each step, we check if the substring of the `haystack` starting at the current index and having the same length as the `needle` matches the `needle`.
+   - If a match is found, we return the current index.
+   - If no match is found after iterating through the entire `haystack`, we return -1.
+
+### Solution Code
+
+#### Python
+
+```
+class Solution(object):
+    def strStr(self, haystack, needle):     
+        if not needle:
+            return 0
+        if len(needle) > len(haystack):
+            return -1
+
+        for i in range(len(haystack) - len(needle) + 1):
+            if haystack[i:i + len(needle)] == needle:
+                return i
+        return -1
+```
+
+#### Java
+
+```
+class Solution {
+    public int strStr(String haystack, String needle) {
+        if (needle.isEmpty()) return 0;
+
+        for (int i = 0; i <= haystack.length() - needle.length(); i++) {
+            if (haystack.substring(i, i + needle.length()).equals(needle)) {
+                return i;
+            }
+        }
+
+        return -1;
+    }
+}
+```
+
+#### C++
+
+```
+class Solution {
+public:
+    int strStr(string haystack, string needle) {
+        if (needle.empty()) return 0;
+
+        for (int i = 0; i <= haystack.length() - needle.length(); i++) {
+            if (haystack.substr(i, needle.length()) == needle) {
+                return i;
+            }
+        }
+
+        return -1;
+    }
+};
+```
+
+### Conclusion
+
+The above solution efficiently finds the index of the first occurrence of a substring `needle` in the string `haystack`. It employs a sliding window technique to traverse the `haystack` string and compare substrings with the `needle`, providing a simple yet effective approach to solving the problem.