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** Link for the Problem** – Implement strStr()– LeetCode Problem

Implement strStr()– LeetCode Problem

**Problem:**

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 * 10`

^{4}`haystack`

and`needle`

consist of only lower-case English characters.

Implement strStr() – LeetCode Solutions

class Solution { public: int strStr(string haystack, string needle) { const int m = haystack.length(); const int n = needle.length(); for (int i = 0; i < m - n + 1; i++) if (haystack.substr(i, n) == needle) return i; return -1; } };

class Solution { public int strStr(String haystack, String needle) { final int m = haystack.length(); final int n = needle.length(); for (int i = 0; i < m - n + 1; ++i) if (haystack.substring(i, i + n).equals(needle)) return i; return -1; } }

class Solution: def strStr(self, haystack: str, needle: str) -> int: m = len(haystack) n = len(needle) for i in range(m - n + 1): if haystack[i:i + n] == needle: return i return -1