Leetcode 28 || Find the Index of the First Occurrence in a String

In this blog, I will explain the mentioned leetcode problem and the solution.

Problem :

Link : https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string/

The problem description says it has two input strings a haystack and a needle. We have to find out if any substring matches the needle string. And the output will be the index of the substring in the haystack string. If there are multiple substrings we need to return the index of the first substring. If no substring matches we need to return -1 as output.

For example,

example 01: input : haystack = 'stringstr' , needle = 'str' output : 0

Here, in haystack at index 0 to 2, there is a match, and another match is found at index 6 to 8. So the output will be 0 as it is the first substring.

example 02 : input : haystack = 'matter' , needle = 'matc' output : -1

Here, there is no substring equal to needle in haystack so the output is -1

Solution :

Let's solve the problem now.

1. First of all, we need to check if the first index of the needle matches any index of the haystack.

2. Then if we find the first matched index at haystack we need to check if the whole needle matches or not

3. If only the first index matches and there are mismatches later, we need to iterate the next index of haystack where the first index of needle matches.

4. If the loop ends without any matches return -1

5. If any substring matches we do not need to iterate further on haystack string.

• One thing is important to discuss here. Do we need to iterate through a loop that ends at the last index of haystack ? What do you think ?

  Let's look at an example,

  input : haystack = 'abcdefghijk' , needle = 'matter' output : -1

  To solve this problem we will iterate through haystack in search of the needle[0] means 'm'. If we iterate on haystack til the last index we will end at 'k'. But do we really need this?

  The length of needle and haystack is 6 and 11 here respectively.

  So if (11-6)= 5th index of haystack does not match needle[0], searching in the rests is meaningless. So we should iterate through

  0 to (length of haystack - length of needle)th index.

• Code :

  Let's jump into code.

    class Solution:
        def strStr(self, haystack: str, needle: str) -> int:
            # default start_index = -1
            start_index = -1
            len_h , len_n = len(haystack), len(needle)
  
            i = 0
            # we will run the loop till len_h-len_n, not till the end
            while i<len_h-len_n+1:
                if haystack[i]==needle[0]:
                    # here we found the first match, so we will change the start_index to i
                    start_index = i
                    j = 0
                    # its time to check if the whole needle matches or not
                    while j<len_n:
                        if haystack[j+start_index]!=needle[j]:
                            # if there is any mismatch , change start_index to -1 and break the loop
                            start_index = -1
                            break
                        j += 1
                    # j == len_n means the needle matches that started at index start_index, so return the start_index
                    if j==len_n:
                        return start_index
                i += 1
  
            return start_index
  

• note: In the code above I've tried to solve in a way so you can understand how to think and solve step by step without using any shortcut tricks or built-in functionality.

Thank you so much for taking the time to read this far! Your thoughts and feedback mean a lot to me, so please don't hesitate to share them with me. Whether you enjoyed it or not, any comments and suggestions you have are most welcome and appreciated