Needle-Haystack

Find the First Occurrence of Needle in Haystack

Problem Statement

Implement a function strStr(haystack, needle) that returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example :

Input:  haystack = "sadbutsad", needle = "sad"
Output: 0
Explanation: "sad" occurs at index 0 and 6. The first occurrence is at 0.

Best Data Structure for Solving the Problem

Since we're dealing with string pattern matching, a simple String traversal with two-pointer or substring comparison is efficient here.

Multiple Approaches

Approach 1: Built-in Method (Quick & Easy)

Idea: Use Java's built-in indexOf() function. It's optimized under the hood and handles all edge cases.


public int strStr(String haystack, String needle) {
    return haystack.indexOf(needle);
}

Approach 2: Manual Sliding Window (Interview-Ready)

Idea: Slide a window of size needle.length() across haystack and compare each substring with needle.


class Solution {
    public int strStr(String haystack, String needle) {
        int n = haystack.length();
        int m = needle.length();

        if (m == 0) return 0;

        for (int i = 0; i <= n - m; i++) {
            if (haystack.substring(i, i + m).equals(needle)) {
                return i;
            }
        }

        return -1;
    }
}

 Pro Tip: haystack.substring(i, i + m) creates a new string—so if you want to save memory, use a custom comparison loop instead.

Approach 3: KMP Algorithm (Optimized for Large Strings)

Idea: Use the Knuth-Morris-Pratt (KMP) algorithm to preprocess the needle for efficient pattern matching in O(n + m) time.

class Solution {

    public int strStr(String haystack, String needle) {

        if (needle.length() == 0) return 0;

        int[] lps = buildLPS(needle);

        int i = 0; // pointer for haystack

        int j = 0; // pointer for needle

        while (i < haystack.length()) {

            if (haystack.charAt(i) == needle.charAt(j)) {

                i++;

                j++;

                if (j == needle.length()) {

                    return i - j; // match found

                }

            } else {

                if (j != 0) {

                    j = lps[j - 1]; // fall back

                } else {

                    i++;

                }

            }

        }

        return -1; // no match

    }


    private int[] buildLPS(String pattern) {

        int[] lps = new int[pattern.length()];

        int len = 0;

        int i = 1;


        while (i < pattern.length()) {

            if (pattern.charAt(i) == pattern.charAt(len)) {

                len++;

                lps[i] = len;

                i++;

            } else {

                if (len != 0) {

                    len = lps[len - 1];

                } else {

                    lps[i] = 0;

                    i++;

                }

            }

        }

        return lps;

    }

}

⚠️ Common Mistakes and Pitfalls

  • Forgetting to return 0 when needle is empty

  • Looping until i < haystack.length() instead of i <= haystack.length() - needle.length()

  • Confusing == with .equals() for strings in Java 

  • Not handling edge cases like empty haystack, needle, or both

📊 Time and Space Complexity Analysis

Step        Time Complexity      Space Complexity
Built-in indexOf()        O(n * m) worst        O(1)
Manual Sliding Window        O(n * m)        O(1)
KMP Algorithm        O(n + m)        O(m)

Where n = haystack.length() and m = needle.length().

🌍 Real-World Applications

  • Searching for a keyword in a large document or dataset

  • Autocomplete or search bar functionality

  • DNA or genome pattern matching

  • Log monitoring for specific strings or patterns


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

Comments

Post a Comment

Popular posts from this blog

Divisible Drama — Who’s In, Who’s Out?

 Divisible Drama — Who’s In, Who’s Out? Problem Statement You’re given two positive integers n and m . num1 = sum of all numbers from 1 to n not divisible by m num2 = sum of all numbers from 1 to n divisible by m Your mission, should you choose to accept it: Return num1 - num2 Example Input: n = 10 , m = 3 Output: 19 Breakdown: Not divisible by 3 → [ 1 , 2 , 4 , 5 , 7 , 8 , 10 ] → num1 = 37 Divisible by 3 → [ 3 , 6 , 9 ] → num2 = 18 Answer = 37 - 18 = 19 Best Data Structure for Solving the Problem This one’s a math.  Why overthink it when Gauss already gave us the formula for sum of 1 to n? Different Approaches (from Brute Force to Optimized) Approach 1: Brute Force (aka List & Sum Vibes) Loop from 1 to n , split the nums based on whether they’re divisible by m , and sum them up. class Solution { public int differenceOfSums ( int n, int m) { int num1 = 0 , num2 = 0 ; for ( int i = 1 ; i <= n; i++)...
Read more

Trailing Zeros in Factorial – How Many and Why?

Trailing Zeros – Counting Zeros in Factorial Factorial-related problems are a fundamental part of Data Structures and Algorithms (DSA). One such classic problem involves counting the number of trailing zeros in N! . In this article, we will explore different approaches to solving this problem, optimize our solution, and examine its real-world applications. Optimal Data Structure for Solving the Problem For this problem, no additional data structures are required. Since factorials grow exponentially, storing the actual factorial value is impractical. Instead, we leverage mathematical properties and division operations to efficiently determine the number of trailing zeros. This approach ensures an optimal solution in terms of both time and space complexity. Approach – From Brute Force to an Optimized Solution Brute Force Approach – Compute and Count A naive approach involves computing the factorial of N and counting the number of trailing zeros. However, this method is highly ineff...
Read more

Minimum equal sum

Minimum Equal Sum – Matching Arrays with Smart Replacements Problem Statement You are given two integer arrays nums1 and nums2 , which may contain zeros. You must replace each 0 with a strictly positive integer (≥ 1) such that the sum of both arrays becomes equal . Your goal is to minimize that common equal sum after all replacements. If there is no way to make the sums equal with any valid replacements, return -1 . Example Input: nums1 = [ 3 , 2 , 0 , 1 , 0 ] nums2 = [ 6 , 5 , 0 ] Output: 12 Explanation: Replace the two zeros in nums1 with 2 and 4 → [3,2,2,1,4] Replace the zero in nums2 with 1 → [6,5,1] Both arrays now sum to 12 . Best Data Structure for the Job This problem is mostly numerical, so there’s no need for complex data structures like trees or heaps. What you do need is: A way to count zeros and compute current sums. Simple math to determine possible ranges. Long data type to handle large sums. Different Approaches Approach 1: Brute Fo...
Read more