Longest Unique Streak – No Repeats Allowed!

Longest Substring Without Repeating Characters – Master the Sliding Window!

Finding a substring without duplicate characters sounds simple—until you try doing it efficiently. Brute force? Too slow. Hashmaps? Maybe. Sliding window? Now we’re talking! Let’s break it down.

Best Data Structure for Solving It

A HashSet or HashMap is the best choice to keep track of unique characters in the substring. The Sliding Window technique helps efficiently expand and contract the substring as needed.

Different Approaches – From Brute Force to Optimized Solutions

1) Brute Force – Checking All Substrings (O(N²) Time Complexity)

  • Generate all substrings and check for uniqueness.

  • Extremely inefficient for large inputs (N = 50,000).

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int maxLength = 0;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                if (isUnique(s, i, j)) {
                    maxLength = Math.max(maxLength, j - i + 1);
                }
            }
        }
        return maxLength;
    }

    private boolean isUnique(String s, int start, int end) {
        boolean[] chars = new boolean[128]; // Assuming ASCII characters
        for (int i = start; i <= end; i++) {
            if (chars[s.charAt(i)]) return false;
            chars[s.charAt(i)] = true;
        }
        return true;
    }
}

2) Optimized Approach – Sliding Window with HashSet (O(N) Time Complexity)

  • Uses a HashSet to track unique characters.

  • Expands the window by adding characters to the set.

  • If a duplicate appears, removes characters from the left.


import java.util.*;

class Solution {
    public int lengthOfLongestSubstring(String s) {
        Set<Character> set = new HashSet<>();
        int left = 0, maxLength = 0;

        for (int right = 0; right < s.length(); right++) {
            while (set.contains(s.charAt(right))) {
                set.remove(s.charAt(left));
                left++;
            }
            set.add(s.charAt(right));
            maxLength = Math.max(maxLength, right - left + 1);
        }
        return maxLength;
    }
}

3) Optimized Approach – Sliding Window with HashMap (O(N) Time Complexity)

  • Uses a HashMap to store the last seen index of each character.

  • If a duplicate is found, move the left pointer to max(previous index +1, current left).

  • Ensures the substring remains unique.


import java.util.*;

class Solution {
    public int lengthOfLongestSubstring(String s) {
        Map<Character, Integer> map = new HashMap<>();
        int left = 0, maxLength = 0;

        for (int right = 0; right < s.length(); right++) {
            if (map.containsKey(s.charAt(right))) {
                left = Math.max(map.get(s.charAt(right)) + 1, left);
            }
            map.put(s.charAt(right), right);
            maxLength = Math.max(maxLength, right - left + 1);
        }
        return maxLength;
    }
}

🛑 Common Mistakes & Pitfalls

  • Confusing substrings with subsequences → The solution must be continuous.

  • Not handling edge cases → Empty strings, single-character strings, or all duplicates.

  • Using nested loops → Leads to unnecessary inefficiency.

📊 Time & Space Complexity Analysis

  • Brute Force: O(N²) time, O(1) space.

  • Sliding Window + HashSet: O(N) time, O(min(N, 256)) space.

  • Sliding Window + HashMap: O(N) time, O(min(N, 256)) space.

🌍 Real-World Applications

  • Text Processing → Detecting unique sequences in strings.

  • Data Compression → Finding the longest unique segment in a dataset.

  • Cybersecurity → Identifying non-repeating patterns in encryption.

Mastering the Sliding Window is a game-changer in DSA! 

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 :)


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