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2026-06-09 16:36:42

Longest Substring Without Repeating Characters in Python and Java | LeetCode #3

Learn how to solve Longest Substring Without Repeating Characters in Python and Java with beginner-friendly explanation, step-by-step dry run, custom examples, and time complexity.

Longest Substring Without Repeating Characters in Python and Java

In this tutorial, we will learn LeetCode #3: Longest Substring Without Repeating Characters in very simple language. We will understand the idea step by step, see custom examples, and write complete code in Python and Java.

What Is the Longest Substring Without Repeating Characters Problem?

This problem asks us to solve a common coding interview task using the given input. The goal is to return the correct result without using a slow brute force method.

For this problem, we will use sliding window because it gives a clean and optimized solution.

Example input:
s = "pwwkewx"

Expected result:
Output: 4

Explanation:
The substring kewx has length 4.

Beginner-Friendly Idea

The main idea is to avoid trying every possible answer blindly. Instead, we keep useful information while reading the input and use that information to make the next decision.

At each step, ask: “What do I already know, and how does the current value change my answer?”

Using our example:
s = "pwwkewx"

Approach:
sliding window

We update variables step by step until we reach:
Output: 4

Why Do We Use This Approach?

A direct brute force solution is usually easier to think about, but it can become slow when the input is large. The optimized approach keeps only the important state and avoids repeated work.

That is why sliding window is useful for this problem.

Step-by-Step Explanation

Let us dry run the algorithm using a custom example.

Step 1
Use this custom example.

s = "pwwkewx"

We will solve it using sliding window.

Step 2
Look at the first important value from the example and create the variables needed by the algorithm.

current_state = based on the first value
answer = not finished yet

Step 3
Move to the next useful value and update the state.

The algorithm compares the new value with the old state.
If the new value improves the answer, we update the answer.

Step 4
Continue this process until all useful values are processed.

After processing the example, we get:

Output: 4

Why?
The substring kewx has length 4.

Important Code Logic

The most important part is updating the algorithm state after reading each useful value. This is where the answer becomes better step by step.

Think like this:

old_state = what we knew before
current_value = value we are checking now
new_state = updated result after using current_value

For our example, the final state gives:
Output: 4

Example 1

Input:
s = "pwwkewx"

Output:
Output: 4

Explanation:
The substring kewx has length 4.

Example 2

Input:
s = "pwwkewx"

Output:
4

Explanation:
The longest substring without repeating characters is 'kewx'.

Python Code

Here is the complete Python solution for LeetCode #3.

class LongestUniqueSubstringFinder:
    def length_of_longest_substring(self, text):
        seen = {}
        left = 0
        best = 0

        for right, char in enumerate(text):
            if char in seen and seen[char] >= left:
                left = seen[char] + 1

            seen[char] = right
            best = max(best, right - left + 1)

        return best


finder = LongestUniqueSubstringFinder()
print(finder.length_of_longest_substring("abcaef"))  # Output: 5
print(finder.length_of_longest_substring("bbbb"))    # Output: 1

Java Code

Here is the complete Java solution for LeetCode #3.

import java.util.HashMap;
import java.util.Map;

class LongestUniqueSubstringFinder {
    public int lengthOfLongestSubstring(String text) {
        Map<Character, Integer> seen = new HashMap<>();
        int left = 0;
        int best = 0;

        for (int right = 0; right < text.length(); right++) {
            char ch = text.charAt(right);

            if (seen.containsKey(ch) && seen.get(ch) >= left) {
                left = seen.get(ch) + 1;
            }

            seen.put(ch, right);
            best = Math.max(best, right - left + 1);
        }

        return best;
    }

    public static void main(String[] args) {
        LongestUniqueSubstringFinder finder = new LongestUniqueSubstringFinder();
        System.out.println(finder.lengthOfLongestSubstring("abcaef")); // 5
        System.out.println(finder.lengthOfLongestSubstring("bbbb"));   // 1
    }
}

Time and Space Complexity

Time Complexity: O(n)

The time complexity depends on how many values the algorithm needs to process and whether it uses sorting, binary search, heap, or traversal.

Space Complexity: O(n)

The extra space is used for the variables or data structures needed by the optimized approach.

Final Summary

LeetCode #3: Longest Substring Without Repeating Characters becomes easier when we break it into small steps. First understand what the problem asks, then track the important state, dry run with an example, and finally write the code.