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In this post, you will find the solution for the **Longest Substring Without Repeating Characters** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Longest Substring Without Repeating Characters– LeetCode Problem

Longest Substring Without Repeating Characters– LeetCode Problem

**Problem:**

Given a string `s`

, find the length of the **longest substring** without repeating characters.

**Example 1:**

Input:s = "abcabcbb"Output:3Explanation:The answer is "abc", with the length of 3.

**Example 2:**

Input:s = "bbbbb"Output:1Explanation:The answer is "b", with the length of 1.

**Example 3:**

Input:s = "pwwkew"Output:3Explanation:The answer is "wke", with the length of 3. Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

**Constraints:**

`0 <= s.length <= 5 * 10`

^{4}`s`

consists of English letters, digits, symbols and spaces.

Longest Substring Without Repeating Characters– LeetCode Solutions

class Solution { public: int lengthOfLongestSubstring(string s) { int ans = 0; vectorcount(128); for (int l = 0, r = 0; r < s.length(); ++r) { ++count[s[r]]; while (count[s[r]] > 1) --count[s[l++]]; ans = max(ans, r - l + 1); } return ans; } };

class Solution { public int lengthOfLongestSubstring(String s) { int ans = 0; int[] count = new int[128]; for (int l = 0, r = 0; r < s.length(); ++r) { ++count[s.charAt(r)]; while (count[s.charAt(r)] > 1) --count[s.charAt(l++)]; ans = Math.max(ans, r - l + 1); } return ans; } }

class Solution: def lengthOfLongestSubstring(self, s: str) -> int: ans = 0 count = Counter() l = 0 for r, c in enumerate(s): count[c] += 1 while count[c] > 1: count[s[l]] -= 1 l += 1 ans = max(ans, r - l + 1) return