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In this post, you will find the solution for the **Longest Substring with At Most Two Distinct 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 with At Most Two Distinct Characters– LeetCode Problem

Longest Substring with At Most Two Distinct Characters– LeetCode Problem

**Problem:**

Given a string ** s** , find the length of the longest substring

**that contains**

*t***at most**2 distinct characters.

**Example 1:**

Input:"eceba"Output:3Explanation:is "ece" which its length is 3.t

**Example 2:**

Input:"ccaabbb"Output:5Explanation:is "aabbb" which its length is 5.t

## Explanation

Two pointers. Keep faster pointer moving to the right until more than 2 distinct characters. Then adjust slower pointer and repeat the same.

Longest Substring with At Most Two Distinct Characters– LeetCode Solutions

Longest Substring with At Most Two Distinct Characters Solution in C++:

class Solution { public: int lengthOfLongestSubstringTwoDistinct(string s) { int ans = 0; int distinct = 0; vector<int> count(128); for (int l = 0, r = 0; r < s.length(); ++r) { if (++count[s[r]] == 1) ++distinct; while (distinct == 3) if (--count[s[l++]] == 0) --distinct; ans = max(ans, r - l + 1); } return ans; } };

Longest Substring with At Most Two Distinct Characters Solution in Java:

class Solution { public int lengthOfLongestSubstringTwoDistinct(String s) { int ans = 0; int distinct = 0; int[] count = new int[128]; for (int l = 0, r = 0; r < s.length(); ++r) { if (++count[s.charAt(r)] == 1) ++distinct; while (distinct == 3) if (--count[s.charAt(l++)] == 0) --distinct; ans = Math.max(ans, r - l + 1); } return ans; } }

Longest Substring with At Most Two Distinct Characters Solution in Python:

class Solution: def lengthOfLongestSubstringTwoDistinct(self, s: str) -> int: ans = 0 distinct = 0 count = [0] * 128 l = 0 for r, c in enumerate(s): count[ord(c)] += 1 if count[ord(c)] == 1: distinct += 1 while distinct == 3: count[ord(s[l])] -= 1 if count[ord(s[l])] == 0: distinct -= 1 l += 1 ans = max(ans, r - l + 1) return ans