Merge pull request #1451 from Hemav009/159

Added 159

Author: ajay-dhangar

dsa-solutions/lc-solutions/0100-0199/0159-longest-substring-with-at.md

@@ -0,0 +1,169 @@
+---
+id: longest-substring-with-at
+title: Longest Substring Given At
+sidebar_label: 0159-Longest Substring Given At
+tags:
+  - Leet code
+description: "Solution to leetocde 159"
+---
+
+### Problem Description
+
+Given a string s , find the length of the longest substring t that contains at most 2 distinct characters.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: "eceba"
+Output: 3
+Explanation: t is "ece" which its length is 3.
+```
+
+**Example 2:**
+
+```
+Input: "ccaabbb"
+Output: 5
+Explanation: t is "aabbb" which its length is 5.
+```
+
+Certainly! Below is the complete algorithm with explanations, followed by the implementations in C++, Python, Java, and JavaScript.
+
+### Algorithm
+
+1. **Initialize**:
+
+   - A hash map to count characters in the current window.
+   - Two pointers, `i` and `j`, to represent the current window. `i` is the right pointer, and `j` is the left pointer.
+   - A variable to keep track of the maximum length of the substring.
+
+2. **Expand the Window**:
+
+   - Move the right pointer `i` to expand the window.
+   - Add the current character to the hash map and update its count.
+
+3. **Shrink the Window if Necessary**:
+
+   - If the number of distinct characters in the hash map exceeds 2, move the left pointer `j` to shrink the window until the number of distinct characters is at most 2.
+   - Decrease the count of the character at the left pointer `j` in the hash map and remove it if its count becomes 0.
+
+4. **Update the Maximum Length**:
+
+   - Update the maximum length of the substring after adjusting the window.
+
+5. **Return the Result**:
+   - Return the maximum length of the substring with at most two distinct characters.
+
+### C++
+
+```cpp
+#include <unordered_map>
+#include <string>
+#include <algorithm>
+
+using namespace std;
+
+class Solution {
+public:
+    int lengthOfLongestSubstringTwoDistinct(string s) {
+        unordered_map<char, int> cnt;
+        int n = s.size();
+        int ans = 0;
+        int j = 0;
+
+        for (int i = 0; i < n; ++i) {
+            cnt[s[i]]++;
+            while (cnt.size() > 2) {
+                cnt[s[j]]--;
+                if (cnt[s[j]] == 0) {
+                    cnt.erase(s[j]);
+                }
+                ++j;
+            }
+            ans = max(ans, i - j + 1);
+        }
+
+        return ans;
+    }
+};
+```
+
+### Python
+
+```python
+class Solution:
+    def lengthOfLongestSubstringTwoDistinct(self, s: str) -> int:
+        from collections import defaultdict
+
+        cnt = defaultdict(int)
+        n = len(s)
+        ans = 0
+        j = 0
+
+        for i in range(n):
+            cnt[s[i]] += 1
+            while len(cnt) > 2:
+                cnt[s[j]] -= 1
+                if cnt[s[j]] == 0:
+                    del cnt[s[j]]
+                j += 1
+            ans = max(ans, i - j + 1)
+
+        return ans
+```
+
+### Java
+
+```java
+import java.util.HashMap;
+
+public class Solution {
+    public int lengthOfLongestSubstringTwoDistinct(String s) {
+        HashMap<Character, Integer> cnt = new HashMap<>();
+        int n = s.length();
+        int ans = 0;
+        int j = 0;
+
+        for (int i = 0; i < n; ++i) {
+            cnt.put(s.charAt(i), cnt.getOrDefault(s.charAt(i), 0) + 1);
+            while (cnt.size() > 2) {
+                cnt.put(s.charAt(j), cnt.get(s.charAt(j)) - 1);
+                if (cnt.get(s.charAt(j)) == 0) {
+                    cnt.remove(s.charAt(j));
+                }
+                ++j;
+            }
+            ans = Math.max(ans, i - j + 1);
+        }
+
+        return ans;
+    }
+}
+```
+
+### JavaScript
+
+```javascript
+var lengthOfLongestSubstringTwoDistinct = function (s) {
+  let cnt = new Map();
+  let n = s.length;
+  let ans = 0;
+  let j = 0;
+
+  for (let i = 0; i < n; ++i) {
+    cnt.set(s[i], (cnt.get(s[i]) || 0) + 1);
+    while (cnt.size > 2) {
+      cnt.set(s[j], cnt.get(s[j]) - 1);
+      if (cnt.get(s[j]) === 0) {
+        cnt.delete(s[j]);
+      }
+      ++j;
+    }
+    ans = Math.max(ans, i - j + 1);
+  }
+
+  return ans;
+};
+```