longest-substring-without-repeating-characters

0003_longest-substring-without-repeating-characters/README.md

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+Given a string `s`, find the length of the **longest substring** without repeating characters.
+
+**Example 1:**
+
+    Input: s = "abcabcbb"
+    Output: 3
+    Explanation: The answer is "abc", with the length of 3.
+    
+
+**Example 2:**
+
+    Input: s = "bbbbb"
+    Output: 1
+    Explanation: The answer is "b", with the length of 1.
+    
+
+**Example 3:**
+
+    Input: s = "pwwkew"
+    Output: 3
+    Explanation: 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 * 104`
+*   `s` consists of English letters, digits, symbols and spaces.
+
+https://leetcode.com/problems/longest-substring-without-repeating-characters

0003_longest-substring-without-repeating-characters/python3/solution.py

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+# Time: O(N)
+# Space: O(N)
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        '''
+        Sliding-window approach
+        '''
+        
+        # Keep a track of unique chars we've seen so far in the substring
+        # we are looking at
+        seen = set()
+        start, end = 0, 0
+        maxl = 0
+        
+        while start <= end < len(s):
+            # Our current substring is denoted by their bounds [start, end]
+            # If we see a char in `end` which is already seen by us, we need
+            # to keep moving `start` until we are left with a new character
+            while s[end] in seen:
+                # Since `start` is moving up, we need to remove existing
+                # `start` char from the set if it exists
+                seen.remove(s[start])
+                start += 1
+            
+            seen.add(s[end])
+            maxl = max(maxl, (end - start) + 1)
+            end += 1
+        
+        return maxl