Added question 003.

Author: luceCoding

leetcode/medium/003_longest_substring_without_repeating_characters.md

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+# 3. Longest Substring Without Repeating Characters
+
+## Sliding Window Solution
+- Run-time: O(N) or 2N
+- Space: O(N)
+- N = Number of characters in string
+
+Using a sliding window and a set, we can identify when we need to move the left side of the sliding window when a duplicate character is found as well as when to stop.
+
+This solution is actually a two pass solution, due to the fact we have to remove elements on the left side from the set, hence, visiting each element twice.
+
+```
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        seen = set()
+        left_idx = longest_substr = 0
+        for idx, ch in enumerate(s):
+            while ch in seen and left_idx <= idx:
+                seen.remove(s[left_idx])
+                left_idx += 1
+            seen.add(ch)
+            longest_substr = max(longest_substr, len(seen))
+        return longest_substr
+```
+
+## One Pass Solution
+- Run-time: O(N)
+- Space: O(N)
+- N = Number of characters in string
+
+To perform a one pass solution, we can use a dictionary where the key is the character and the value is its index.
+Similar to the set, we can use the dictionary to check if we have a duplicate in our sliding window.
+If that is true, we can immediately move the left side to the previous duplicated character's index + 1.
+
+```
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        seen = dict()
+        left_idx = longest_substr = 0
+        for idx, ch in enumerate(s):
+            if ch in seen:
+                left_idx = seen[ch] + 1
+            seen[ch] = idx
+            longest_substr = max(longest_substr, idx - left_idx + 1)
+        return longest_substr
+```