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+# 0128. Longest Consecutive Sequence
+
+- Difficulty: medium
+- Link: https://leetcode.com/problems/longest-consecutive-sequence/
+- Topics: Array-String
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT: List[int]
+    - OUTPUT: integer
+2. Check the main goal
+    - time complexity $O(n)$
+
+# Naive Solution
+
+> in $O(nlogn)$
+> 
+
+### Thought Process
+
+1. sort Array
+2. iterate
+    1. if 下一個 = 當前 + 1：count ++
+    2. if 下一個 = 當前：往下一個 loop
+    3. else：
+        - 比對 maxCount 與 count，更新 maxCount
+        - 將 count 重置回 1
+- Implement
+    
+    ```python
+    class Solution:
+        def longestConsecutive(self, nums: List[int]) -> int:
+            """
+            1. sort array
+            2. iterate it
+            """
+            
+            if len(nums) == 0:
+                return 0
+            if len(nums) == 1:
+                return 1
+            
+            nums.sort()        
+            maxCount = 1
+            count = 1
+            for i in range(0, len(nums) - 1):
+                if nums[i+1] == nums[i]:
+                    continue
+                if nums[i+1] == nums[i] + 1:
+                    count += 1
+                else:
+                    count = 1
+                if count > maxCount:
+                    maxCount = count
+            return maxCount
+    ```
+    
+
+### Complexity
+
+- Time complexity: $O(nlogn)$
+    
+    ![Untitled](./Untitled.png)
+    
+- Space complexity:$O(1)$
+
+### Problems & Improvement
+
+- time complexity $O(n)$
+
+# Improvement
+
+> $O(n)$
+> 
+
+### Thought Process
+
+1. use a set
+2. iterate the array
+    1. 該個 num 是否為 sequence 的起始點 ( num - 1 是否在 set 之中)
+        
+        [為起始點 (num - 1不在set中)]
+        
+        - 紀錄新的 sequence
+        - 並向後尋找，後面的值是否在 set 之中
+        
+        [非起始點 (num - 1 在 set 中)]： continue
+        
+3. 計算每個 sequence 的數量
+4. 回傳最長者
+- Implement
+    
+    ```python
+    class Solution:
+        def longestConsecutive(self, nums: List[int]) -> int:
+            """
+            1. use a set
+            2. iterate the array
+                1. 該個 num 是否為 sequence 的起始點 ( num - 1 是否在 set 之中)
+    
+                    [為起始點 (num - 1不在set中)]
+    
+                    - 紀錄新的 sequence
+                    - 並向後尋找，後面的值是否在 set 之中
+    
+                    [非起始點 (num - 1 在 set 中)]： continue
+    
+            3. 計算每個 sequence 的數量
+            4. 回傳最長者
+            """
+            numSet = set(nums)
+            sequence = []
+            for num in numSet:
+                if num - 1 in numSet: continue
+                newSequence = []
+                newSequence.append(num)
+                nextNum = num + 1
+                while nextNum in numSet:
+                    newSequence.append(nextNum)
+                    nextNum += 1
+                sequence.append(newSequence)
+            
+            maxCount = 0
+            for i in range(len(sequence)):
+                if len(sequence[i]) > maxCount:
+                    maxCount = len(sequence[i])
+            
+            return maxCount
+    ```
+    
+
+### Complexity
+
+- Time complexity: $O(n)$
+    
+    ![Untitled](./Untitled%201.png)
+    
+- Space complexity: $O(n)$
+    
+    ![Untitled](./Untitled%202.png)
+    
+
+# Check special cases, check error
+
+- 只需要記住數量，不需要記得實際數字，所以可以不用記 sequence
+    - Space Complexity: $O(1)$
+    - implement
+        
+        ```python
+        class Solution:
+            def longestConsecutive(self, nums: List[int]) -> int:
+                """
+                1. use a set
+                2. iterate the array
+                    1. 該個 num 是否為 sequence 的起始點 ( num - 1 是否在 set 之中)
+        
+                        [為起始點 (num - 1不在set中)]
+                        
+                        - 並向後尋找，後面的值是否在 set 之中
+                        - 記住找到多長
+                        - 比較當前最長
+        
+                        [非起始點 (num - 1 在 set 中)]： continue
+        
+                4. 回傳最長者
+                """
+                numSet = set(nums)
+                maxCount = 0
+                for num in numSet:
+                    if num - 1 in numSet: continue
+                    nextNum = num + 1
+                    while nextNum in numSet:
+                        nextNum += 1
+                    count = nextNum - num
+                    maxCount = max(maxCount, count)
+                return maxCount
+        ```
+        
+- 算 nextNum 時就像是在加長度了，所以不用多一個變數去存 nextNum
+    
+    ```python
+    class Solution:
+        def longestConsecutive(self, nums: List[int]) -> int:
+            numSet = set(nums)
+            longest = 0
+            for num in numSet:
+                if num - 1 in numSet: continue
+                length = 0
+                while num + length in numSet:
+                    length += 1
+                longest = max(longest, length)
+            return longest
+    ```
+    
+
+# Note
+
+- [Leetcode 128 - LONGEST CONSECUTIVE SEQUENCE](https://www.youtube.com/watch?v=P6RZZMu_maU)
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