feat: [Week 01-4] solve longest-consecutive-sequence

Author: Chaedie

longest-consecutive-sequence/Chaedie.py

@@ -0,0 +1,46 @@
+"""
+Solution: 
+    0. i will use prev, cur pointers in for loop, 
+        so i have to get rid of edge case when len of nums is at most 1
+    1. use hash set for remove duplicate elements 
+    2. sort list
+    3. iterate sorted_nums and use two pointers (prev, cur) 
+        compare prev and cur and count if the difference is 1
+    4. use max method, memorize maxCount
+    5. return maxCount
+
+Time Complexity: 
+    1. remove duplicate by hash set -> O(n)
+    2. sort the list -> O(n log n)
+    3. iterate sorted_nums -> O(n)
+
+    so time complexity of this solution will be O(n log n)
+
+Space Complexity:
+    1. set() -> O(n)
+    2. sorted_nums -> O(n)
+    3. count , maxCount -> O(1)
+
+    space complexity of this solution will be O(n)
+"""
+
+
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        if len(nums) <= 1:
+            return len(nums)
+
+        sorted_nums = sorted(list(set(nums)))
+
+        count = 1
+        maxCount = 1
+        for i in range(1, len(sorted_nums)):
+            prev = sorted_nums[i - 1]
+            cur = sorted_nums[i]
+            if prev + 1 == cur:
+                count += 1
+                maxCount = max(count, maxCount)
+            else:
+                count = 1
+
+        return maxCount