Merge pull request #2365 from yerim01/main

[yerim01] WEEK 01 solutions

Author: yerim01

contains-duplicate/yerim01.py

@@ -0,0 +1,22 @@
+# Leetcode 217: https://leetcode.com/problems/contains-duplicate/description/
+# Goal: Given an array of integers,
+# return True if the array has duplicates or return False if all elements are distinct.
+# Approach:
+# - Use a hash set to track elements we have already seen.
+# - Iterate through the array and if the current number already exists in the set, return True.
+# - Otherwise, add the current number to the set.
+# Time complexity: O(n)
+# - We iterate through the array once.
+# Space complexity: O(n)
+# - We store all elements in the set in the worst case.
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        h = set()
+
+        for n in nums:
+            if n not in h:
+                h.add(n)
+            else:
+                return True
+        return False

longest-consecutive-sequence/yerim01.py

@@ -0,0 +1,46 @@
+# Leetcode 128: https://leetcode.com/problems/longest-consecutive-sequence/
+# Goal: Given an unsorted array of integers nums, 
+# return the length of the longest consecutive elements sequence.
+# Constraints:
+# - Unsorted array
+# - Must run in O(n) time
+# - Duplicates exist
+# Approach:
+# - Use a variable `longest=0` to track the length of the longest sequence.
+# - Convert nums into a hash set to check numbers in O(1).
+# - Iterate the nums.
+# - Use variables `currentL=1` to track the length of current sequence
+#   and `j=1` to represent the distance from the current number.
+# - For checking larger numbers, do current num + j
+#   and for checking smaller numbers, do current num - j
+# - Iterate through the set while consecutive number exists in the set.
+#   If it founds the number, remove it from the set and update currentL & j.
+# - Update longest with the mximum length.
+# - Return longest.
+# Time complexity: O(n)
+# - Each number is removed from the set
+# Space complexity: O(n)
+# - Using a set. n -> size of the nums
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        longest = 0
+        s = set(nums)
+
+        for c in nums:
+            currentL = 1
+            j = 1
+
+            while c+j in s:
+                s.remove(c+j)
+                currentL += 1
+                j += 1
+
+            j = 1
+            while c-j in s:
+                s.remove(c-j)
+                currentL += 1
+                j += 1
+            
+            longest = max(longest, currentL)
+        
+        return longest