Merge pull request #2401 from hyeri0903/main

[hyeri0903] WEEK 02 solutions

Author: SamTheKorean

3sum/hyeri0903.py

@@ -0,0 +1,39 @@
+from itertools import combinations
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        """
+        time complexity : O(n^2)
+        space complexity : O(1)
+        """
+        answer = []
+        nums.sort()
+        n = len(nums)
+
+        for i in range(n):
+            #skipped if nums[i] == nums[i-1] to avoid duplicate triplets
+            if i > 0 and nums[i] == nums[i-1]:
+                continue
+
+            #search with two pointer
+            left, right = i+1, n-1
+
+            while left < right:
+                total = nums[left] + nums[i] + nums[right]
+                if total == 0:
+                    answer.append([nums[left], nums[i], nums[right]])
+                    
+                    #move the pointers past duplicates
+                    while left < right and nums[left] == nums[left+1]:
+                        left += 1
+                    while left < right and nums[right] == nums[right-1]:
+                        right -= 1
+
+                    left += 1
+                    right -= 1
+                elif total < 0:
+                    left += 1
+                else:
+                    right -= 1
+
+        return answer

climbing-stairs/hyeri0903.py

@@ -0,0 +1,17 @@
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        """
+        To find the number of distinct ways clibe to the top
+
+        time complexity: O(n)
+        space complexity: O(n)
+        """
+        dp = [0] * (n+1)
+
+        for i in range(n+1):
+            if i == 0 or i == 1 or i == 2:
+                dp[i] = i
+            else:
+                dp[i] = dp[i-1] + dp[i-2]
+
+        return dp[n]

product-of-array-except-self/hyeri0903.py

@@ -0,0 +1,21 @@
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        """
+        time complexity : O(n)
+        space complexity : O(n)
+        """
+        n = len(nums)
+        prefix = [1] * n
+        suffix = [1] * n
+        answer = [1] * n
+
+        for i in range(1, n):
+            prefix[i] = prefix[i-1] * nums[i-1]
+
+        for i in range(n-2, -1, -1):
+            suffix[i] = suffix[i+1] * nums[i+1]
+        
+        for i in range(n):
+            answer[i] = prefix[i] * suffix[i]
+        
+        return answer

valid-anagram/hyeri0903.py

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+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        """
+        Determine whether two strings are anagrams.
+
+        Time Complexity: O(n)
+        Space Complexity: O(k)
+        - n: length of the string
+        - k: number of unique characters
+        """
+        if len(s) != len(t):
+            return False
+
+        dic_s = {}
+        dic_t = {}
+
+        for i in s:
+            if i in dic_s:
+                dic_s[i] += 1
+            else:
+                dic_s[i] = 1
+        
+        for j in t:
+            if j in dic_t:
+                dic_t[j] += 1
+            else:
+                dic_t[j] = 1
+        
+        return dic_s == dic_t

validate-binary-search-tree/hyeri0903.py

@@ -0,0 +1,35 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
+        """
+        checking validation using inorder traversal
+        always greater than previous node's key
+
+        time complexity: O(n)
+
+        """
+
+        prev = [None]
+
+        def inorder(node):
+            if not node:
+                return True
+
+            #search left sub tree first
+            if not inorder(node.left):
+                return False
+            
+            if prev[0] is not None and node.val <= prev[0]:
+                return False
+
+            #set current node value
+            prev[0] = node.val
+            #search right sub tree
+            return inorder(node.right)
+
+        return inorder(root)