Merge pull request #355 from kjb512/main

[kayden] Week 02 Solutions

Author: ijlijijij

construct-binary-tree-from-preorder-and-inorder-traversal/kayden.py

@@ -0,0 +1,31 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(N)
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+
+        mapping = {}
+
+        for i in range(len(inorder)):
+            mapping[inorder[i]] = i
+
+        preorder = collections.deque(preorder)
+
+        def build(start, end):
+            if start > end: return None
+
+            root = TreeNode(preorder.popleft())
+            mid = mapping[root.val]
+
+            root.left = build(start, mid - 1)
+            root.right = build(mid + 1, end)
+
+            return root
+
+        return build(0, len(preorder) - 1)

counting-bits/kayden.py

@@ -0,0 +1,10 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(N)
+class Solution:
+
+    def countBits(self, n: int) -> List[int]:
+        ans = [0 for _ in range(n + 1)]
+        for num in range(1, n + 1):
+            ans[num] = ans[num >> 1] + (num & 1)
+
+        return ans

decode-ways/kayden.py

@@ -0,0 +1,21 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(N)
+class Solution:
+    def numDecodings(self, s: str) -> int:
+        if s[0] == "0":
+            return 0
+
+        n = len(s)
+        dp = [0] * (n + 1)
+        dp[0] = 1
+        dp[1] = 1
+
+        for i in range(2, n + 1):
+
+            if int(s[i - 1]) != 0:
+                dp[i] += dp[i - 1]
+
+            if 10 <= int(s[i - 2:i]) <= 26:
+                dp[i] += dp[i - 2]
+
+        return dp[n]

encode-and-decode-strings/kayden.py

@@ -0,0 +1,29 @@
+class Solution:
+
+    # 시간복잡도: O(N)
+    # 공간복잡도: O(1)
+    def encode(self, strs):
+        res = ""
+        for str in strs:
+            size = len(str)
+            res += str(size)
+            res += str
+
+        return res
+
+    # 시간복잡도: O(N)
+    # 공간복잡도: O(N)
+    def decode(self, str):
+        idx = 0
+        limit = len(str)
+        res = []
+
+        while idx < limit:
+            num = str[idx]
+            text = ""
+            for _ in range(num):
+                text += str[idx]
+                idx+=1
+            res.append(text)
+
+        return res

valid-anagram/kayden.py

@@ -0,0 +1,20 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(N)
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+
+        counter_s = {}
+        for letter in s:
+            counter_s[letter] = counter_s.get(letter, 0) + 1
+
+        counter_t = {}
+        for letter in t:
+            counter_t[letter] = counter_t.get(letter, 0) + 1
+
+        for letter, cnt in counter_s.items():
+            if counter_t.get(letter, 0) != cnt:
+                return False
+
+        return True