Merge pull request #414 from kjb512/main

[kayden] Week 04 Solutions

Author: ijlijijij

longest-common-subsequence/kayden.py

@@ -0,0 +1,17 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(N)
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        nums = set(nums)
+        answer = 0
+
+        for num in nums:
+            if num - 1 not in nums:
+                length = 1
+
+                while num + length in nums:
+                    length += 1
+
+                answer = max(answer, length)
+
+        return answer

maximum-product-subarray/kayden.py

@@ -0,0 +1,33 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(1)
+class Solution:
+    def maxProduct(self, nums: List[int]) -> int:
+        n = len(nums)
+        positive, negative = 0, 0
+
+        if nums[0] > 0:
+            positive = nums[0]
+        else:
+            negative = nums[0]
+
+        answer = max(nums)
+
+        for i in range(1, n):
+            if nums[i] >= 0:
+                positive *= nums[i]
+                negative *= nums[i]
+
+                if positive == 0:
+                    positive = nums[i]
+
+            else:
+                temp = positive
+                positive = negative * nums[i]
+                negative = temp * nums[i]
+
+                if negative == 0:
+                    negative = nums[i]
+
+            answer = max(answer, positive)
+
+        return answer

missing-number/kayden.py

@@ -0,0 +1,7 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(1)
+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        n = len(nums)
+        total = n*(n+1)//2
+        return total - sum(nums)

valid-palindrome/kayden.py

@@ -0,0 +1,25 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(N)
+class Solution:
+
+    def isPalindrome(self, s: str) -> bool:
+        string = ""
+
+        for letter in s:
+            if letter.isalnum(): # if ('a' <= char <= 'z') or ('A' <= char <= 'Z') or ('0' <= char <= '9'):
+                string += letter.lower()
+
+
+        def valid(s):
+            start, end = 0, len(s)-1
+
+            while start < end:
+                if s[start] != s[end]:
+                    return False
+
+                start += 1
+                end -= 1
+
+            return True
+
+        return valid(string)

word-search/kayden.py

@@ -0,0 +1,38 @@
+# 시간복잡도: O(N*M*4^limit) limit: word의 길이
+# 공간복잡도: O(N*M)
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        m = len(board)
+        n = len(board[0])
+        limit = len(word)
+        visited = [[False for _ in range(n)] for _ in range(m)]
+        dx = [0, 0, -1, 1]
+        dy = [-1, 1, 0, 0]
+
+        target = 0
+
+        def dfs(x, y, idx):
+
+            if idx == limit - 1:
+                return True
+
+            for i in range(4):
+                nx, ny = x + dx[i], y + dy[i]
+
+                if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny] and board[nx][ny] == word[idx + 1]:
+                    visited[nx][ny] = True
+                    if dfs(nx, ny, idx + 1):
+                        return True
+                    visited[nx][ny] = False
+
+            return False
+
+        for i in range(m):
+            for j in range(n):
+                if board[i][j] == word[target]:
+                    visited[i][j] = True
+                    if dfs(i, j, 0):
+                        return True
+                    visited[i][j] = False
+
+        return False