Merge pull request #448 from kjb512/main

[kayden] Week 05 Solutions

Author: ijlijijij

3sum/kayden.py

@@ -0,0 +1,62 @@
+class Solution:
+
+    # 해시맵
+    # 시간복잡도: O(N^2)
+    # 공간복잡도: O(N)
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        n = len(nums)
+        nums.sort()
+
+        check = {}
+        for idx, num in enumerate(nums):
+            check[num] = idx
+
+        answer = set()
+        for i in range(n-2):
+            if nums[i] > 0:
+                break
+            if i > 0 and nums[i] == nums[i-1]:
+                continue
+            for j in range(i + 1, n):
+                target = -(nums[i] + nums[j])
+                if not check.get(target):
+                    continue
+                if j >= check[target]:
+                    continue
+
+                answer.add((nums[i], nums[j], target))
+
+        return list(answer)
+
+    # 투포인터
+    # 시간복잡도: O(N^2)
+    # 공간복잡도: O(N)
+    def threeSum2(self, nums: List[int]) -> List[List[int]]:
+        n = len(nums)
+        nums.sort()
+
+        answer = set()
+        for i in range(n-2):
+            if nums[i] > 0:
+                break
+            if i > 0 and nums[i] == nums[i-1]:
+                continue
+            l, r = i+1, n-1
+            while l<r:
+                if nums[l] + nums[r] == -nums[i]:
+                    answer.add((nums[i], nums[l], nums[r]))
+                    l += 1
+                    r -= 1
+                    while nums[l] == nums[l - 1] and l < r:
+                        l += 1
+
+                    while nums[r] == nums[r + 1] and l < r:
+                        r -= 1
+
+                if nums[l] + nums[r] < -nums[i]:
+                    l += 1
+
+                if nums[l] + nums[r] > -nums[i]:
+                    r -= 1
+
+        return list(answer)

best-time-to-buy-and-sell-stock/kayden.py

@@ -0,0 +1,15 @@
+# 시간복잡도: O(N)
+# 공간복잡도: O(1)
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+
+        answer = 0
+        cur = float(inf)
+        for price in prices:
+            if cur < price:
+                answer = max(answer, price - cur)
+
+            if price < cur:
+                cur = price
+
+        return answer

group-anagrams/kayden.py

@@ -0,0 +1,10 @@
+# 시간복잡도: O(N*AlogA) A: 0<= strs[i]의 길이 <= 100
+# 공간복잡도: O(N)
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        groups = {}
+        for anagram in strs:
+            key = str(sorted(anagram))
+            groups.setdefault(key, []).append(anagram)
+
+        return list(groups.values())

implement-trie-prefix-tree/kayden.py

@@ -0,0 +1,40 @@
+class Node:
+
+    def __init__(self, ending=False):
+        self.children = {}
+        self.ending = ending
+
+# 공간복잡도: O(w*l) w: 단어 수 l: 단어의 평균 길이
+class Trie:
+
+    def __init__(self):
+        self.head = Node(ending=True)
+
+    # 시간복잡도: O(N)
+    def insert(self, word: str) -> None:
+        node = self.head
+        for ch in word:
+            if ch not in node.children:
+                node.children.setdefault(ch, Node())
+            node = node.children[ch]
+        node.ending = True
+
+    # 시간복잡도: O(N)
+    def search(self, word: str) -> bool:
+        node = self.head
+        for ch in word:
+            if ch not in node.children:
+                return False
+            node = node.children[ch]
+
+        return node.ending
+
+    # 시간복잡도: O(N)
+    def startsWith(self, prefix: str) -> bool:
+        node = self.head
+        for ch in prefix:
+            if ch not in node.children:
+                return False
+            node = node.children[ch]
+
+        return True

word-break/kayden.py

@@ -0,0 +1,25 @@
+# 시간복잡도: O(S*W)
+#   S: s의 길이 300 W: worDict 각 단어의 총 길이 20*1000
+#       300 * 20*1000 = 6*1e6 (600만)
+# 공간복잡도: O(S)
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        memo = {}
+        def dfs(idx):
+            if idx in memo:
+                return memo[idx]
+
+            if idx == len(s):
+                return True
+
+            for word in wordDict:
+                l = len(word)
+                if s[idx:idx + l] == word:
+                    if dfs(idx + l):
+                        memo[idx] = True
+                        return True
+
+            memo[idx] = False
+            return False
+
+        return dfs(0)