Merge pull request #863 from heypaprika/main

[croucs] Week : 5

Author: TonyKim9401

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

@@ -0,0 +1,15 @@
+# Big-O 예측
+# Time : O(n)
+# Space : O(1)
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        min_price = prices[0]
+        profit = 0
+
+        for price in prices[1:]:
+            if min_price > price:
+                min_price = price
+            
+            profit = max(profit, price - min_price)
+        return profit
+

encode-and-decode-strings/heypaprika.py

@@ -0,0 +1,21 @@
+# Big-O 예측
+# Time : O(n)
+# Space : O(1)
+class Codec:
+    def encode(self, strs: List[str]) -> str:
+        """Encodes a list of strings to a single string.
+        """
+        new_str = "-!@$@#!_".join(strs)
+        return new_str
+
+    def decode(self, s: str) -> List[str]:
+        """Decodes a single string to a list of strings.
+        """
+        return s.split("-!@$@#!_")
+
+
+
+strs = ["Hello","World"]
+codec = Codec()
+codec.decode(codec.encode(strs))
+

group-anagrams/heypaprika.py

@@ -0,0 +1,23 @@
+# Big-O 예측
+# Time : O(k * nlog(n)) (k : strs 배열 수, n : strs 원소의 unique 단어 개수)
+# Space : O(k + n)
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        count_dict = {}
+        num = 0
+        dicted_list = []
+        for s in strs:
+            cur = str(sorted(Counter(s).items()))
+            dicted_list.append(cur)
+            if cur in count_dict:
+                continue
+            count_dict[cur] = num
+            num += 1
+        
+        ans_list = [[] for _ in range(len(count_dict))]
+        for k, v in count_dict.items():
+            for i, dicted in enumerate(dicted_list):
+                if dicted == k:
+                    ans_list[v].append(strs[i])
+        return ans_list
+

implement-trie-prefix-tree/heypaprika.py

@@ -0,0 +1,37 @@
+# Big-O 예측
+# Time : O(n) (n : 단어 개수)
+# Space : O(n) 
+class Trie:
+
+    def __init__(self):
+        self.trie_dict = {}
+        self.trie_dict_list = [{}]
+
+    def insert(self, word: str) -> None:
+        self.trie_dict[word] = 1
+        for i in range(1, len(word) + 1):
+            if i > len(self.trie_dict_list) and i > 1:
+                self.trie_dict_list.append({})
+            self.trie_dict_list[i-1][word[:i]] = 1
+
+    def search(self, word: str) -> bool:
+        return word in self.trie_dict
+
+    def startsWith(self, prefix: str) -> bool:
+        if len(prefix) > len(self.trie_dict_list):
+            return False
+        return prefix in self.trie_dict_list[len(prefix)-1]
+
+
+# Your Trie object will be instantiated and called as such:
+trie = Trie()
+trie.insert("apple")
+trie.search("apple")
+trie.search("app")
+trie.startsWith("app")
+trie.insert("app")
+trie.search("app")
+
+# param_2 = trie.search(word)
+# param_3 = trie.startsWith(prefix)
+

word-break/heypaprika.py

@@ -0,0 +1,31 @@
+# Big-O 예측
+# Time : O(n^2) 최악의 경우 wordDict의 개수의 제곱만큼
+# Space : O(n^2) 최악의 경우 a에 wordDict의 개수의 제곱만큼
+
+# 쉽게 접근한 풀이
+# 맨 앞의 것을 처리할 수 있는 것을 다 고르기 (startswith)
+# 그들을 했을 때의 결과를 가지고 다음 것으로 동일하게 반복 진행하기.
+
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        a = [[s]]
+        i = 1
+        s_list = None
+        duplicate_dict = {}
+        while True:
+            s_list = a[-1]
+            a.append([])
+            for s in s_list:
+                for word in wordDict:
+                    if s.startswith(word):
+                        surplus_word = s[len(word):]
+                        if surplus_word not in duplicate_dict:
+                            a[-1].append(surplus_word)
+                            duplicate_dict[surplus_word] = 1
+            # a[-1] = list(set(a[-1]))
+            if "" in a[-1]:
+                return True
+            if not a[-1]:
+                return False
+            i += 1
+