Merge pull request #456 from lymchgmk/feat/week5

[EGON] Week 5 Solutions

Author: SamTheKorean

3sum/EGON.py

@@ -0,0 +1,76 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        return self.solveWithTwoPointer(nums)
+
+    """
+    Runtime: 691 ms (Beats 62.42%)
+    Time Complexity: O(n^2)
+        - nums를 정렬하는데 O(n * log n)
+        - 첫 index를 정하기 위해 range(len(nums) - 2) 조회하는데 O(n - 2)
+        - i + 1 부터 n - 1까지 lo, hi 투포인터 조회하는데, i가 최소값인 0인 경우를 upper bound로 계산하면 O(n - 1)
+        > O(n * log n) + O(n - 2) * O(n - 1) ~= O(n * log n) + O(n^2) ~= O(n^2)
+
+    Memory: 20.71 MB (Beats 30.94%)
+    Space Complexity:
+        - num는 정렬하긴 했는데 자기자신 그대로 사용하므로 계산 외
+        > lo나 hi나 triplet_sum은 input에 영향없는 크기의 메모리를 사용하므로 O(1)
+    """
+
+    def solveWithTwoPointer(self, nums: List[int]) -> List[List[int]]:
+        nums.sort()
+        triplets = []
+
+        for i in range(len(nums) - 2):
+            if 1 <= i and nums[i] == nums[i - 1]:
+                continue
+
+            lo, hi = i + 1, len(nums) - 1
+            while lo < hi:
+                triplet_sum = nums[i] + nums[lo] + nums[hi]
+
+                if triplet_sum < 0:
+                    lo += 1
+                elif triplet_sum > 0:
+                    hi -= 1
+                else:
+                    triplets.append([nums[i], nums[lo], nums[hi]])
+
+                    while lo < hi and nums[lo] == nums[lo + 1]:
+                        lo += 1
+                    while lo < hi and nums[hi] == nums[hi - 1]:
+                        hi -= 1
+
+                    lo += 1
+                    hi -= 1
+
+        return triplets
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        nums = [-1, 0, 1, 2, -1, -4]
+        output = [[-1, -1, 2], [-1, 0, 1]]
+        self.assertEqual(Solution.threeSum(Solution(), nums), output)
+
+    def test_2(self):
+        nums = [0, 1, 1]
+        output = []
+        self.assertEqual(Solution.threeSum(Solution(), nums), output)
+
+    def test_3(self):
+        strs = [0, 0, 0]
+        output = [[0, 0, 0]]
+        self.assertEqual(Solution.threeSum(Solution(), strs), output)
+
+    def test_4(self):
+        strs = [0, 0, 0, 0, 0, 0, 0, 0]
+        output = [[0, 0, 0]]
+        self.assertEqual(Solution.threeSum(Solution(), strs), output)
+
+
+if __name__ == '__main__':
+    main()

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

@@ -0,0 +1,55 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        return self.solveWithStack(prices)
+
+    """
+    Runtime: 749 ms (Beats 46.22%)
+    Time Complexity: O(n)
+        - prices의 길이 n 만큼 조회에 O(n)
+        - 조회하며 수행하는 연산들은 .pop, .append, 고정된 크기에 대한 max 이므로 O(1)
+        > O(n) * O(1) ~= O(n) 
+        
+    Memory: 27.33 MB (Beats 91.14%)
+    Space Complexity: O(1)
+        - 크기가 1로 유지되는 stack: List[int] 사용
+        - max_profit: int 사용
+        > O(1)
+    """
+    def solveWithStack(self, prices: List[int]) -> int:
+        max_profit = 0
+        stack = []
+        for price in prices:
+            if not stack:
+                stack.append(price)
+                continue
+
+            min_price = stack.pop()
+            if min_price < price:
+                max_profit = max(price - min_price, max_profit)
+                stack.append(min_price)
+            elif min_price > price:
+                stack.append(price)
+            else:
+                stack.append(min_price)
+
+        return max_profit
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        prices = [7,1,5,3,6,4]
+        output = 5
+        self.assertEqual(Solution.maxProfit(Solution(), prices), output)
+
+    def test_2(self):
+        prices = [7,6,4,3,1]
+        output = 0
+        self.assertEqual(Solution.maxProfit(Solution(), prices), output)
+
+
+if __name__ == '__main__':
+    main()

group-anagrams/EGON.py

@@ -0,0 +1,52 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        return self.solveWithHashableKey(strs)
+
+    """
+    Runtime: 88 ms (Beats 65.68%)
+    Time Complexity: 
+        - strs의 길이 n 만큼 조회에 O(n)
+        - string을 key로 변환하는데 string의 길이 L에 대해 O(L * log L) * O(L)
+            - string의 최대 길이는 100이므로 O(100 * log 100) * O(100)가 upper bound
+        - anagram_dict 갱신에 O(1)
+        - 최대 크기가 n인 anagram_dict.values()를 list로 변환하는데 O(n^2)
+        > O(n) * O(L * log L) * O(L) + O(n^2) ~= O(n^2)
+
+    Memory: 19.32 MB (Beats 95.12%)
+    Space Complexity: O(n * L)
+        - 최대 n개의 key-value pair를 가질 수 있는 anagram_dict 사용, upper bound
+          anagram_dict의 value는 최대 길이 L인 List[str] 이므로 O(n * L)
+        > O(n * L)
+    """
+    def solveWithHashableKey(self, strs: List[str]) -> List[List[str]]:
+        anagram_dict = {}
+        for string in strs:
+            key = ''.join(sorted(string))
+            anagram_dict[key] = anagram_dict.get(key, []) + [string]
+
+        return list(anagram_dict.values())
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        strs = ["eat","tea","tan","ate","nat","bat"]
+        output = [["bat"],["nat","tan"],["ate","eat","tea"]]
+        self.assertEqual(Solution.groupAnagrams(Solution(), strs), output)
+
+    def test_2(self):
+        strs = [""]
+        output = [[""]]
+        self.assertEqual(Solution.groupAnagrams(Solution(), strs), output)
+
+    def test_3(self):
+        strs = ["a"]
+        output = [["a"]]
+        self.assertEqual(Solution.groupAnagrams(Solution(), strs), output)
+
+
+if __name__ == '__main__':
+    main()

implement-trie-prefix-tree/EGON.py

@@ -0,0 +1,77 @@
+from unittest import TestCase, main
+
+
+"""
+Runtime: 136 ms (Beats 33.86%)
+Time Complexity:
+    > insert: word의 각 문자마다 조회하므로 O(L)
+    > search: 최대 word의 모든 문자를 조회하므로 O(L), upper bound
+    > startsWith: 최대 prefix의 모든 문자를 조회하므로 O(L'), upper bound
+
+Memory: 32.60 MB (Beats 16.08%)
+Space Complexity: O(n * L), upper bound
+    - 최악의 경우 삽입하는 모든 word가 공통점이 하나도 없는 경우를 상정해보면, 삽입하는 word의 갯수를 n, word의 최대 길이를 L
+    > O(n * L), upper bound
+"""
+
+
+class Node:
+    def __init__(self, key, data=None):
+        self.key = key
+        self.data = data
+        self.children = {}
+
+
+class Trie:
+
+    def __init__(self):
+        self.root = Node(None)
+
+    def insert(self, word: str) -> None:
+        curr_node = self.root
+        for char in word:
+            if char not in curr_node.children:
+                curr_node.children[char] = Node(char)
+
+            curr_node = curr_node.children[char]
+
+        curr_node.data = word
+
+    def search(self, word: str) -> bool:
+        curr_node = self.root
+        for char in word:
+            if char in curr_node.children:
+                curr_node = curr_node.children[char]
+            else:
+                return False
+
+        return curr_node.data == word
+
+    def startsWith(self, prefix: str) -> bool:
+        curr_node = self.root
+        for char in prefix:
+            if char in curr_node.children:
+                curr_node = curr_node.children[char]
+            else:
+                return False
+        else:
+            return True
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        # commands = ["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
+        # words = [[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
+        # output = [None, None, True, False, True, None, True]
+
+        trie = Trie()
+        trie.insert("apple")
+        self.assertEqual(trie.search("apple"), True)
+        self.assertEqual(trie.search("app"), False)
+        self.assertEqual(trie.startsWith("app"), True)
+        trie.insert("app")
+        self.assertEqual(trie.search("app"), True)
+
+
+if __name__ == '__main__':
+    main()

word-break/EGON.py

@@ -0,0 +1,122 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Node:
+    def __init__(self, key, data=None):
+        self.key = key
+        self.data = data
+        self.children = {}
+
+
+class Trie:
+
+    def __init__(self):
+        self.root = Node(None)
+
+    def insert(self, word: str) -> None:
+        curr_node = self.root
+        for char in word:
+            if char not in curr_node.children:
+                curr_node.children[char] = Node(char)
+
+            curr_node = curr_node.children[char]
+
+        curr_node.data = word
+
+    def search(self, word: str) -> bool:
+        curr_node = self.root
+        for char in word:
+            if char in curr_node.children:
+                curr_node = curr_node.children[char]
+            else:
+                return False
+
+        return curr_node.data == word
+
+    def startsWith(self, prefix: str) -> bool:
+        curr_node = self.root
+        for char in prefix:
+            if char in curr_node.children:
+                curr_node = curr_node.children[char]
+            else:
+                return False
+        else:
+            return True
+
+
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        return self.solveWithTrieDFSMemo(s, wordDict)
+
+    """
+    Runtime: 38 ms (Beats 70.76%)
+    Time Complexity: O(n * L + s^2)
+        - s의 길이를 S, wordDict의 길이를 L, word의 최대 길이를 W라 하면
+            - trie에 insert하는데 O(W * L), upper bound
+            - DFS에서 curr_s의 모든 문자 조회에 최대 O(S), 접두사를 제거하는 curr_s.removeprefix에서 최대 O(S)
+        > O(W * L) + O(S) * O(S) = O(W * L + S^2) upper bound
+
+    Memory: 17.70 MB (Beats 5.25%)
+    Space Complexity:
+         - trie 생성 및 갱신에 O(W * L)
+         - visited는 최악의 경우 s의 모든 부분 문자열을 저장하므로 O(S^2) upper bound
+         > O(W * L) + O(S^2) = O(W * L + S^2) upper bound
+    """
+    def solveWithTrieDFSMemo(self, s: str, wordDict: List[str]) -> bool:
+        trie = Trie()
+        for word in wordDict:
+            trie.insert(word)
+
+        stack = [s]
+        visited = set()
+        while stack:
+            curr_s = stack.pop()
+            if curr_s in visited:
+                continue
+
+            curr_node = trie.root
+            for char in curr_s:
+                if char in curr_node.children:
+                    curr_node = curr_node.children[char]
+                    if curr_node.data is not None:
+                        post_s = curr_s.removeprefix(curr_node.data)
+                        if not post_s:
+                            return True
+                        else:
+                            stack.append(post_s)
+                else:
+                    visited.add(curr_s)
+                    break
+
+        return False
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        s = "leetcode"
+        wordDict = ["leet","code"]
+        output = True
+        self.assertEqual(Solution.wordBreak(Solution(), s, wordDict), output)
+
+    def test_2(self):
+        s = "applepenapple"
+        wordDict = ["apple","pen"]
+        output = True
+        self.assertEqual(Solution.wordBreak(Solution(), s, wordDict), output)
+
+    def test_3(self):
+        s = "catsandog"
+        wordDict = ["cats","dog","sand","and","cat"]
+        output = False
+        self.assertEqual(Solution.wordBreak(Solution(), s, wordDict), output)
+
+    def test_4(self):
+        s = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
+        wordDict = ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]
+        output = False
+        self.assertEqual(Solution.wordBreak(Solution(), s, wordDict), output)
+
+
+if __name__ == '__main__':
+    main()