Merge pull request #467 from lymchgmk/feat/weak6

[EGON] Week 6 Solutions

container-with-most-water/EGON.py

@@ -0,0 +1,47 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        return self.solveWithTwoPointer(height)
+
+    """
+    Runtime: 527 ms (Beats 47.12%)
+    Time Complexity: O(n)
+        - 투 포인터의 총합 조회 범위가 height의 길이와 같으므로 O(n)
+        - area 갱신을 위한 계산에서 항이 2개인 max와 항이 2개인 min 중첩에 O(2) * O(2)
+        > O(n) * O(4) ~= O(n)
+
+    Memory: 29.61 MB (Beats 38.93%)
+    Space Complexity: O(1)
+        > 정수형 변수들만 사용했으므로 O(1)
+    """
+    def solveWithTwoPointer(self, height: List[int]) -> int:
+        left, right = 0, len(height) - 1
+        area = 0
+        while left < right:
+            area = max(area, (right - left) * min(height[right], height[left]))
+
+            if height[left] <= height[right]:
+                left += 1
+            else:
+                right -= 1
+
+        return area
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        height = [1, 8, 6, 2, 5, 4, 8, 3, 7]
+        output = 49
+        self.assertEqual(Solution.maxArea(Solution(), height), output)
+
+    def test_2(self):
+        height = [1, 1]
+        output = 1
+        self.assertEqual(Solution.maxArea(Solution(), height), output)
+
+
+if __name__ == '__main__':
+    main()

design-add-and-search-words-data-structure/EGON.py

@@ -0,0 +1,86 @@
+from unittest import TestCase, main
+
+
+class Node:
+
+    def __init__(self, key):
+        self.key = key
+        self.isWordEnd = False
+        self.children = {}
+
+
+class Trie:
+    WILD_CARD = '.'
+
+    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.isWordEnd = True
+
+    def search(self, node: Node, word: str, idx: int) -> bool:
+        if idx == len(word):
+            return node.isWordEnd
+
+        for idx in range(idx, len(word)):
+            if word[idx] == self.WILD_CARD:
+                for child in node.children.values():
+                    if self.search(child, word, idx + 1) is True:
+                        return True
+                else:
+                    return False
+
+            if word[idx] in node.children:
+                return self.search(node.children[word[idx]], word, idx + 1)
+            else:
+                return False
+
+
+"""
+Runtime: 1810 ms (Beats 22.46%)
+Time Complexity:
+    > addWord: word의 길이만큼 순회하므로 O(L)
+    > search:
+        - word의 평균 길이를 W이라하면,
+        - '.'가 포함되어 있지 않는 경우 O(W), early return 가능하므로 upper bound
+        - '.'가 포함되어 있는 경우, 해당 노드의 child만큼 재귀, trie의 평균 자식 수를 C라 하면 O(W) * O(C), early return 가능하므로 upper bound
+            - trie의 평균 자식 수는 addWord의 실행횟수 C'에 선형적으로 비레(겹치는 char가 없는 경우 upper bound)
+        > O(W) * O(C) ~= O(W) * O(C') ~= O(W * C'), upper bound
+
+Memory: 66.78 MB (Beats 12.26%)
+Space Complexity: O(1)
+    > addWord: 
+        - 삽입한 word의 평균 길이 L만큼 Node가 생성 및 Trie에 추가, O(L)
+        - addWord의 실행횟수 C'에 비례, O(C')
+        > O(L) * O(C') ~= O(L * C')
+    > search:
+        > 만들어진 Trie와 패러미터 word, 정수 변수 idx를 사용하므로 O(1)
+    
+"""
+class WordDictionary:
+
+    def __init__(self):
+        self.trie = Trie()
+
+    def addWord(self, word: str) -> None:
+        self.trie.insert(word)
+
+    def search(self, word: str) -> bool:
+        return self.trie.search(self.trie.root, word, 0)
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        wordDictionary = WordDictionary()
+        wordDictionary.addWord("at")
+        self.assertEqual(wordDictionary.search(".at"), False)
+
+
+if __name__ == '__main__':
+    main()

longest-increasing-subsequence/EGON.py

@@ -0,0 +1,55 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def lengthOfLIS(self, nums: List[int]) -> int:
+        return self.solve_with_Memo_BS(nums)
+
+    """
+    Runtime: 68 ms (Beats 86.42%)
+    Time Complexity: O(n * log n)
+        - nums 배열 조회에 O(n)
+        - 최악의 경우 num의 모든 원소에 대해 bisect_left 실행가능, O(log n) upper bound
+        > O(n) * O(log n) ~= O(n * log n)
+
+    Memory: 16.92 MB (Beats 29.49%)
+    Space Complexity: O(n)
+        > 최대 크기가 n인 lis 배열 사용에 O(n)
+    """
+    def solve_with_Memo_BS(self, nums: List[int]) -> int:
+
+        def bisect_left(lis: List[int], target: int):
+            lo, hi = 0, len(lis)
+            while lo < hi:
+                mid = lo + (hi - lo) // 2
+                if lis[mid] < target:
+                    lo = mid + 1
+                else:
+                    hi = mid
+
+            return lo
+
+        lis = []
+        for num in nums:
+            if not lis:
+                lis.append(num)
+                continue
+
+            if lis[-1] < num:
+                lis.append(num)
+            else:
+                lis[bisect_left(lis, num)] = num
+
+        return len(lis)
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        prices = [10,9,2,5,3,7,101,18]
+        output = 4
+        self.assertEqual(Solution.lengthOfLIS(Solution(), prices), output)
+
+
+if __name__ == '__main__':
+    main()

spiral-matrix/EGON.py

@@ -0,0 +1,49 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        return self.solve(matrix)
+
+    """
+    Runtime: 37 ms (Beats 44.53%)
+    Time Complexity: O(MAX_R * MAX_C)
+
+    Memory: 16.56 MB (Beats 43.42%)
+    Space Complexity: O(1)
+        - result를 제외하고 matrix의 값을 변경해서 visited 분기
+    """
+    def solve(self, matrix: List[List[int]]) -> List[int]:
+        MAX_R, MAX_C = len(matrix), len(matrix[0])
+        DIRS = ((0, 1), (1, 0), (0, -1), (-1, 0))
+        result = []
+        r, c, dir_idx = 0, -1, 0
+        for _ in range(MAX_R * MAX_C):
+            r += DIRS[dir_idx][0]
+            c += DIRS[dir_idx][1]
+
+            if 0 <= r < MAX_R and 0 <= c < MAX_C and matrix[r][c] is not None:
+                result.append(matrix[r][c])
+                matrix[r][c] = None
+            else:
+                r -= DIRS[dir_idx][0]
+                c -= DIRS[dir_idx][1]
+                dir_idx = (dir_idx + 1) % len(DIRS)
+                r += DIRS[dir_idx][0]
+                c += DIRS[dir_idx][1]
+                result.append(matrix[r][c])
+                matrix[r][c] = None
+
+        return result
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
+        output = [1,2,3,4,8,12,11,10,9,5,6,7]
+        self.assertEqual(Solution.spiralOrder(Solution(), matrix), output)
+
+
+if __name__ == '__main__':
+    main()

valid-parentheses/EGON.py

@@ -0,0 +1,58 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def isValid(self, s: str) -> bool:
+        return self.solveWithStack(s)
+
+    """
+    Runtime: 34 ms (Beats 66.54%)
+    Time Complexity: O(n)
+        > 문자열 s의 길이 L만큼 조회하므로 O(n), early return이 있으므로 upper bound
+
+    Memory: 16.59 MB (Beats 50.97%)
+    Space Complexity: O(n)
+        - close_p_dict의 크기는 상수로 처리
+        > 최대 길이가 L인 배열 stack을 사용했으므로 O(n)
+    """
+    def solveWithStack(self, s: str) -> bool:
+        close_p_dict = {')': '(', '}': '{', ']': '['}
+        stack = []
+        for curr_p in s:
+            if not stack or curr_p not in close_p_dict:
+                stack.append(curr_p)
+                continue
+
+            if close_p_dict[curr_p] != stack[-1]:
+                return False
+            else:
+                stack.pop()
+
+        return not stack
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        s = "()"
+        output = True
+        self.assertEqual(Solution.isValid(Solution(), s), output)
+
+    def test_2(self):
+        s = "()[]{}"
+        output = True
+        self.assertEqual(Solution.isValid(Solution(), s), output)
+
+    def test_3(self):
+        s = "(]"
+        output = False
+        self.assertEqual(Solution.isValid(Solution(), s), output)
+
+    def test_4(self):
+        s = "([])"
+        output = True
+        self.assertEqual(Solution.isValid(Solution(), s), output)
+
+
+if __name__ == '__main__':
+    main()