feat : solve with python

Author: changmuk.im

longest-substring-without-repeating-characters/EGON.py

@@ -0,0 +1,80 @@
+from unittest import TestCase, main
+
+
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        return self.solve_sliding_window(s)
+
+    """
+    Runtime: 313 ms (Beats 8.97%)
+    Time Complexity:
+        - left가 0에서 len(s)까지 조회, right가 left + 1 부터 len(s)까지 조회하므로 O(n * (n + 1) / 2)
+            - left가 조회할 때마다, 2항 max 연산하므로 * 2
+        > O((n * (n + 1) / 2) * 2) ~= O(n ^ 2)
+
+    Memory: 16.51 (Beats 81.63%)
+    Space Complexity: O(n)
+        > checker가 최대 s의 길이만큼 커질 수 있으므로 O(n), upper bound
+    """
+    def solve_two_pointer(self, s: str) -> int:
+        if not s:
+            return 0
+
+        max_length = 1
+        for left in range(len(s)):
+            if len(s) - left + 1 < max_length:
+                return max_length
+
+            right = left + 1
+            checker = set(s[left])
+            while right < len(s):
+                if s[right] in checker:
+                    break
+
+                checker.add(s[right])
+                right += 1
+
+            max_length = max(max_length, len(checker))
+
+        return max_length
+
+    """
+    Runtime: 58 ms (Beats 46.47%)
+    Time Complexity:
+        - 중복 검사는 set을 사용하므로 O(1)
+        - right가 len(s)까지 조회하므로 O(n)
+            - right가 조회한 뒤 2항 max 연산하는데 O(2)
+        - left가 최대 right까지 조회하고 right < len(s) 이므로 O(n), upper bound
+        > O(n) * O(2) + O(n) ~= O(n)
+
+    Memory: 16.60 (Beats 41.73%)
+    Space Complexity: O(n)
+        > checker가 최대 s의 길이만큼 커질 수 있으므로 O(n), upper bound
+    """
+    def solve_sliding_window(self, s: str) -> int:
+        max_length = 0
+        left = right = 0
+        checker = set()
+        while left < len(s) and right < len(s):
+            while right < len(s) and s[right] not in checker:
+                checker.add(s[right])
+                right += 1
+
+            max_length = max(max_length, len(checker))
+
+            while left < len(s) and (right < len(s) and s[right] in checker):
+                checker.remove(s[left])
+                left += 1
+
+        return max_length
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        s = "pwwkew"
+        output = 3
+        self.assertEqual(Solution.lengthOfLongestSubstring(Solution(), s), output)
+
+
+if __name__ == '__main__':
+    main()

number-of-islands/EGON.py

@@ -0,0 +1,172 @@
+from collections import deque
+from typing import Generic, List, Optional, TypeVar
+from unittest import TestCase, main
+
+
+T = TypeVar('T')
+
+
+class Node(Generic[T]):
+    def __init__(self, value: T):
+        self.value = value
+        self.prev = None
+        self.post = None
+
+
+class Deque(Generic[T]):
+    def __init__(self):
+        self.head: Optional[Node[T]] = None
+        self.tail: Optional[Node[T]] = None
+        self.size = 0
+
+    def is_empty(self):
+        return self.size == 0
+
+    def appendright(self, value: T):
+        node = Node(value)
+        if self.is_empty():
+            self.head = self.tail = node
+        else:
+            node.prev = self.tail
+            if self.tail:
+                self.tail.post = node
+            self.tail = node
+        self.size += 1
+
+    def appendleft(self, value: T):
+        node = Node(value)
+        if self.is_empty():
+            self.head = self.tail = node
+        else:
+            node.post = self.head
+            if self.head:
+                self.head.prev = node
+            self.head = node
+        self.size += 1
+
+    def popright(self) -> T:
+        if self.is_empty():
+            raise IndexError("Deque is empty!")
+
+        value = self.tail.value
+        self.tail = self.tail.prev
+        if self.tail:
+            self.tail.post = None
+        else:
+            self.head = None
+        self.size -= 1
+        return value
+
+    def popleft(self) -> T:
+        if self.is_empty():
+            raise IndexError("Deque is empty!")
+
+        value = self.head.value
+        self.head = self.head.post
+        if self.head:
+            self.head.prev = None
+        else:
+            self.tail = None
+        self.size -= 1
+        return value
+
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        return self.solve_bfs_custom(grid)
+
+    """
+    Runtime: 243 ms (Beats 60.00%)
+    Time Complexity: O(MAX_R * MAX_C)
+        - 2차원 배열 grid를 조회하며 deq에 enqueue하는데 O(MAX_R * MAX_C)
+            - deq의 원소 하나 당 DIRS 크기만큼 탐색에 O(4), 원소에 해당하는 grid의 값이 "0"인 경우 탐색하지 않으므로 upper bound
+            - deque의 appendleft, popleft에 O(1)
+        > O(MAX_R * MAX_C) * O(4) ~= O(MAX_R * MAX_C)
+
+    Memory: 18.82 (Beats 83.56%)
+    Space Complexity: O(MAX_R * MAX_C)
+        - visited 역할을 grid의 원소의 값을 변경하여 사용하였으므로 무시
+        - deque의 최대 크기는 MAX_R * MAX_C 이므로 O(MAX_R * MAX_C), upper bound
+        > O(MAX_R * MAX_C)
+    """
+    def solve_bfs(self, grid: List[List[str]]) -> int:
+
+        def is_island(r: int, c: int) -> bool:
+            nonlocal grid
+
+            if grid[r][c] != "1":
+                return False
+
+            deq = deque([(r, c)])
+            deq.appendleft((r, c))
+            while deq:
+                curr_r, curr_c = deq.popleft()
+                grid[r][c] = "0"
+                for dir_r, dir_c in DIRS:
+                    post_r, post_c = curr_r + dir_r, curr_c + dir_c
+                    if 0 <= post_r < MAX_R and 0 <= post_c < MAX_C and grid[post_r][post_c] == "1":
+                        grid[post_r][post_c] = "0"
+                        deq.appendleft((post_r, post_c))
+
+            return True
+
+        MAX_R, MAX_C = len(grid), len(grid[0])
+        DIRS = ((-1, 0), (1, 0), (0, -1), (0, 1))
+        count = 0
+        for r in range(MAX_R):
+            for c in range(MAX_C):
+                count += 1 if is_island(r, c) else 0
+        return count
+
+    """
+    Runtime: 283 ms (Beats 23.05%)
+    Time Complexity: O(MAX_R * MAX_C)
+        > solve_bfs와 동일
+    Memory: 19.98 (Beats 44.38%)
+    Space Complexity: O(MAX_R * MAX_C)
+        > solve_bfs와 동일
+    """
+    def solve_bfs_custom(self, grid: List[List[str]]) -> int:
+
+        def is_island(r: int, c: int) -> bool:
+            nonlocal grid
+
+            if grid[r][c] != "1":
+                return False
+
+            deq = Deque()
+            deq.appendleft((r, c))
+            while not deq.is_empty():
+                curr_r, curr_c = deq.popleft()
+                grid[r][c] = "0"
+                for dir_r, dir_c in DIRS:
+                    post_r, post_c = curr_r + dir_r, curr_c + dir_c
+                    if 0 <= post_r < MAX_R and 0 <= post_c < MAX_C and grid[post_r][post_c] == "1":
+                        grid[post_r][post_c] = "0"
+                        deq.appendleft((post_r, post_c))
+
+            return True
+
+        MAX_R, MAX_C = len(grid), len(grid[0])
+        DIRS = ((-1, 0), (1, 0), (0, -1), (0, 1))
+        count = 0
+        for r in range(MAX_R):
+            for c in range(MAX_C):
+                count += 1 if is_island(r, c) else 0
+        return count
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        grid = [
+            ["1", "1", "0", "0", "0"],
+            ["1", "1", "0", "0", "0"],
+            ["0", "0", "1", "0", "0"],
+            ["0", "0", "0", "1", "1"]
+        ]
+        output = 3
+        self.assertEqual(Solution.numIslands(Solution(), grid), output)
+
+
+if __name__ == '__main__':
+    main()

reverse-linked-list/EGON.py

@@ -0,0 +1,88 @@
+from typing import List, Optional
+from unittest import TestCase, main
+
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+    def description(self) -> List[int]:
+        desc = []
+        node = self
+        while node:
+            desc.append(node.val)
+            node = node.next
+
+        return desc
+
+
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        return self.solve_recursively(head)
+
+    """
+    Runtime: 41 ms (Beats 28.37%)
+    Time Complexity: O(n)
+        - Linked List의 모든 node를 순회하는데 O(n)
+        - post 변수 할당, curr.next, prev와 curr의 참조 변경은 O(1)
+        > O(n) * O(1) ~= O(n)
+
+    Memory: 17.66 (Beats 85.13%)
+    Space Complexity: O(1)
+        - Linked List의 크기와 무관한 ListNode 타입의 prev, curr, post 변수를 할당하여 사용
+        > O(1)
+    """
+    def solve_iteratively(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        prev, curr = None, head
+        while curr:
+            post = curr.next
+            curr.next = prev
+            prev, curr = curr, post
+
+        return prev
+
+    """
+    Runtime: 38 ms (Beats 50.06%)
+    Time Complexity: O(n)
+        - Linked List의 모든 node 만큼 재귀를 호출하므로 O(n)
+        - post 변수 할당, curr.next의 참조 변경은 O(1)
+        > O(n) * O(1) ~= O(n)
+
+    Memory: 17.78 (Beats 27.88%)
+    Space Complexity: O(n)
+        > Linked List의 모든 node 만큼 reverse 함수가 스택에 쌓이므로 O(n)
+    """
+    def solve_recursively(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        def reverse(prev: Optional[ListNode], curr: Optional[ListNode]) -> Optional[ListNode]:
+            if not curr:
+                return prev
+
+            post = curr.next
+            curr.next = prev
+            return reverse(curr, post)
+
+        return reverse(None, head)
+
+
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        node1 = ListNode(val=1)
+        node2 = ListNode(val=2)
+        node3 = ListNode(val=3)
+        node4 = ListNode(val=4)
+        node5 = ListNode(val=5)
+        node1.next = node2
+        node2.next = node3
+        node3.next = node4
+        node4.next = node5
+
+        output = [5, 4, 3, 2, 1]
+        self.assertEqual(Solution.reverseList(Solution(), node1).description(), output)
+
+
+if __name__ == '__main__':
+    main()