Improved tasks 236-347

Author: javadev

src/main/python/g0201_0300/s0236_lowest_common_ancestor_of_a_binary_tree/readme.md

@@ -46,6 +46,12 @@ According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/
 ## Solution
 
 ```python
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):

src/main/python/g0201_0300/s0238_product_of_array_except_self/readme.md

@@ -34,6 +34,8 @@ You must write an algorithm that runs in `O(n)` time and without using the divis
 ## Solution
 
 ```python
+from typing import List
+
 class Solution:
     def productExceptSelf(self, nums: List[int]) -> List[int]:
         product = 1

src/main/python/g0201_0300/s0239_sliding_window_maximum/readme.md

@@ -59,6 +59,9 @@ Return _the max sliding window_.
 ## Solution
 
 ```python
+from typing import List
+from collections import deque
+
 class Solution:
     def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
         q = deque()

src/main/python/g0201_0300/s0240_search_a_2d_matrix_ii/readme.md

@@ -39,6 +39,8 @@ Write an efficient algorithm that searches for a `target` value in an `m x n` in
 ## Solution
 
 ```python
+from typing import List
+
 class Solution:
     def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
         if not matrix or not matrix[0]:

src/main/python/g0201_0300/s0283_move_zeroes/readme.md

@@ -31,6 +31,8 @@ Given an integer array `nums`, move all `0`'s to the end of it while maintaining
 ## Solution
 
 ```python
+from typing import List
+
 class Solution:
     def moveZeroes(self, nums: List[int]) -> None:
         """

src/main/python/g0201_0300/s0287_find_the_duplicate_number/readme.md

@@ -50,6 +50,8 @@ You must solve the problem **without** modifying the array `nums` and uses only
 ## Solution
 
 ```python
+from typing import List
+
 class Solution:
     def findDuplicate(self, nums: List[int]) -> int:
         arr = [0] * (len(nums) + 1)

src/main/python/g0201_0300/s0295_find_median_from_data_stream/readme.md

@@ -48,6 +48,8 @@ Implement the MedianFinder class:
 ## Solution
 
 ```python
+from typing import List
+
 import heapq
 
 class MedianFinder:

src/main/python/g0201_0300/s0300_longest_increasing_subsequence/readme.md

@@ -39,6 +39,7 @@ A **subsequence** is a sequence that can be derived from an array by deleting so
 ## Solution
 
 ```python
+from typing import List
 import bisect
 
 class Solution:

src/main/python/g0301_0400/s0322_coin_change/readme.md

@@ -40,6 +40,8 @@ You may assume that you have an infinite number of each kind of coin.
 ## Solution
 
 ```python
+from typing import List
+
 class Solution:
     def coinChange(self, coins: List[int], amount: int) -> int:
         dp = [0] * (amount + 1)

src/main/python/g0301_0400/s0338_counting_bits/readme.md

@@ -46,6 +46,8 @@ Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for
 ## Solution
 
 ```python
+from typing import List
+
 class Solution:
     def countBits(self, num: int) -> List[int]:
         result = [0] * (num + 1)