answer:class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def add_trees_in_reverse_level_order(root1, root2): Returns the sum of two binary trees represented as numbers in reversed level-order format. def add_lists_as_numbers(l1, l2): carry, head = 0, None current = None while l1 or l2 or carry: val1 = l1.pop(0) if l1 else 0 val2 = l2.pop(0) if l2 else 0 total = val1 + val2 + carry carry = total // 10 new_node = TreeNode(total % 10) if not head: head = new_node current = head else: current.left = new_node current = current.left return head def tree_to_list(tree): q, result = [tree], [] while q: node = q.pop(0) if node: result.append(node.val) q.append(node.left) q.append(node.right) return result list1 = tree_to_list(root1) list2 = tree_to_list(root2) return add_lists_as_numbers(list1, list2)

answer:def find_subarray_with_target_sum(nums, target): Find a contiguous subarray whose sum is exactly equal to the target. # Dictionary to store cumulative sum and corresponding index cum_sum_map = {} current_sum = 0 for i, num in enumerate(nums): # Add the current number to the running sum current_sum += num # Check if the current running sum is equal to the target if current_sum == target: return nums[:i+1] # Check if any subarray sums up to the target by checking if # (current_sum - target) is present in the cumulative sum map if (current_sum - target) in cum_sum_map: return nums[cum_sum_map[current_sum - target] + 1:i + 1] # Store the current running sum with its index in the map cum_sum_map[current_sum] = i return []

answer:from typing import List def numIslands(mat: List[List[int]]) -> int: Returns the number of islands in the given binary matrix. An island is a group of connected 1's surrounded by 0's. if not mat: return 0 m, n = len(mat), len(mat[0]) visited = [[False] * n for _ in range(m)] def dfs(i, j): if i < 0 or i >= m or j < 0 or j >= n or mat[i][j] == 0 or visited[i][j]: return visited[i][j] = True dfs(i+1, j) # down dfs(i-1, j) # up dfs(i, j+1) # right dfs(i, j-1) # left count = 0 for i in range(m): for j in range(n): if mat[i][j] == 1 and not visited[i][j]: dfs(i, j) count += 1 return count

answer:def longest_valid_parentheses(s): Finds the length of the longest valid parentheses substring. Parameters: s (str): A string consisting of '(' and ')'. Returns: int: The length of the longest valid parentheses substring. max_len = 0 stack = [] start = -1 for i, char in enumerate(s): if char == '(': stack.append(i) else: if stack: stack.pop() if stack: max_len = max(max_len, i - stack[-1]) else: max_len = max(max_len, i - start) else: start = i return max_len