K0000. 圈复杂度计算

小慕正在参与一个代码复杂度分析项目，他需要计算以下三份代码的圈复杂度。 第1份代码：

def find_paths(maze, x, y, path=None, visited=None):
    if path is None:
        path = []
    if visited is None:
        visited = set()

    rows, cols = len(maze), len(maze[0])

    # 检查是否到达终点
    if x == rows - 1 and y == cols - 1:
        return [path + [(x, y)]]

    # 标记当前点为已访问
    visited.add((x, y))
    path.append((x, y))

    # 定义方向：右、下、左、上
    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
    paths = []

    for dx, dy in directions:
        nx, ny = x + dx, y + dy
        if 0 <= nx < rows and 0 <= ny < cols and (nx, ny) not in visited and maze[nx][ny] == 0:
            paths.extend(find_paths(maze, nx, ny, path[:], visited.copy()))

    return paths

第2份代码：

def coin_change(coins, amount):
    # dp[i] 表示组成金额 i 所需的最少硬币数
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0  # 组成金额 0 不需要任何硬币

    for coin in coins:
        for i in range(coin, amount + 1):
            dp[i] = min(dp[i], dp[i - coin] + 1)

    return dp[amount] if dp[amount] != float('inf') else -1

第3份代码：

def validate_and_process_parentheses(s):
    stack = []  # 用于存储索引
    matched = [-1] * len(s)  # 标记每个位置是否匹配

    for i, char in enumerate(s):
        if char == '(':
            stack.append(i)
        elif char == ')' and stack:
            match_index = stack.pop()
            matched[match_index] = i  # 记录匹配的位置
            matched[i] = match_index

    is_balanced = all(matched[i] != -1 for i in range(len(s)) if s[i] in "()")

    # 寻找最长