738. Monotone Increasing Digits 968. Binary Tree Cameras

738. Monotone Increasing Digits

An integer has monotone increasing digits单调递增数字 if and only if each pair of adjacent digits x and y satisfy x <= y.

Given an integer n, return the largest number that is less than or equal to n with monotone increasing digits.

 violent solution:

Time complexity: O(n × m)      m is the length of the number n
Space complexity: O(1)

class Solution:
    def monotoneIncreasingDigits(self, n: int) -> int:
        for i in range(n, -1, -1):
            if self.check_num(i):
                return i
        #return 0

    def check_num(self, n):
        max = 10
        while n:
            t = n % 10
            if max >= t:
                max = t
            else:
                return False
            n = n//10
        return True

greedy:

1. A local optimum leads to a global

2. traversal from right to the left

class Solution:
    def monotoneIncreasingDigits(self, n: int) -> int:
        n_str = list(str(n))

        for i in range(len(n_str) - 1, 0, -1):
            if n_str[i] < n_str[i - 1]: #string can compare the value, but you can still use int()
                n_str[i - 1] = str(int(n_str[i - 1]) - 1)

                for j in range(i, len(n_str)):
                    n_str[j] = '9'
        
        return int(''.join(n_str))

Time complexity: O(n),   n is the length of the number
Space complexity: O(n), need a string, it is more convenient to convert to string operation

968. Binary Tree Cameras

You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor监控 its parent, itself, and its immediate children.

Return the minimum number of cameras needed to monitor all nodes of the tree.

Local optimization: let the parent of a leaf node plant a camera, the least number of cameras used.
Overall optimization: minimize the number of cameras used for all. 

Case 1: Both left and right nodes are covered

 Case 2: At least one of the left and right nodes is uncovered

 Case 3: At least one of the left and right nodes has a camera

 Case 4: Header node not covered

 

class Solution:
         # Greedy Algo:
        # 从下往上安装摄像头：跳过leaves这样安装数量最少，局部最优 -> 全局最优
        # 先给leaves的父节点安装，然后每隔两层节点安装一个摄像头，直到Head
        # 0: 该节点未覆盖
        # 1: 该节点有摄像头
        # 2: 该节点有覆盖
    def minCameraCover(self, root: TreeNode) -> int:
        # 定义递归函数
        result = [0]  # 用于记录摄像头的安装数量
        if self.traversal(root, result) == 0:
            result[0] += 1

        return result[0]

        
    def traversal(self, cur: TreeNode, result: List[int]) -> int:
        if not cur:
            return 2

        left = self.traversal(cur.left, result)
        right = self.traversal(cur.right, result)

        # 情况1: 左右节点都有覆盖
        if left == 2 and right == 2:
            return 0

        # 情况2:
        # left == 0 && right == 0 左右节点无覆盖
        # left == 1 && right == 0 左节点有摄像头，右节点无覆盖
        # left == 0 && right == 1 左节点无覆盖，右节点有摄像头
        # left == 0 && right == 2 左节点无覆盖，右节点覆盖
        # left == 2 && right == 0 左节点覆盖，右节点无覆盖
        if left == 0 or right == 0:
            result[0] += 1
            return 1

        # 情况3:
        # left == 1 && right == 2 左节点有摄像头，右节点有覆盖
        # left == 2 && right == 1 左节点有覆盖，右节点有摄像头
        # left == 1 && right == 1 左右节点都有摄像头
        if left == 1 or right == 1:
            return 2