[数据结构与算法]Leetcode 2100. Find Good Days to Rob the Bank [Python]

设置两个状态array，分别记录当前的位置之前，有多少人头非增加的天数，和此位置之后人头非减少的天数。处理后一个array时，反过来处理，其实也就是反着看security在当前位置，有多少非增加的天数。然后，再次遍历security位置，看每个位置是否满足三条件，后两个条件可以直接查两个状态array得到。

class Solution:
    def goodDaysToRobBank(self, security: List[int], time: int) -> List[int]:
        increasing = [0 for _ in range(len(security))]
        decreasing = [0 for _ in range(len(security))]
        
        for idx in range(1, len(security)):
            if security[idx-1] >= security[idx]:
                increasing[idx] = increasing[idx-1] + 1
            else:
                increasing[idx] = 0
        
        for idx in range(len(security)-2, -1, -1):
            if security[idx] <= security[idx+1]:
                decreasing[idx] = decreasing[idx+1] + 1
            else:
                decreasing[idx] = 0
        res = []       
        for idx in range(len(security)):
            if idx - time >= 0 and len(security) -1 - idx >= time and increasing[idx]>= time and decreasing[idx]>= time:
                res.append(idx)
                
        print(increasing)
        print(decreasing)
        return res