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2100. Find Good Days to Rob the Bank

Difficulty Topics

Description

You and a gang of thieves are planning on robbing a bank. You are given a 0-indexed integer array security, where security[i] is the number of guards on duty on the ith day. The days are numbered starting from 0. You are also given an integer time.

The ith day is a good day to rob the bank if:

  • There are at least time days before and after the ith day,
  • The number of guards at the bank for the time days before i are non-increasing, and
  • The number of guards at the bank for the time days after i are non-decreasing.

More formally, this means day i is a good day to rob the bank if and only if security[i - time] >= security[i - time + 1] >= ... >= security[i] <= ... <= security[i + time - 1] <= security[i + time].

Return a list of all days (0-indexed) that are good days to rob the bank. The order that the days are returned in does not matter.

 

Example 1:

Input: security = [5,3,3,3,5,6,2], time = 2
Output: [2,3]
Explanation:
On day 2, we have security[0] >= security[1] >= security[2] <= security[3] <= security[4].
On day 3, we have security[1] >= security[2] >= security[3] <= security[4] <= security[5].
No other days satisfy this condition, so days 2 and 3 are the only good days to rob the bank.

Example 2:

Input: security = [1,1,1,1,1], time = 0
Output: [0,1,2,3,4]
Explanation:
Since time equals 0, every day is a good day to rob the bank, so return every day.

Example 3:

Input: security = [1,2,3,4,5,6], time = 2
Output: []
Explanation:
No day has 2 days before it that have a non-increasing number of guards.
Thus, no day is a good day to rob the bank, so return an empty list.

 

Constraints:

  • 1 <= security.length <= 105
  • 0 <= security[i], time <= 105

Solution

find-good-days-to-rob-the-bank.py
class Solution:
    def goodDaysToRobBank(self, security: List[int], time: int) -> List[int]:
        n = len(security)
        if time == 0: return list(range(n))
        if n <= time * 2: return []

        res = []
        inc = [1] * n
        dec = [1] * n

        for i, x in enumerate(security):
            if i == 0: continue

            if x <= security[i - 1]:
                inc[i] += inc[i - 1]

        for i in range(n - 2, -1, -1):
            if security[i] <= security[i + 1]:
                dec[i] += dec[i + 1]

        for i, x in enumerate(security[time: n - time], time):
            if x <= security[i - 1] and inc[i - 1] >= time and x <= security[i + 1] and dec[i + 1] >= time:
                res.append(i)



        return res