2100. Find Good Days to Rob the Bank

Problem

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 <= 10^5

• 0 <= security[i], time <= 10^5

Solution (Java)

class Solution {
    public List<Integer> goodDaysToRobBank(int[] security, int time) {
        int n = security.length;
        // dec: # of non-increasing elements before [i]
        // inc: # of non-decreasing elements after [i]
        int[] dec = new int[n];
        int[] inc = new int[n];
        for (int i = 1; i < n; i++) {
            if (security[i] <= security[i - 1]) {
                dec[i] = dec[i - 1] + 1;
            }
            // no need for else, because array elements are inited as 0
        }
        for (int i = n - 2; i >= 0; i--) {
            if (security[i] <= security[i + 1]) {
                inc[i] = inc[i + 1] + 1;
            }
            // no need for else, because array elements are inited as 0
        }
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (dec[i] >= time && inc[i] >= time) {
                res.add(i);
            }
        }
        return res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).