Skip to content

Latest commit

 

History

History
68 lines (56 loc) · 2.8 KB

README.md

File metadata and controls

68 lines (56 loc) · 2.8 KB

2100. Find Good Days to Rob the Bank

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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn good_days_to_rob_bank(security: Vec<i32>, time: i32) -> Vec<i32> {
        let mut prefix_l = vec![0; security.len()];
        let mut prefix_r = vec![0; security.len()];

        for i in 1..security.len() {
            if security[i] <= security[i - 1] {
                prefix_l[i] = prefix_l[i - 1] + 1;
            }
            if security[security.len() - i - 1] <= security[security.len() - i] {
                prefix_r[security.len() - i - 1] = prefix_r[security.len() - i] + 1;
            }
        }

        (time..security.len() as i32 - time)
            .filter(|&i| prefix_l[i as usize] >= time && prefix_r[i as usize] >= time)
            .collect()
    }
}