LeetCode in Kotlin

2100. Find Good Days to Rob the Bank

Medium

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

class Solution {
    fun goodDaysToRobBank(security: IntArray, time: Int): List<Int> {
        val n = security.size
        // dec: # of non-increasing elements before [i]
        // inc: # of non-decreasing elements after [i]
        val dec = IntArray(n)
        val inc = IntArray(n)
        for (i in 1 until n) {
            if (security[i] <= security[i - 1]) {
                dec[i] = dec[i - 1] + 1
            }
            // no need for else, because array elements are inited as 0
        }
        for (i in n - 2 downTo 0) {
            if (security[i] <= security[i + 1]) {
                inc[i] = inc[i + 1] + 1
            }
            // no need for else, because array elements are inited as 0
        }
        val res: MutableList<Int> = ArrayList()
        for (i in 0 until n) {
            if (dec[i] >= time && inc[i] >= time) {
                res.add(i)
            }
        }
        return res
    }
}

This site is open source. Improve this page.