LeetCode in Kotlin

2483. Minimum Penalty for a Shop

Medium

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y':

• if the ith character is 'Y', it means that customers come at the ith hour
• whereas 'N' indicates that no customers come at the ith hour.

If the shop closes at the jth hour (0 <= j <= n), the penalty is calculated as follows:

• For every hour when the shop is open and no customers come, the penalty increases by 1.
• For every hour when the shop is closed and customers come, the penalty increases by 1.

Return the earliest hour at which the shop must be closed to incur a minimum penalty.

Note that if a shop closes at the jth hour, it means the shop is closed at the hour j.

Example 1:

Input: customers = “YYNY”

Output: 2

Explanation:

• Closing the shop at the 0^th hour incurs in 1+1+0+1 = 3 penalty.
• Closing the shop at the 1^st hour incurs in 0+1+0+1 = 2 penalty.
• Closing the shop at the 2^nd hour incurs in 0+0+0+1 = 1 penalty.
• Closing the shop at the 3^rd hour incurs in 0+0+1+1 = 2 penalty.
• Closing the shop at the 4^th hour incurs in 0+0+1+0 = 1 penalty.

Closing the shop at 2^nd or 4^th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.

Example 2:

Input: customers = “NNNNN”

Output: 0

Explanation: It is best to close the shop at the 0^th hour as no customers arrive.

Example 3:

Input: customers = “YYYY”

Output: 4

Explanation: It is best to close the shop at the 4^th hour as customers arrive at each hour.

Constraints:

• 1 <= customers.length <= 105
• customers consists only of characters 'Y' and 'N'.

Solution

class Solution {
    fun bestClosingTime(customers: String): Int {
        val yes = IntArray(customers.length + 1)
        val no = IntArray(customers.length + 1)
        var count = 0
        for (i in customers.length - 1 downTo 0) {
            if (customers[i] == 'Y') {
                count++
            }
            yes[i] = count
        }
        count = 0
        for (i in 0 until customers.length) {
            if (customers[i] == 'N') {
                count++
            }
            no[i + 1] = count
        }
        var min = Int.MAX_VALUE
        var res = 0
        for (i in yes.indices) {
            val sum = yes[i] + no[i]
            if (min > sum) {
                min = sum
                res = i
            }
        }
        return res
    }
}

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