LeetCode in Kotlin

3952. Maximum Total Value of Covered Indices

Medium

You are given an integer array nums of length n and a binary string s of length n, where s[i] == '1' means index i initially contains a token and s[i] == '0' means it does not.

You may perform the following operation any number of times:

• Choose a token currently located at index i, where i > 0, such that this token has not been moved before.
• Move this token from index i to index i - 1.

An index is considered covered if it contains a token after all moves.

Return an integer denoting the maximum total value of nums at the covered indices after optimally performing the operations.

Example 1:

Input: nums = [9,2,6,1], s = “0101”

Output: 15

Explanation:

• Initially, indices 1 and 3 contain tokens.
• Move the token from index 3 to index 2.
• Move the token from index 1 to index 0.
• The covered indices are [0, 2], so the total value is nums[0] + nums[2] = 9 + 6 = 15.

Example 2:

Input: nums = [5,1,4], s = “001”

Output: 4

Explanation:

• Initially, only index 2 contains a token.
• It is optimal to leave the token at index 2.
• The covered index is [2], so the total value is nums[2] = 4.

Example 3:

Input: nums = [9,3,5], s = “011”

Output: 14

Explanation:

• Initially, indices 1 and 2 contain tokens.
• Move the token from index 1 to index 0.
• The covered indices are [0, 2], so the total value is nums[0] + nums[2] = 9 + 5 = 14.

Constraints:

• 1 <= n == nums.length == s.length <= 105
• 1 <= nums[i] <= 105
• s[i] is either '0' or '1'

Solution

import kotlin.math.min

class Solution {
    fun maxTotal(nums: IntArray, s: String): Long {
        val n = nums.size
        var mn = 100000
        var res: Long = 0
        for (i in n - 1 downTo 0) {
            res += nums[i].toLong()
            mn = min(mn, nums[i])
            if (s[i] == '0') {
                res -= mn.toLong()
                mn = 100000
            }
        }
        return res
    }
}

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