LeetCode in Kotlin

3784. Minimum Deletion Cost to Make All Characters Equal

Medium

You are given a string s of length n and an integer array cost of the same length, where cost[i] is the cost to delete the ith character of s.

You may delete any number of characters from s (possibly none), such that the resulting string is non-empty and consists of equal characters.

Return an integer denoting the minimum total deletion cost required.

Example 1:

Input: s = “aabaac”, cost = [1,2,3,4,1,10]

Output: 11

Explanation:

Deleting the characters at indices 0, 1, 2, 3, 4 results in the string "c", which consists of equal characters, and the total cost is cost[0] + cost[1] + cost[2] + cost[3] + cost[4] = 1 + 2 + 3 + 4 + 1 = 11.

Example 2:

Input: s = “abc”, cost = [10,5,8]

Output: 13

Explanation:

Deleting the characters at indices 1 and 2 results in the string "a", which consists of equal characters, and the total cost is cost[1] + cost[2] = 5 + 8 = 13.

Example 3:

Input: s = “zzzzz”, cost = [67,67,67,67,67]

Output: 0

Explanation:

All characters in s are equal, so the deletion cost is 0.

Constraints:

• n == s.length == cost.length
• 1 <= n <= 105
• 1 <= cost[i] <= 109
• s consists of lowercase English letters.

Solution

class Solution {
    fun minCost(s: String, cost: IntArray): Long {
        val arr = LongArray(26)
        var m = Int.MIN_VALUE.toLong()
        var sum: Long = 0
        for (i in 0..<s.length) {
            arr[s[i].code - 'a'.code] += cost[i].toLong()
        }
        for (i in arr) {
            if (i > m) {
                m = i
            }
            sum += i
        }
        return sum - m
    }
}

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