LeetCode in Kotlin

2976. Minimum Cost to Convert String I

Medium

You are given two 0-indexed strings source and target, both of length n and consisting of lowercase English letters. You are also given two 0-indexed character arrays original and changed, and an integer array cost, where cost[i] represents the cost of changing the character original[i] to the character changed[i].

You start with the string source. In one operation, you can pick a character x from the string and change it to the character y at a cost of z if there exists any index j such that cost[j] == z, original[j] == x, and changed[j] == y.

Return the minimum cost to convert the string source to the string target using any number of operations. If it is impossible to convert source to target, return -1.

Note that there may exist indices i, j such that original[j] == original[i] and changed[j] == changed[i].

Example 1:

Input: source = “abcd”, target = “acbe”, original = [“a”,”b”,”c”,”c”,”e”,”d”], changed = [“b”,”c”,”b”,”e”,”b”,”e”], cost = [2,5,5,1,2,20]

Output: 28

Explanation: To convert the string “abcd” to string “acbe”:

• Change value at index 1 from ‘b’ to ‘c’ at a cost of 5.
• Change value at index 2 from ‘c’ to ‘e’ at a cost of 1.
• Change value at index 2 from ‘e’ to ‘b’ at a cost of 2.
• Change value at index 3 from ‘d’ to ‘e’ at a cost of 20.

The total cost incurred is 5 + 1 + 2 + 20 = 28.

It can be shown that this is the minimum possible cost.

Example 2:

Input: source = “aaaa”, target = “bbbb”, original = [“a”,”c”], changed = [“c”,”b”], cost = [1,2]

Output: 12

Explanation: To change the character ‘a’ to ‘b’ change the character ‘a’ to ‘c’ at a cost of 1, followed by changing the character ‘c’ to ‘b’ at a cost of 2, for a total cost of 1 + 2 = 3. To change all occurrences of ‘a’ to ‘b’, a total cost of 3 * 4 = 12 is incurred.

Example 3:

Input: source = “abcd”, target = “abce”, original = [“a”], changed = [“e”], cost = [10000]

Output: -1

Explanation: It is impossible to convert source to target because the value at index 3 cannot be changed from ‘d’ to ‘e’.

Constraints:

• 1 <= source.length == target.length <= 105
• source, target consist of lowercase English letters.
• 1 <= cost.length == original.length == changed.length <= 2000
• original[i], changed[i] are lowercase English letters.
• 1 <= cost[i] <= 106
• original[i] != changed[i]

Solution

import kotlin.math.min

class Solution {
    fun minimumCost(
        inputText: String,
        desiredText: String,
        fromLetters: CharArray,
        toLetters: CharArray,
        transformationCost: IntArray,
    ): Long {
        val alphabetSize = 26
        val transformationMatrix = Array(alphabetSize) { IntArray(alphabetSize) }
        for (idx in 0 until alphabetSize) {
            transformationMatrix[idx].fill(Int.MAX_VALUE)
            transformationMatrix[idx][idx] = 0
        }
        var i = 0
        while (i < fromLetters.size) {
            val origChar = fromLetters[i].code - 'a'.code
            val newChar = toLetters[i].code - 'a'.code
            val changeCost = transformationCost[i]
            transformationMatrix[origChar][newChar] =
                min(transformationMatrix[origChar][newChar], changeCost)
            i++
        }
        var k = 0
        do {
            for (row in 0 until alphabetSize) {
                for (col in 0 until alphabetSize) {
                    if (transformationMatrix[row][k] != Int.MAX_VALUE &&
                        transformationMatrix[k][col] != Int.MAX_VALUE
                    ) {
                        transformationMatrix[row][col] = min(
                            transformationMatrix[row][col],
                            (
                                transformationMatrix[row][k] +
                                    transformationMatrix[k][col]
                                ),
                        )
                    }
                }
            }
            k++
        } while (k < alphabetSize)
        var totalCost: Long = 0
        for (pos in 0 until inputText.length) {
            val startChar = inputText[pos].code - 'a'.code
            val endChar = desiredText[pos].code - 'a'.code
            if (startChar == endChar) {
                continue
            }
            if (transformationMatrix[startChar][endChar] == Int.MAX_VALUE) {
                return -1
            } else {
                totalCost += transformationMatrix[startChar][endChar].toLong()
            }
        }
        return totalCost
    }
}

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