LeetCode-in-Java

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 java.util.Arrays;

public class Solution {
    public long minimumCost(
            String inputText,
            String desiredText,
            char[] fromLetters,
            char[] toLetters,
            int[] transformationCost) {
        int alphabetSize = 26;
        int[][] transformationMatrix = new int[alphabetSize][alphabetSize];
        for (int idx = 0; idx < alphabetSize; idx++) {
            Arrays.fill(transformationMatrix[idx], Integer.MAX_VALUE);
            transformationMatrix[idx][idx] = 0;
        }
        int i = 0;
        while (i < fromLetters.length) {
            int origChar = fromLetters[i] - 'a';
            int newChar = toLetters[i] - 'a';
            int changeCost = transformationCost[i];
            transformationMatrix[origChar][newChar] =
                    Math.min(transformationMatrix[origChar][newChar], changeCost);
            i++;
        }
        int k = 0;
        do {
            for (int row = 0; row < alphabetSize; row++) {
                for (int col = 0; col < alphabetSize; col++) {
                    if (transformationMatrix[row][k] != Integer.MAX_VALUE
                            && transformationMatrix[k][col] != Integer.MAX_VALUE) {
                        transformationMatrix[row][col] =
                                Math.min(
                                        transformationMatrix[row][col],
                                        transformationMatrix[row][k]
                                                + transformationMatrix[k][col]);
                    }
                }
            }
            k++;
        } while (k < alphabetSize);
        long totalCost = 0;
        for (int pos = 0; pos < inputText.length(); pos++) {
            int startChar = inputText.charAt(pos) - 'a';
            int endChar = desiredText.charAt(pos) - 'a';
            if (startChar == endChar) {
                continue;
            }
            if (transformationMatrix[startChar][endChar] == Integer.MAX_VALUE) {
                return -1;
            } else {
                totalCost += transformationMatrix[startChar][endChar];
            }
        }
        return totalCost;
    }
}

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