2976. Minimum Cost to Convert String I

Problem Link

Description

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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

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C++

class Solution {
public:
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
        vector<vector<long long>> dist(26, vector<long long>(26, INT_MAX));
 
        for (int i = 0; i < original.size(); i++) {
            int s = original[i] - 'a', t = changed[i] - 'a';
            dist[s][t] = min(dist[s][t], (long long)cost[i]);
        }
 
        for (int k = 0; k < 26; k++)
            for (int i = 0; i < 26; i++)
                for (int j = 0; j < 26; j++)
                    dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
 
        long long ans = 0;
 
        for (int i = 0; i < source.size(); i++) {
            if (source[i] != target[i]) {
                int s = source[i] - 'a', t = target[i] - 'a';
 
                if (dist[s][t] >= INT_MAX)
                    return -1;
                
                ans += dist[s][t];
            }
        }
 
        return ans;
    }
};

Python3

class Solution:
    def minimumCost(self, source: str, target: str, original: List[str], changed: List[str], cost: List[int]) -> int:
        change = [[inf] * 26 for _ in range(26)]
        
        def f(x):
            return ord(x) - ord('a')
        
        for x, y, cost in zip(original, changed, cost):
            change[f(x)][f(y)] = min(change[f(x)][f(y)], cost)
        
        for k in range(26):
            for i in range(26):
                for j in range(26):
                    change[i][j] = min(change[i][j], change[i][k] + change[k][j])
        
        res = 0
        for x, y in zip(source, target):
            if x == y: continue
            
            if change[f(x)][f(y)] == inf: return -1
            
            res += change[f(x)][f(y)]
        
        return res