Description
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
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