# 2976. Minimum Cost to Convert String I

## 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]

## Solutions

Solution 1: Floyd Algorithm

According to the problem description, we can consider each letter as a node, and the conversion cost between each pair of letters as a directed edge. We first initialize a $26 \times 26$ two-dimensional array $g$, where $g[i][j]$ represents the minimum cost of converting letter $i$ to letter $j$. Initially, $g[i][j] = \infty$, and if $i = j$, then $g[i][j] = 0$.

Next, we traverse the arrays $original$, $changed$, and $cost$. For each index $i$, we update the cost $cost[i]$ of converting $original[i]$ to $changed[i]$ to $g[original[i]][changed[i]]$, taking the minimum value.

Then, we use the Floyd algorithm to calculate the minimum cost between any two nodes in $g$. Finally, we traverse the strings $source$ and $target$. If $source[i] \neq target[i]$ and $g[source[i]][target[i]] \geq \infty$, it means that the conversion cannot be completed, so we return $-1$. Otherwise, we add $g[source[i]][target[i]]$ to the answer.

After the traversal ends, we return the answer.

The time complexity is $O(m + n + |\Sigma|^3)$, and the space complexity is $O(|\Sigma|^2)$. Where $m$ and $n$ are the lengths of the arrays $original$ and $source$ respectively; and $|\Sigma|$ is the size of the alphabet, that is, $|\Sigma| = 26$.

• class Solution {
public long minimumCost(
String source, String target, char[] original, char[] changed, int[] cost) {
final int inf = 1 << 29;
int[][] g = new int[26][26];
for (int i = 0; i < 26; ++i) {
Arrays.fill(g[i], inf);
g[i][i] = 0;
}
for (int i = 0; i < original.length; ++i) {
int x = original[i] - 'a';
int y = changed[i] - 'a';
int z = cost[i];
g[x][y] = Math.min(g[x][y], z);
}
for (int k = 0; k < 26; ++k) {
for (int i = 0; i < 26; ++i) {
for (int j = 0; j < 26; ++j) {
g[i][j] = Math.min(g[i][j], g[i][k] + g[k][j]);
}
}
}
long ans = 0;
int n = source.length();
for (int i = 0; i < n; ++i) {
int x = source.charAt(i) - 'a';
int y = target.charAt(i) - 'a';
if (x != y) {
if (g[x][y] >= inf) {
return -1;
}
ans += g[x][y];
}
}
return ans;
}
}

• class Solution {
public:
long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
const int inf = 1 << 29;
int g[26][26];
for (int i = 0; i < 26; ++i) {
fill(begin(g[i]), end(g[i]), inf);
g[i][i] = 0;
}

for (int i = 0; i < original.size(); ++i) {
int x = original[i] - 'a';
int y = changed[i] - 'a';
int z = cost[i];
g[x][y] = min(g[x][y], z);
}

for (int k = 0; k < 26; ++k) {
for (int i = 0; i < 26; ++i) {
for (int j = 0; j < 26; ++j) {
g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
}
}
}

long long ans = 0;
int n = source.length();
for (int i = 0; i < n; ++i) {
int x = source[i] - 'a';
int y = target[i] - 'a';
if (x != y) {
if (g[x][y] >= inf) {
return -1;
}
ans += g[x][y];
}
}
return ans;
}
};

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


• func minimumCost(source string, target string, original []byte, changed []byte, cost []int) (ans int64) {
const inf = 1 << 29
g := make([][]int, 26)
for i := range g {
g[i] = make([]int, 26)
for j := range g[i] {
if i == j {
g[i][j] = 0
} else {
g[i][j] = inf
}
}
}

for i := 0; i < len(original); i++ {
x := int(original[i] - 'a')
y := int(changed[i] - 'a')
z := cost[i]
g[x][y] = min(g[x][y], z)
}

for k := 0; k < 26; k++ {
for i := 0; i < 26; i++ {
for j := 0; j < 26; j++ {
g[i][j] = min(g[i][j], g[i][k]+g[k][j])
}
}
}
n := len(source)
for i := 0; i < n; i++ {
x := int(source[i] - 'a')
y := int(target[i] - 'a')
if x != y {
if g[x][y] >= inf {
return -1
}
ans += int64(g[x][y])
}
}
return
}

• function minimumCost(
source: string,
target: string,
original: string[],
changed: string[],
cost: number[],
): number {
const g: number[][] = Array.from({ length: 26 }, () => Array(26).fill(Infinity));
for (let i = 0; i < 26; ++i) {
g[i][i] = 0;
}
for (let i = 0; i < original.length; ++i) {
let x: number = original[i].charCodeAt(0) - 'a'.charCodeAt(0);
let y: number = changed[i].charCodeAt(0) - 'a'.charCodeAt(0);
let z: number = cost[i];
g[x][y] = Math.min(g[x][y], z);
}

for (let k = 0; k < 26; ++k) {
for (let i = 0; i < 26; ++i) {
for (let j = 0; j < 26; ++j) {
g[i][j] = Math.min(g[i][j], g[i][k] + g[k][j]);
}
}
}

let ans: number = 0;
let n: number = source.length;
for (let i = 0; i < n; ++i) {
let x: number = source.charCodeAt(i) - 'a'.charCodeAt(0);
let y: number = target.charCodeAt(i) - 'a'.charCodeAt(0);
if (x !== y) {
if (g[x][y] >= Infinity) {
return -1;
}
ans += g[x][y];
}
}
return ans;
}