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 <= 105source,targetconsist of lowercase English letters.1 <= cost.length == original.length == changed.length <= 2000original[i],changed[i]are lowercase English letters.1 <= cost[i] <= 106original[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, $|\Sigma| = 26$.
Python3
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
Java
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;
}
}
C++
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;
}
};
Go
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
}
TypeScript
function minimumCost(
source: string,
target: string,
original: string[],
changed: string[],
cost: number[],
): number {
const [n, m, MAX] = [source.length, original.length, Number.POSITIVE_INFINITY];
const g: number[][] = Array.from({ length: 26 }, () => Array(26).fill(MAX));
const getIndex = (ch: string) => ch.charCodeAt(0) - 'a'.charCodeAt(0);
for (let i = 0; i < 26; ++i) g[i][i] = 0;
for (let i = 0; i < m; ++i) {
const x = getIndex(original[i]);
const y = getIndex(changed[i]);
const z = 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; g[i][k] < MAX && j < 26; j++) {
if (g[k][j] < MAX) {
g[i][j] = Math.min(g[i][j], g[i][k] + g[k][j]);
}
}
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
const x = getIndex(source[i]);
const y = getIndex(target[i]);
if (x === y) continue;
if (g[x][y] === MAX) return -1;
ans += g[x][y];
}
return ans;
}
JavaScript
/**
* @param {string} source
* @param {string} target
* @param {character[]} original
* @param {character[]} changed
* @param {number[]} cost
* @return {number}
*/
var minimumCost = function (source, target, original, changed, cost) {
const [n, m, MAX] = [source.length, original.length, Number.POSITIVE_INFINITY];
const g = Array.from({ length: 26 }, () => Array(26).fill(MAX));
const getIndex = ch => ch.charCodeAt(0) - 'a'.charCodeAt(0);
for (let i = 0; i < 26; ++i) g[i][i] = 0;
for (let i = 0; i < m; ++i) {
const x = getIndex(original[i]);
const y = getIndex(changed[i]);
const z = 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; g[i][k] < MAX && j < 26; j++) {
if (g[k][j] < MAX) {
g[i][j] = Math.min(g[i][j], g[i][k] + g[k][j]);
}
}
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
const x = getIndex(source[i]);
const y = getIndex(target[i]);
if (x === y) continue;
if (g[x][y] === MAX) return -1;
ans += g[x][y];
}
return ans;
};