Welcome to Subscribe On Youtube
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, $|\Sigma| = 26$.
-
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; }