Today I have done leetcode's July LeetCoding Challenge with cpp.

July LeetCoding Challenge 27

Description

Minimum Cost to Convert String I

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

template<typename T>
std::ostream& operator<<(std::ostream &out, const std::vector<T> &v) {
  if(v.size() == 0) {
    out << "[]" << std::endl;
    return out;
  }
  out << '[' << v[0];
  for(int i = 1; i < v.size(); ++i) {
    out << ", " << v[i];
  }
  out << ']';
  return out;
}
class Solution {
  using ll = long long;
public:
  long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
    vector<vector<ll>> graph(26, vector<ll>(26, INT_MAX));
    for(int i = 0; i < 26; ++i) {
      graph[i][i] = 0;
    }
    for(int i = 0; i < changed.size(); ++i) {
      graph[original[i] - 'a'][changed[i] - 'a'] = min<ll>(graph[original[i] - 'a'][changed[i] - 'a'], cost[i]);
    }

    for(int k = 0; k < 26; ++k) {
      for(int i = 0; i < 26; ++i) {
        for(int j = 0; j < 26; ++j) {
          graph[i][j] = min(graph[i][j], graph[i][k] + graph[k][j]);
        }
      }
    }
    int answer = 0;
    for(int i = 0; i < source.length(); ++i) {
      if(graph[source[i] - 'a'][target[i] - 'a'] >= INT_MAX) return -1;
      answer += graph[source[i] - 'a'][target[i] - 'a'];
    }
    return answer;
  }
};