Problem
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]
Solution
/**
* @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) {
var adjacentMap = {};
for (var i = 0; i < cost.length; i++) {
adjacentMap[original[i]] = adjacentMap[original[i]] || {};
adjacentMap[original[i]][changed[i]] = Math.min(
adjacentMap[original[i]][changed[i]] || Number.MAX_SAFE_INTEGER,
cost[i],
);
}
var res = 0;
var minCostMap = {};
for (var j = 0; j < source.length; j++) {
minCostMap[source[j]] = minCostMap[source[j]] || dijkstra(source[j], adjacentMap);
var costMap = minCostMap[source[j]];
if (costMap[target[j]] === undefined) return -1;
res += costMap[target[j]];
}
return res;
};
var dijkstra = function(s, adjacentMap) {
var queue = new MinPriorityQueue();
var minCostMap = {};
queue.enqueue(s, 0);
while (queue.size()) {
var { element, priority } = queue.dequeue();
if (minCostMap[element] <= priority) continue;
minCostMap[element] = priority;
var arr = Object.keys(adjacentMap[element] || {});
for (var i = 0; i < arr.length; i++) {
queue.enqueue(arr[i], priority + adjacentMap[element][arr[i]]);
}
}
return minCostMap;
};
Explain:
nope.
Complexity:
- Time complexity : O(m * n).
- Space complexity : O(m).
Solution 2
/**
* @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) {
var minCostMap = Array(26).fill(0).map(() => Array(26).fill(Number.MAX_SAFE_INTEGER));
var a = 'a'.charCodeAt(0);
for (var i = 0; i < cost.length; i++) {
var o = original[i].charCodeAt(0) - a;
var c = changed[i].charCodeAt(0) - a;
minCostMap[o][c] = Math.min(
minCostMap[o][c],
cost[i],
);
}
for (var k = 0; k < 26; k++) {
for (var i = 0; i < 26; i++) {
for (var j = 0; j < 26; j++) {
minCostMap[i][j] = Math.min(
minCostMap[i][j],
minCostMap[i][k] + minCostMap[k][j],
);
}
}
}
var res = 0;
for (var j = 0; j < source.length; j++) {
var s = source[j].charCodeAt(0) - a;
var t = target[j].charCodeAt(0) - a;
if (s === t) continue;
if (minCostMap[s][t] === Number.MAX_SAFE_INTEGER) return -1;
res += minCostMap[s][t];
}
return res;
};
Explain:
nope.
Complexity:
- Time complexity : O(m * n).
- Space complexity : O(1).