feat: solve No.2976

Author: BaffinLee

2901-3000/2976. Minimum Cost to Convert String I.md

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+# 2976. Minimum Cost to Convert String I
+
+- Difficulty: Medium.
+- Related Topics: Array, String, Graph, Shortest Path.
+- Similar Questions: Can Convert String in K Moves, Minimum Moves to Convert String.
+
+## 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 <= 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
+
+```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) {
+    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
+
+```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) {
+    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).
+