feat: add solutions to lc problems: No.2976,2977 (#2161)

Author: yanglbme

solution/2900-2999/2976.Minimum Cost to Convert String I/README.md

@@ -69,6 +69,18 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：Floyd 算法**
+
+根据题目描述，我们可以将每个字母看作一个节点，每对字母的转换成本看作一条有向边。那么我们先初始化一个 $26 \times 26$ 的二维数组 $g$，其中 $g[i][j]$ 表示字母 $i$ 转换成字母 $j$ 的最小成本。初始时 $g[i][j] = \infty$，如果 $i = j$，那么 $g[i][j] = 0$。
+
+然后我们遍历数组 $original$、$changed$ 和 $cost$，对于每个下标 $i$，我们将 $original[i]$ 转换成 $changed[i]$ 的成本 $cost[i]$ 更新到 $g[original[i]][changed[i]]$ 中，取最小值。
+
+接下来，我们使用 Floyd 算法计算出 $g$ 中任意两个节点之间的最小成本。最后，我们遍历字符串 $source$ 和 $target$，如果 $source[i] \neq target[i]$，并且 $g[source[i]][target[i]] \geq \infty$，那么说明无法完成转换，返回 $-1$。否则，我们将 $g[source[i]][target[i]]$ 累加到答案中。
+
+遍历结束后，返回答案即可。
+
+时间复杂度 $O(m + n + |\Sigma|^3)$，空间复杂度 $O(|\Sigma|^2)$。其中 $m$ 和 $n$ 分别是数组 $original$ 和 $source$ 的长度；而 $|\Sigma|$ 是字母表的大小，即 $|\Sigma| = 26$。
+
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solution/2900-2999/2976.Minimum Cost to Convert String I/README_EN.md

@@ -57,6 +57,18 @@ It can be shown that this is the minimum possible cost.
 
 ## 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, $|\Sigma| = 26$.
+
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 ### **Python3**