feat: add solutions to lc problem: No.3253 (#3415)

No.3253.Construct String with Minimum Cost (Easy)

Author: yanglbme

solution/3200-3299/3253.Construct String with Minimum Cost (Easy)/README.md

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+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3253.Construct%20String%20with%20Minimum%20Cost%20%28Easy%29/README.md
+---
+
+<!-- problem:start -->
+
+# [3253. Construct String with Minimum Cost (Easy) 🔒](https://leetcode.cn/problems/construct-string-with-minimum-cost-easy)
+
+[English Version](/solution/3200-3299/3253.Construct%20String%20with%20Minimum%20Cost%20%28Easy%29/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>You are given a string <code>target</code>, an array of strings <code>words</code>, and an integer array <code>costs</code>, both arrays of the same length.</p>
+
+<p>Imagine an empty string <code>s</code>.</p>
+
+<p>You can perform the following operation any number of times (including <strong>zero</strong>):</p>
+
+<ul>
+	<li>Choose an index <code>i</code> in the range <code>[0, words.length - 1]</code>.</li>
+	<li>Append <code>words[i]</code> to <code>s</code>.</li>
+	<li>The cost of operation is <code>costs[i]</code>.</li>
+</ul>
+
+<p>Return the <strong>minimum</strong> cost to make <code>s</code> equal to <code>target</code>. If it&#39;s not possible, return -1.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">target = &quot;abcdef&quot;, words = [&quot;abdef&quot;,&quot;abc&quot;,&quot;d&quot;,&quot;def&quot;,&quot;ef&quot;], costs = [100,1,1,10,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">7</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The minimum cost can be achieved by performing the following operations:</p>
+
+<ul>
+	<li>Select index 1 and append <code>&quot;abc&quot;</code> to <code>s</code> at a cost of 1, resulting in <code>s = &quot;abc&quot;</code>.</li>
+	<li>Select index 2 and append <code>&quot;d&quot;</code> to <code>s</code> at a cost of 1, resulting in <code>s = &quot;abcd&quot;</code>.</li>
+	<li>Select index 4 and append <code>&quot;ef&quot;</code> to <code>s</code> at a cost of 5, resulting in <code>s = &quot;abcdef&quot;</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">target = &quot;aaaa&quot;, words = [&quot;z&quot;,&quot;zz&quot;,&quot;zzz&quot;], costs = [1,10,100]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">-1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>It is impossible to make <code>s</code> equal to <code>target</code>, so we return -1.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= target.length &lt;= 2000</code></li>
+	<li><code>1 &lt;= words.length == costs.length &lt;= 50</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= target.length</code></li>
+	<li><code>target</code> and <code>words[i]</code> consist only of lowercase English letters.</li>
+	<li><code>1 &lt;= costs[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：字典树 + 记忆化搜索
+
+我们首先创建一个字典树 $\textit{trie}$，字典树的每个节点包含一个长度为 $26$ 的数组 $\textit{children}$，数组中的每个元素都是一个指向下一个节点的指针。字典树的每个节点还包含一个 $\textit{cost}$ 变量，表示从根节点到当前节点的最小花费。
+
+我们遍历 $\textit{words}$ 数组，将每个单词插入到字典树中，同时更新每个节点的 $\textit{cost}$ 变量。
+
+接下来，我们定义一个记忆化搜索函数 $\textit{dfs}(i)$，表示从 $\textit{target}[i]$ 开始构造字符串的最小花费。那么答案就是 $\textit{dfs}(0)$。
+
+函数 $\textit{dfs}(i)$ 的计算过程如下：
+
+-   如果 $i \geq \textit{len}(\textit{target})$，表示已经构造完整个字符串，返回 $0$。
+-   否则，我们从 $\textit{trie}$ 的根节点开始，遍历 $\textit{target}[i]$ 开始的所有后缀，找到最小花费，即 $\textit{trie}$ 中的 $\textit{cost}$ 变量，加上 $\textit{dfs}(j+1)$ 的结果，其中 $j$ 是 $\textit{target}[i]$ 开始的后缀的结束位置。
+
+最后，如果 $\textit{dfs}(0) < \textit{inf}$，返回 $\textit{dfs}(0)$，否则返回 $-1$。
+
+时间复杂度 $O(n^2 + L)$，空间复杂度 $O(n + L)$。其中 $n$ 是 $\textit{target}$ 的长度，而 $L$ 是 $\textit{words}$ 数组中所有单词的长度之和。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Trie:
+    def __init__(self):
+        self.children: List[Optional[Trie]] = [None] * 26
+        self.cost = inf
+
+    def insert(self, word: str, cost: int):
+        node = self
+        for c in word:
+            idx = ord(c) - ord("a")
+            if node.children[idx] is None:
+                node.children[idx] = Trie()
+            node = node.children[idx]
+        node.cost = min(node.cost, cost)
+
+
+class Solution:
+    def minimumCost(self, target: str, words: List[str], costs: List[int]) -> int:
+        @cache
+        def dfs(i: int) -> int:
+            if i >= len(target):
+                return 0
+            ans = inf
+            node = trie
+            for j in range(i, len(target)):
+                idx = ord(target[j]) - ord("a")
+                if node.children[idx] is None:
+                    return ans
+                node = node.children[idx]
+                ans = min(ans, node.cost + dfs(j + 1))
+            return ans
+
+        trie = Trie()
+        for word, cost in zip(words, costs):
+            trie.insert(word, cost)
+        ans = dfs(0)
+        return ans if ans < inf else -1
+```
+
+#### Java
+
+```java
+class Trie {
+    public final int inf = 1 << 29;
+    public Trie[] children = new Trie[26];
+    public int cost = inf;
+
+    public void insert(String word, int cost) {
+        Trie node = this;
+        for (char c : word.toCharArray()) {
+            int idx = c - 'a';
+            if (node.children[idx] == null) {
+                node.children[idx] = new Trie();
+            }
+            node = node.children[idx];
+        }
+        node.cost = Math.min(node.cost, cost);
+    }
+}
+
+class Solution {
+    private Trie trie = new Trie();
+    private char[] target;
+    private Integer[] f;
+
+    public int minimumCost(String target, String[] words, int[] costs) {
+        for (int i = 0; i < words.length; ++i) {
+            trie.insert(words[i], costs[i]);
+        }
+        this.target = target.toCharArray();
+        f = new Integer[target.length()];
+        int ans = dfs(0);
+        return ans < trie.inf ? ans : -1;
+    }
+
+    private int dfs(int i) {
+        if (i >= target.length) {
+            return 0;
+        }
+        if (f[i] != null) {
+            return f[i];
+        }
+        f[i] = trie.inf;
+        Trie node = trie;
+        for (int j = i; j < target.length; ++j) {
+            int idx = target[j] - 'a';
+            if (node.children[idx] == null) {
+                return f[i];
+            }
+            node = node.children[idx];
+            f[i] = Math.min(f[i], node.cost + dfs(j + 1));
+        }
+        return f[i];
+    }
+}
+```
+
+#### C++
+
+```cpp
+const int inf = 1 << 29;
+
+class Trie {
+public:
+    Trie* children[26]{};
+    int cost = inf;
+
+    void insert(string& word, int cost) {
+        Trie* node = this;
+        for (char c : word) {
+            int idx = c - 'a';
+            if (!node->children[idx]) {
+                node->children[idx] = new Trie();
+            }
+            node = node->children[idx];
+        }
+        node->cost = min(node->cost, cost);
+    }
+};
+
+class Solution {
+public:
+    int minimumCost(string target, vector<string>& words, vector<int>& costs) {
+        Trie* trie = new Trie();
+        for (int i = 0; i < words.size(); ++i) {
+            trie->insert(words[i], costs[i]);
+        }
+        int n = target.length();
+        int f[n];
+        memset(f, 0, sizeof(f));
+        auto dfs = [&](auto&& dfs, int i) -> int {
+            if (i >= n) {
+                return 0;
+            }
+            if (f[i]) {
+                return f[i];
+            }
+            f[i] = inf;
+            Trie* node = trie;
+            for (int j = i; j < n; ++j) {
+                int idx = target[j] - 'a';
+                if (!node->children[idx]) {
+                    return f[i];
+                }
+                node = node->children[idx];
+                f[i] = min(f[i], node->cost + dfs(dfs, j + 1));
+            }
+            return f[i];
+        };
+        int ans = dfs(dfs, 0);
+        return ans < inf ? ans : -1;
+    }
+};
+```
+
+#### Go
+
+```go
+const inf = 1 << 29
+
+type Trie struct {
+	children [26]*Trie
+	cost     int
+}
+
+func NewTrie() *Trie {
+	return &Trie{cost: inf}
+}
+
+func (t *Trie) insert(word string, cost int) {
+	node := t
+	for _, c := range word {
+		idx := c - 'a'
+		if node.children[idx] == nil {
+			node.children[idx] = NewTrie()
+		}
+		node = node.children[idx]
+	}
+	node.cost = min(node.cost, cost)
+}
+
+func minimumCost(target string, words []string, costs []int) int {
+	trie := NewTrie()
+	for i, word := range words {
+		trie.insert(word, costs[i])
+	}
+
+	n := len(target)
+	f := make([]int, n)
+	var dfs func(int) int
+	dfs = func(i int) int {
+		if i >= n {
+			return 0
+		}
+		if f[i] != 0 {
+			return f[i]
+		}
+		f[i] = inf
+		node := trie
+		for j := i; j < n; j++ {
+			idx := target[j] - 'a'
+			if node.children[idx] == nil {
+				return f[i]
+			}
+			node = node.children[idx]
+			f[i] = min(f[i], node.cost+dfs(j+1))
+		}
+		return f[i]
+	}
+	if ans := dfs(0); ans < inf {
+		return ans
+	}
+	return -1
+}
+```
+
+#### TypeScript
+
+```ts
+const inf = 1 << 29;
+
+class Trie {
+    children: (Trie | null)[];
+    cost: number;
+
+    constructor() {
+        this.children = Array(26).fill(null);
+        this.cost = inf;
+    }
+
+    insert(word: string, cost: number): void {
+        let node: Trie = this;
+        for (const c of word) {
+            const idx = c.charCodeAt(0) - 97;
+            if (!node.children[idx]) {
+                node.children[idx] = new Trie();
+            }
+            node = node.children[idx]!;
+        }
+        node.cost = Math.min(node.cost, cost);
+    }
+}
+
+function minimumCost(target: string, words: string[], costs: number[]): number {
+    const trie = new Trie();
+    for (let i = 0; i < words.length; ++i) {
+        trie.insert(words[i], costs[i]);
+    }
+
+    const n = target.length;
+    const f: number[] = Array(n).fill(0);
+    const dfs = (i: number): number => {
+        if (i >= n) {
+            return 0;
+        }
+        if (f[i]) {
+            return f[i];
+        }
+        f[i] = inf;
+        let node: Trie | null = trie;
+        for (let j = i; j < n; ++j) {
+            const idx = target.charCodeAt(j) - 97;
+            if (!node?.children[idx]) {
+                return f[i];
+            }
+            node = node.children[idx];
+            f[i] = Math.min(f[i], node!.cost + dfs(j + 1));
+        }
+        return f[i];
+    };
+
+    const ans = dfs(0);
+    return ans < inf ? ans : -1;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->