Construct String with Minimum Cost (Easy)
<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>
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
}
