Construct String with Minimum Cost
<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 <code>-1</code>.</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;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words.length == costs.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= target.length</code></li>
	<li>The total sum of <code>words[i].length</code> is less than or equal to <code>5 * 10<sup>4</sup></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>4</sup></code></li>
</ul>
Array; String; Dynamic Programming; Suffix Array
class Hashing {
private:
    vector<long> p, h;
    long mod;

public:
    Hashing(const string& word, long base, long mod)
        : p(word.size() + 1, 1)
        , h(word.size() + 1, 0)
        , mod(mod) {
        for (int i = 1; i <= word.size(); ++i) {
            p[i] = p[i - 1] * base % mod;
            h[i] = (h[i - 1] * base + word[i - 1]) % mod;
        }
    }

    long query(int l, int r) {
        return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
    }
};

class Solution {
public:
    int minimumCost(string target, vector<string>& words, vector<int>& costs) {
        const int base = 13331;
        const int mod = 998244353;
        const int inf = INT_MAX / 2;

        int n = target.size();
        Hashing hashing(target, base, mod);

        vector<int> f(n + 1, inf);
        f[0] = 0;

        set<int> ss;
        for (const string& w : words) {
            ss.insert(w.size());
        }

        unordered_map<long, int> d;
        for (int i = 0; i < words.size(); ++i) {
            long x = 0;
            for (char c : words[i]) {
                x = (x * base + c) % mod;
            }
            d[x] = d.find(x) == d.end() ? costs[i] : min(d[x], costs[i]);
        }

        for (int i = 1; i <= n; ++i) {
            for (int j : ss) {
                if (j > i) {
                    break;
                }
                long x = hashing.query(i - j + 1, i);
                if (d.contains(x)) {
                    f[i] = min(f[i], f[i - j] + d[x]);
                }
            }
        }

        return f[n] >= inf ? -1 : f[n];
    }
};