Delete Operation for Two Strings
<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of <strong>steps</strong> required to make</em> <code>word1</code> <em>and</em> <code>word2</code> <em>the same</em>.</p>

<p>In one <strong>step</strong>, you can delete exactly one character in either string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;sea&quot;, word2 = &quot;eat&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You need one step to make &quot;sea&quot; to &quot;ea&quot; and another step to make &quot;eat&quot; to &quot;ea&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;leetcode&quot;, word2 = &quot;etco&quot;
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code> and <code>word2</code> consist of only lowercase English letters.</li>
</ul>
String; Dynamic Programming
function minDistance(word1: string, word2: string): number {
    const m = word1.length;
    const n = word2.length;
    const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    for (let i = 1; i <= m; ++i) {
        f[i][0] = i;
    }
    for (let j = 1; j <= n; ++j) {
        f[0][j] = j;
    }
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            if (word1[i - 1] === word2[j - 1]) {
                f[i][j] = f[i - 1][j - 1];
            } else {
                f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + 1;
            }
        }
    }
    return f[m][n];
}
