<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>

class Solution {
public:
    int minDistance(string word1, string word2) {
        int width = word1.size();
        int height = word2.size();
        short arr[(width + 1) * 2];
        iota(&arr[0], &arr[1] + width, 0);
        
        auto prev_line = &arr[0];
        auto cur_line = prev_line + width + 1;
        
        for (int i1 = 0, i2 = 1; i1 < height; ++i1, ++i2) {
            cur_line[0] = i2;
            for (int j1 = 0, j2 = 1; j1 < width; ++j1, ++j2) {
                cur_line[j2] = word2[i1] == word1[j1] ?
                    prev_line[j1] : min(cur_line[j1], prev_line[j2]) + 1;
            }
            cur_line = exchange(prev_line, cur_line);
        }
        return prev_line[width];
    }
};
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