<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of operations required to convert <code>word1</code> to <code>word2</code></em>.</p>

<p>You have the following three operations permitted on a word:</p>

<ul>
	<li>Insert a character</li>
	<li>Delete a character</li>
	<li>Replace a character</li>
</ul>

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

<pre>
<strong>Input:</strong> word1 = &quot;horse&quot;, word2 = &quot;ros&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
horse -&gt; rorse (replace &#39;h&#39; with &#39;r&#39;)
rorse -&gt; rose (remove &#39;r&#39;)
rose -&gt; ros (remove &#39;e&#39;)
</pre>

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

<pre>
<strong>Input:</strong> word1 = &quot;intention&quot;, word2 = &quot;execution&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> 
intention -&gt; inention (remove &#39;t&#39;)
inention -&gt; enention (replace &#39;i&#39; with &#39;e&#39;)
enention -&gt; exention (replace &#39;n&#39; with &#39;x&#39;)
exention -&gt; exection (replace &#39;n&#39; with &#39;c&#39;)
exection -&gt; execution (insert &#39;u&#39;)
</pre>

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

<ul>
	<li><code>0 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code> and <code>word2</code> consist of lowercase English letters.</li>
</ul>

// DFS + DP
class Solution {
    int M;
    int N;

    vector<vector<int>> dp;

    int dfs(int i1, int i2, string& word1, string word2) {
        if (i1 == -1 && i2 == -1) {
            return 0;
        }
        if (i1 == -1) {
            return i2 + 1;
        }
        if (i2 == -1) {
            return i1 + 1;
        }

        if (dp[i1][i2] != -1) {
            return dp[i1][i2];
        }

        if (word1[i1] == word2[i2]) {
            return dp[i1][i2] = dfs(i1 - 1, i2 - 1, word1, word2);
        } else {
            int a = 1 + dfs(i1 - 1, i2, word1, word2);
            int b = 1 + dfs(i1, i2 - 1, word1, word2);
            int c = 1 + dfs(i1 - 1, i2 - 1, word1, word2);
            return dp[i1][i2] = min(a, min(b, c));
        }
    }
public:
    int minDistance(string word1, string word2) {
        M = word1.size();
        N = word2.size();
        dp = vector(M, vector<int>(N, -1));
        return dfs(M - 1, N - 1, word1, word2);
    }
};
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