<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>

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size();
        int n = word2.size();

        // vector<vector<int>> dp(m + 1, vector<int>(n + 1, -1));

        // for (int i = 0; i < m; i++) {
        //     for (int j = 0; j < n; j++) {
        //         if (i == 0 && j == 0) {
        //             dp[i][j] = 0
        //         } else if (i == 0) {
        //             dp[i][j] = j;
        //         } else if (j == 0) {
        //             dp[i][j] = i;
        //         } else {
        //             dp[i][j] = min(
        //                 dp[i-1][j],
        //                 dp[i][j-1]
        //             )
        //         }
        //     }
        // }
        return dfs(word1, word2, 0, 0);
    }
private:
    unordered_map<int, unordered_map<int, int>> memo;
    
    int dfs(const string& s1, const string& s2, int i, int j) {
        // Base Cases: we have an empty string
        if (i == s1.size() && j == s2.size()) {
            return 0;
        }
        if (i == s1.size()) {
            return s2.size() - j;
        }
        if (j == s2.size()) {
            return s1.size() - i;
        }

        // Recursive Case:
        if (memo[i].count(j)) {
            return memo[i][j];
        }

        if (s1[i] == s2[j]) {
            return memo[i][j] = dfs(s1, s2, i + 1, j + 1);
        }

        int insertOp = dfs(s1, s2, i, j + 1);
        int deleteOp = dfs(s1, s2, i + 1, j);
        int replaceOp = dfs(s1, s2, i + 1, j + 1);

        return memo[i][j] = min({insertOp, deleteOp, replaceOp}) + 1;
    }
};

/*
insert
delete
replace

R(s1, s2) =
    R(s1[1:], s2[1:]) if s1[0] == s2[0]
    min (
        R(s2[0] + s1, s2),      // insert
        R(s1[1:], s2),          // delete
        R(s1', s2),             // replace
    )
*/
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