<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 find(int i, int j, int m, int n, string word1, string word2,
             vector<vector<int>>& dp) {
        if ((i >= m && j >= n))
            return 0;

        if (i >= m)
            return n - j;
        if (j >= n)
            return m - i;

        if (dp[i][j] != -1)
            return dp[i][j];

        if (word1[i] == word2[j])
            return dp[i][j] = find(i + 1, j + 1, m, n, word1, word2, dp);
        else
            return dp[i][j] = 1 + min(find(i, j + 1, m, n, word1, word2, dp),
                                      find(i + 1, j, m, n, word1, word2, dp));
    }
    int minDistance(string word1, string word2) {
        int m = word1.length(), n = word2.length();

        vector<vector<int>> dp(m, vector<int>(n, -1));
        return find(0, 0, m, n, word1, word2, dp);
    }
};
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