<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 solve(string& word1, string& word2, int index1, int index2, int del,
              vector<vector<vector<int>>>& dp) {
        if (index1 == word1.length() && index2 == word2.length()) {
            return 0;
        }
        if (index1 == word1.length()) {
            return word2.length() - index2;
        }
        if (index2 == word2.length()) {
            return word1.length() - index1;
        }

        if (dp[index1][index2][del] != -1) {
            return dp[index1][index2][del];
        }
        int candelete = INT_MAX;
        int insert = INT_MAX;
        int include = INT_MAX;
        if (word1[index1] == word2[index2]) {
            include = solve(word1, word2, index1 + 1, index2 + 1, 0, dp);
        } else {
            if (del != 1) {
                candelete =
                    1 + min(solve(word1, word2, index1 + 1, index2, 1, dp),
                            solve(word1, word2, index1, index2 + 1, 1, dp));
            } else {
                insert =
                    1 + min(solve(word1, word2, index1 + 1, index2, 0, dp),
                            solve(word1, word2, index1, index2 + 1, 0, dp));
            }
        }

        return dp[index1][index2][del] = min(min(candelete, insert), include);
    }

    int minDistance(string word1, string word2) {
        int n = word1.length();
        int m = word2.length();
        vector<vector<vector<int>>> dp(
            n, vector<vector<int>>(m, vector<int>(2, -1)));
        return solve(word1, word2, 0, 0, 0, dp);
    }
};
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