<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 a, string b, int i, int j, vector<vector<int>>& dp) {
        if (i == a.size() || j == b.size()) return 0;
        if (dp[i][j] != -1) return dp[i][j];
        int ans = 0;
        if (a[i] == b[j]) ans = 1 + solve(a, b, i + 1, j + 1, dp);
        else ans = max(solve(a, b, i, j + 1, dp), solve(a, b, i + 1, j, dp));
        return dp[i][j] = ans;
    }

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

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