<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 n, int m, vector <vector <int>> &dp){
        if(n == 0 || m == 0)
            return 0;

        if(dp[n-1][m-1] != -1)
            return dp[n-1][m-1];
        
        if(a[n-1] == b[m-1])
            return dp[n-1][m-1] = 1 + solve(a, b, n-1, m-1, dp);
        
        return dp[n-1][m-1] = max(solve(a, b, n, m-1, dp), solve(a, b, n-1, m, dp));
    }

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