<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(string text1, string text2, int i1, int i2, vector<vector<int>> &dp) {
        if(text1.size() == i1 || text2.size() == i2) return 0;
        if(dp[i1][i2] != -1) return dp[i1][i2];
        if(text1[i1] == text2[i2]) {
            return dp[i1][i2] = 1 + find(text1, text2, i1+1, i2+1, dp);
        } else {
            return dp[i1][i2] = max(find(text1, text2, i1, i2+1, dp), find(text1, text2, i1+1, i2, dp));
        }
    }
    int minDistance(string text1, string text2) {
        int n = text1.size();
        int m = text2.size();
        vector<vector<int>> dp(n, vector<int> (m, -1));
        int longestSub = find(text1, text2, 0, 0, dp);
        return n+m-(2*longestSub);
    }    
};
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