<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>

int t[501][501];
class Solution {
public:
Solution()
{
    memset(t,-1,sizeof(t));
}
int solve(string x,string y,int n,int m)
{
    if(n==0 || m==0)return 0;

    if(t[n][m]!=-1)return t[n][m];

    if(x[n-1]==y[m-1])
    {
        return t[n][m]= 1+solve(x,y,n-1,m-1);
    }
    else
    {
        return t[n][m]= max(solve(x,y,n-1,m),solve(x,y,n,m-1));
    }
}
    int minDistance(string word1, string word2) {
        
        int n = word1.length();
        int m = word2.length();

        return (n + m - 2*solve(word1,word2,n,m));
    }
};
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