<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 {
    int lcs(string s1,string s2,int m,int n){
        int static t[1001][1001];
        memset(t,-1,sizeof(t));
    for (int i = 0; i < m+1; i++)
    {
        for (int j = 0; j < n+1; j++)
        {
            if(i==0 || j==0)  t[i][j]=0;
        }
        
    }
    
    for (int i = 1; i < m+1; i++)
    {
        for (int j = 1; j < n+1; j++)
        {
            if(s1[i-1]==s2[j-1]) t[i][j]=1+t[i-1][j-1];
            else t[i][j]=max(t[i-1][j],t[i][j-1]);
        }
        
    }
    return t[m][n];
    }
public:
    int minDistance(string s1, string s2) {
        int n1=s1.length(),n2=s2.length();
        int k=lcs(s1,s2,n1,n2);
        return n1+n2-2*k;

    }
};
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