<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 word1,string word2,int n, int m,vector<vector<int>>&dp)
    {
        if(n==0||m==0)
        {
            return 0;
        }
        if(dp[n][m]!=-1)
        {
            return dp[n][m];
        }
        if(word1[n-1]==word2[m-1])
        {
            return dp[n][m]=1+solve(word1,word2,n-1,m-1,dp);
        }
        else
        {
            return dp[n][m]=max(solve(word1,word2,n,m-1,dp),solve(word1,word2,n-1,m,dp));
        }
    }
    int minDistance(string word1, string word2) {
        int n=word1.size();
        int m=word2.size();
        vector<vector<int>>dp(n+1,vector<int>(m+1,-1));

        int lcs=solve(word1,word2,n,m,dp);
        return m+n-2*lcs;
        
    }
};
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