<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(int i,int j,int n,int m,string word1,string word2,vector<vector<int>>&dp){
        
        if(i==n){
            return m-j;
        }
        if(j==m) return n-i;
        if(dp[i][j]!=-1) return dp[i][j];
        if(dp[i][j]!=-1) return dp[i][j];
        int delete1=0;
        int delete2=0;
        if(word1[i]==word2[j]){
            return dp[i][j]=solve(i+1,j+1,n,m,word1,word2,dp);
        }
        else{
            delete1=1+solve(i+1,j,n,m,word1,word2,dp);
            delete2=1+solve(i,j+1,n,m,word1,word2,dp);
        }
        return dp[i][j]=min(delete2,delete1);

    }
    int minDistance(string word1, string word2) {
        int i=0;
        int j=0;
        int n=word1.length();
        int m=word2.length();
        vector<vector<int>>dp(n,vector<int>(m,-1));
        return solve(i,j,n,m,word1,word2,dp);
    }
};
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