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

        return minDist(word1,word2,n-1,m-1,dp);
    }








//recursion
    // int minDist(string word1, string word2,int n,int m){
    //     if(n==-1)return m+1;
    //     if(m==-1)return n+1;
    //     if(word1[n]==word2[m])
    //     return 0+minDist(word1,word2,n-1,m-1);
    //     else
    //     return 1+min(minDist(word1,word2,n,m-1),minDist(word1,word2,n-1,m));
    // }
    // int minDistance(string word1, string word2) {
    //     int n=word1.size();
    //     int m=word2.size();

    //     return minDist(word1,word2,n-1,m-1);
    // }
};
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