<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 dp[501][501];
int helper(int idx1,int idx2,string word1,string word2){
    if(idx1<0){
        return idx2+1;
    }
    else if(idx2<0)return idx1+1;
    if(dp[idx1][idx2]!=-1)return dp[idx1][idx2];

    if(word1[idx1]==word2[idx2]){
        return dp[idx1][idx2]= helper(idx1-1,idx2-1,word1,word2);
    }
    else return dp[idx1][idx2]=min(1+helper(idx1-1,idx2,word1,word2),1+helper(idx1,idx2-1,word1,word2));
}
    int minDistance(string word1, string word2) {
        if(word1==word2)return 0;
        memset(dp,-1,sizeof(dp));
        return helper(word1.size()-1,word2.size()-1,word1,word2);
        
    }
};
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