<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 lcs(string &s1, string s2, int n , int m){
        if(n==0 || m==0)return dp[n][m]= 0;
        if(dp[n][m]!=-1)return dp[n][m];

        if(s1[n-1]==s2[m-1])return dp[n][m] = 1 + lcs(s1,s2,n-1,m-1);
        else {
           return dp[n][m] = max( lcs(s1,s2,n,m-1), lcs(s1,s2,n-1,m));
        }
    }
    int minDistance(string word1, string word2) {
        int n = word1.size() ,m = word2.size();
        for(int i =0;i<n+1;i++)for(int j =0;j<m+1;j++)dp[i][j]=-1;

        int ans  = lcs(word1,word2,n,m);

        return n+m - 2*ans;

    }
};
90106