<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,string word1,string word2,int s1,int s2,vector<vector<int>>& dp){ 

        if(i>= s1 || j>=s2){
            if(i<s1) return s1-i;
            if(j<s2) return s2-j;
            return 0;
        }
        //cout<<i<<" "<<j<<endl;
        if(dp[i][j]!=-1) return dp[i][j];

        if(word1[i] == word2[j]){
            return dp[i][j] = solve(i+1,j+1,word1,word2,s1,s2,dp);
        }
        else{
            return dp[i][j] = 1+min(solve(i+1,j,word1,word2,s1,s2,dp),solve(i,j+1,word1,word2,s1,s2,dp));
        }
    }
    int minDistance(string word1, string word2) {
        int n = word1.size() , m = word2.size();
        vector<vector<int>> dp(n+2,vector<int>(m+2,-1));

        // for(int i=0;i<=n;i++){
        //     dp[i][m] = n-i;
        // }

        // for(int i=0;i<=m;i++){
        //     dp[n][i] = m-i;
        // }

        // for(int i=n-1;i>=1;i--){
        //     for(int j=m-1;j>=1;j--){
        //         if(word1[i] == word2[j]){
        //             dp[i][j] = dp[i+1][j+1];
        //         }
        //         else{
        //             dp[i][j] = 1+min(dp[i+1][j],dp[i][j+1]);
        //         }
        //     }
        // }

        return solve(0,0,word1,word2,n,m,dp);
    }
};
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