<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(string s1,string s2,int i,int j,vector<vector<int>>&dp){
        if(i == s1.length() || j == s2.length()){
            return 0;
        }

        if(dp[i][j] != -1){
            return dp[i][j];
        }

        if(s1[i] == s2[j]){
            int ans1 = 1 + solve(s1,s2,i+1,j+1,dp);
            
            dp[i][j] = ans1;

            return dp[i][j];
        }
        else{
            int ans2 = max(solve(s1,s2,i,j+1,dp), solve(s1,s2,i+1,j,dp));

            dp[i][j] = ans2;

            return dp[i][j];
        }
    }

    int minDistance(string word1, string word2) {
        vector<vector<int>>dp(word1.length(),vector<int>(word2.length(),-1));
        int maxSubarray = solve(word1,word2,0,0,dp);

        return word1.length() + word2.length() - 2*maxSubarray;
    }
};
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