<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, vector<vector<int>> &dp){
        if(i == 0 || j == 0) return 0;

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

        if(word1[i-1] == word2[j-1]){
            return dp[i][j] = 2 + solve(i-1, j-1, word1, word2, dp);
        }
        else{
            return dp[i][j] = max(solve(i-1, j, word1, word2, dp), solve(i, j-1, word1, word2, dp));
        }

    }

    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        vector<vector<int>> dp(n+1, vector<int>(m+1, -1));
        return n+m - solve(n, m, word1, word2, dp);
    }
};
94683