<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>= word1.size()) {
            return word2.size()-j;
        }
        else if(j>=word2.size()) {
            return word1.size()-i;
        }
        if(dp[i][j] != -1) {
            return dp[i][j];
        }
        int ans = 500;
        if(word1[i] == word2[j]) {
            ans = min(ans,solve(i+1,j+1,word1,word2,dp));
        }
        else {
            ans = min(ans,1+solve(i+1,j,word1,word2,dp));
            ans = min(ans,1+solve(i,j+1,word1,word2,dp));
        }
        return dp[i][j] = ans;
    }
    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size(),vector<int>(word2.size(),-1));
        return solve(0,0,word1,word2,dp);
    }
};
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