<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:
    vector<vector<int>>dp;
    int solve(string s1, string s2, int n, int m){
        if(n == 0){
            return m;
        }

        if(m == 0){
            return n;
        }

        if(dp[n][m] != -1){
            return dp[n][m];
        }

        if(s1[n-1] == s2[m-1]){
            dp[n][m] = 0 + solve(s1, s2, n-1, m-1);
        }

        else{
            dp[n][m] = 1 + min(solve(s1, s2, n, m-1), solve(s1, s2, n-1, m)) ;
        }

        return dp[n][m];
    }
    int minDistance(string word1, string word2) {
        int n = word1.size();
        int m = word2.size();

        dp.resize(n+1, vector<int>(m+1, -1));
        return solve(word1, word2, n, m);
    }
};
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