<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 {
    private:
    int solve(string w1, string w2, int i, int j,  vector<vector<int>> &dp) {
        // Base Case
        if(i == w1.length()) {
            return(w2.length()-j);
        }
        if(j == w2.length()) {
            return(w1.length()-i);
        }

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

        int ans = INT_MAX;

        if(w1[i] == w2[j]) {
            ans = solve(w1, w2, i+1, j+1, dp);
        }
        else {
            ans = 1 + min(solve(w1, w2, i+1, j, dp), solve(w1, w2, i, j+1,  dp));
        }
        return  dp[i][j] = ans;


    }
public:
    int minDistance(string word1, string word2) {

        vector<vector<int>> dp(word1.length()+1, vector<int>(word2.length()+1, -1));
        return solve(word1, word2, 0, 0, dp);
        
    }
};
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