<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 {
    int memo[501][501];
    int LCS(string s1,string s2, int n, int m){
        // if any of the string is zero then no common subsequence
        if(n == 0 || m == 0) return 0;
        
        if(memo[n][m] != -1) return memo[n][m];
        
        if(s1[n-1] == s2[m-1]){
            return memo[n][m] = 1 + LCS(s1,s2,n-1,m-1);
        } 
        else{
            return memo[n][m] = max(LCS(s1,s2,n-1,m),LCS(s1,s2,n,m-1));
        }
    }

public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();

        memset(memo,-1,sizeof(memo));
        int lcs = LCS(word1,word2,n,m);

        int ans = abs(lcs-n) + abs(lcs-m);
        return ans;
    }
    // TC = SC = O(n*m)
};
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