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