Delete Operation for Two Strings : memoisation array value meaning

Question

Given two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.

In one step, you can delete exactly one character in either string.

Leetcode question

I am able to write coorrect code and memoise it.But i have a doubt. When i print the memo[][] array ,i get the below value for input Input: word1 = "sea", word2 = "eat" Output: 2

Memo Array Pic

What i understand memo[i][j] number of min steps to make both words same ( 0,0 is Empty string ) so value of memo[0][0] should be 0 since it represent empty strings. So why memo[0][0] ==2 and not 0?

public int minDistance(String word1, String word2) {


     int[][] memo=new int[word1.length() + 1 ][word2.length() + 1 ];
    
    for( int i =0 ; i< word1.length() + 1  ;i++)
    {
        for( int j =0 ; j< word2.length() + 1 ;j++)
        {
           memo[i][j]=-1; 
        }
    }
    
    int ans = solver(word1,word2,0,0,memo) ;
    printMemo(memo);
    
    return ans;
    
}


  public int solver(String word1, String word2,int i ,int j,int[][] memo) {
      
         if( memo[i][j]!=-1)
        return memo[i][j];
      
      if( i==word1.length() && j==word2.length() )
      {
          return 0;
      }
      
        if( i==word1.length()  && j !=word2.length() )
      {
          return word2.length() -j;
      }
      
      
          if( j==word2.length()  && i !=word1.length() )
      {
          return word1.length()-i;
      }
    
    
      int del1=0;
      int del2=0;
      if( word1.charAt(i)==word2.charAt(j))
      {
          return  memo[i][j]=solver(word1,word2,i+1,j+1,memo);
      } else
      {
          del1= 1+solver(word1,word2,i+1,j,memo);
          del2= 1+ solver(word1,word2,i,j+1,memo);
          
      }
      
        int ans=Math.min(del1,del2);
      
       memo[i][j]= ans;
      
      return ans ;
    
  }

`

Answer 1

In this particular solution, memo[i][j] means the cost for word1[i..end] and word2[j..end]. So, the base case is memo[len1][len2] = 0, and the answer is at memo[0][0]. Perhaps your understanding of the solution suggested to make it the other way around, but with this code, this is it.