input: 
  def editDistance(string1: str, string2: str) -> int:
      """ Compute the Levenshtein (edit) distance between two strings.
      >>> editDistance("kitten", "sitting")
      3
      >>> editDistance("kitten", "kitten")
      0
      >>> editDistance("intention", "execution")
      5
      """
      m = len(string1)
      n = len(string2)
      
      # Create a matrix of size (m+1)*(n+1)
      dp = [[0 for x in range(n+1)] for x in range(m+1)]
      
      # Fill the matrix based on the characters of the strings
      for i in range(m+1):
          for j in range(n+1):
              if i == 0:
                  dp[i][j] = j  # If first string is empty, insert all chars of second string
              elif j == 0:
                  dp[i][j] = i  # If second string is empty, remove all chars of first string
              elif string1[i-1] == string2[j-1]:
                  dp[i][j] = dp[i-1][j-1]  # If the last characters are the same, ignore it and recur for the remaining
              else:
                  dp[i][j] = 1 + min(dp[i][j-1],      # Insert
                                     dp[i-1][j],      # Remove
                                     dp[i-1][j-1])    # Replace
      
      return dp[m][n]
  
  # Example case
  print(editDistance("kitten", "sitting"))  # Should print 3
  
output:     m = len(string1)
      n = len(string2)
      
      # Create a matrix of size (m+1)*(n+1)
      dp = [[0 for x in range(n+1)] for x in range(m+1)]
      
      # Fill the matrix based on the characters of the strings
      for i in range(m+1):
          for j in range(n+1):
              if i == 0:
                  dp[i][j] = j  # If first string is empty, insert all chars of second string
              elif j == 0:
                  dp[i][j] = i  # If second string is empty, remove all chars of first string
              elif string1[i-1] == string2[j-1]:
                  dp[i][j] = dp[i-1][j-1]  # If the last characters are the same, ignore it and recur for the remaining
              else:
                  dp[i][j] = 1 + min(dp[i][j-1],      # Insert
                                     dp[i-1][j],      # Remove
                                     dp[i-1][j-1])    # Replace
      
      return dp[m][n]