I think the answer is a but I am not for sure
Answer:
a lan party
Explanation:
because it's your personal party
For applying the filter you need to select the data on which you want to apply it so it will allow you to look at only data you select from the filter options.
Answer:
see explaination
Explanation:
import java.io.*;
import java.util.Scanner;
public class Winners {
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(new File("WorldSeriesWinners.txt"));
String commands[] = new String[100000];
int c = 0;
while (sc.hasNextLine()) {
String input = sc.nextLine();
System.out.println(input);
if (input.isEmpty())
continue;
commands[c++] = input;
}
sc.close();
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter the name of a team: ");
String name = keyboard.nextLine();
int count = 0;
for (int i = 0; i < c; i++) {
if (commands[i] != null) {
if (commands[i].equals(name)) {
++count;
}
}
}
if(count!=0)
System.out.println(name + " has won the World Series in the time period from 1903 through 2018 " +count + " number of times" );
else
System.out.println("Team with name "+name+ " does not exists");
}
}
Answer:
See explaination for the code
Explanation:
def wordsOfFrequency(words, freq):
d = {}
res = []
for i in range(len(words)):
if(words[i].lower() in d):
d[words[i].lower()] = d[words[i].lower()] + 1
else:
d[words[i].lower()] = 1
for word in words:
if d[word.lower()]==freq:
res.append(word)
return res
Note:
First a dictionary is created to keep the count of the lowercase form of each word.
Then, using another for loop, each word count is matched with the freq, if it matches, the word is appended to the result list res.
Finally the res list is appended.
