CodeGym/Help with Java Tasks/What's wrong?

Level 22

Baku

11.04.2019
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11comments

What's wrong?

Question about the task

Resolved

I do not really get what problem is , that's looks like solution , but I think something wrong with my understanding exactly to conditions.

Enter 10 strings from the keyboard and count the number of different letters in them (for the 26 lowercase letters of the English alphabet). Display the results on the screen in alphabetical order.

Example output:
a 5 b 8 c 3 d 7 e 0 … z 9

Requirements:

The program must read data from the keyboard 10 times.
The program should display text on the screen.
The displayed text should contain 26 lines.
Each line of output should contain a letter of the English alphabet, a space, and the number of times the letter was found in the entered lines.

package com.codegym.task.task10.task1012; import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.HashMap; import java.util.Map; /* Number of letters */ public class Solution { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); // Alphabet String abc = "abcdefghijklmnopqrstuvwxyz"; char[] abcArray = abc.toCharArray(); ArrayList<Character> alphabet = new ArrayList<>(); for (char letter : abcArray) { alphabet.add(letter); } // Read in strings ArrayList<String> list = new ArrayList<>(); for (int i = 0; i < 10; i++) { String s = reader.readLine(); list.add(s.toLowerCase()); } Map<String , Integer> map = new HashMap<>(); for (int i = 0 ; i < list.size() ; i++){ String s = list.get(i); if (!map.containsKey(s)){ map.put(s, 1); } else map.put(s, map.get(s) + 1); } for (Map.Entry<String , Integer> entry : map.entrySet()){ System.out.println(entry.getKey() + " " + entry.getValue()); } } }

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Henrique

Level 41 , São Paulo, Brazil

20 May 2020, 19:23

Hey there! Could somebody clarify to me the logic behind the code bellow?

if (!map.containsKey(s)){
    map.put(s, 1);

Why is the value of this map 1? Thanks!

Adrian

Level 27 , Spanish Town, Jamaica

25 May 2020, 07:48

that is the counter. you get the String entered (list.get(i)); split it into characters. (toCharArray(); the if statement reads: if that character is not already in the map, put it there with a value of 1(its frequency). the else block says: if you find that character again, retrieve the value and increment it.

Henrique

Level 41 , São Paulo, Brazil

25 May 2020, 14:49

Thank you for your time answering, Adrian! But could you clarify just one more thing? I get what you said, but not completely. Because when there is not a certain letter ("e", for example), the program shows the number "0" beside the letter, right? For example, "e 0". My question is: where does this zero come from, since the initial value is "1"? Thanks!

Adrian

Level 27 , Spanish Town, Jamaica

25 May 2020, 17:52

I completely understand that question, I had it myself. so, the if block will put a letter in the map only if it is present. for instance: with the input string at list.get(i) being : "Did" the first iteration of the if block looks if the character 'd' is present in the map, since it isn't it gets added with a count of 1. on the next iteration, it adds 'i' also with a count of 1. when it comes across the last 'd' it runs the else block. the code: map.get(Key) finds the value mapped to that key, so it would find the key 'd' has a count of 1 and increment it to 2. if we should now print this map in a foreach loop, only two letters would be printed: 'i ' and 'd'. you could use the array of letters already created to print the ordered alphabet: this way, every time you print a letter, you retrieve the value mapped to that letter key.

for(char x : abcArray)
    System.out.println(x + " " + map.get(x)):

This is the important detail however: any letter that is not in the map will return a value of null, not zero. to solve this issue, i wrote this:

for(Character ch : alphabet ){
            if(!map.containsKey(ch)){
                map.put(ch,0);
            }
            System.out.println(ch + " " + map.get(ch));
        }

which just means that any letter not found in the map will be put in with a value of 0 since they are not present in the input list. i hope this helps

Henrique

Level 41 , São Paulo, Brazil

25 May 2020, 18:55

Ohh now everything is much clearer! Thanks a lot, Adrian! :D

Turner

Level 11 , Pittsburgh, United States

18 October 2019, 20:22

I am pleased to see that using HashMap merge does not trip the verification/validation. Cool!

JeRiF94

Level 22 , Baku, Azerbaijan

12 April 2019, 11:53

DONE

Guadalupe Gagnon

Level 37 , Tampa, United States

12 April 2019, 14:13

I really like using HashMaps, and HashSets. You did a really good job with setting up this solve, I especially liked the logic between lines 37-42. Computational wise HashSets/Maps are quite a bit faster than Lists,the only problem is that they do not maintain order. If you have a way to order them, like the alphabet array used in your code, or order is not important for the task, then you should definitely used them when you need to store data. Good luck with your continued learning!

JeRiF94

Level 22 , Baku, Azerbaijan

13 April 2019, 22:56

Thank you very much one more time , if we start talk about order , than a would like to say , that I do it throw TreeMap , but compili do not agree with me , afterwards I pass this taks because I decide to make it clear and correct , Some guys tell me the TreeMap take a big memory , and I am agree with them. So good luck you to , I think you also studying , good luck you too , see you soon here.

Guadalupe Gagnon

Level 37 , Tampa, United States

11 April 2019, 23:55

solution

#1 Line 35, your HashMap should be <Character, Integer> since you are counting individual characters and not strings, which leads to.... #2 for each word contained in the String List (the loop at line 36) you need to split each word into characters. You can do this by calling .toCharArray() on the variable. You can use a for each loop like so:

String s = list.get(i);
for (char x : s.toCharArray){
    // put lines 38-42 here and change them to work with 'x' instead of 's'
}

#3 Then last you can't iterate through a map like that to pass this task because the output is not guaranteed to be in the order that you need it to be, however at line 20 you have the entire alphabet in order. Loop through your map by replacing lines 45- 47 like this:

for(char x : abcArray)
    System.out.println(x + " " + map.get(x)):

#4 Now you can remove any code not needed. Lines 22 - 24 fit this.

Darko Jakimovski

Level 18 , Kriva Palanka, Macedonia, The Former Yugoslav Republic of

23 April 2019, 08:58

Thought you might like this dude new Java 8 feature