The algorithm works on paper but running it gives different result.

Question about the task

Under discussion

Please advise, what am I doing wrong?

Read a list of 20 words from the keyboard.
You need to count how many times each word appears in the list.
The result should be represented as a Map<String, Integer>, where the key is a unique word, and the value is the number of times the word appears in the list.

Display the contents of the map.
The case (uppercase/lowercase) affects the results.

Requirements:

The countWords method must declare and initialize a HashMap<String, Integer>.
The countWords method must return the created map.
The countWords method should add keys that correspond to unique words, and values that correspond to how many times those words occur.
The program should display the resulting map.
The main method must call the countWords method.

package com.codegym.task.task10.task1016; import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.HashMap; import java.util.Map; /* Identical words in a list */ public class Solution { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); ArrayList<String> words = new ArrayList<String>(); for (int i = 0; i < 20; i++) { words.add(reader.readLine()); } Map<String, Integer> map = countWords(words); for (Map.Entry<String, Integer> pair : map.entrySet()) { System.out.println(pair.getKey() + " " + pair.getValue()); } } public static Map<String, Integer> countWords(ArrayList<String> list) { HashMap<String, Integer> result = new HashMap<String, Integer>(); int[] count = new int[list.size()]; for (int i = 1; i < list.size(); i++) { if (list.get(1).compareTo(list.get(i-1)) == 0) { if (result.containsKey(list.get(i))) { result.replace(list.get(i), count[i]+= 1); } else result.put(list.get(i), count[i] = 1); } else { if (result.containsKey(list.get(i))) {result.replace(list.get(i), count[i]+= 1);} else result.put(list.get(i), count[i] = 1); } } return result; } }

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Misiu

Level 41 , Gdansk, Poland

19 November 2020, 21:24

Lines 38-40 and 44-46 are the same. Strange instruction: do A or do A No repetitions at all - "count" should be 1. If IF in line 36 is true then "count" should be 2 (or more).

Andrei

Level 41

20 November 2020, 06:51

Hi Misiu, here's the logic behind the identical blocks of code (38-40 & 44-46) First if the items compared are identical then check map to see if the element is already existent (this is for later on in the loop), if it is existent then replace the current value with +1; If it is not existent in the map, add it to the map with value 1. Then, if the 2 values are not identical, take the first value and check if it exists already in the map. It if exists add +1 to the value of this key. If not add it to the map, with value 1. Proceed to the next word and compare it with the previous. I feel the logic behind this is good. It is not clear to me where I am wrong. Count is for each item, not a general count as it is an array[] of ints and the repetition is in the for loop.

Misiu

Level 41 , Gdansk, Poland

20 November 2020, 13:02

I still can't see the logic. Why 1 in line 36: list.get(1)? For each next "i" count[i] = 0, so you can't get "count[i]" bigger then 1 Try: - it is not important if words are one next to another - put to "result" map all unique words with "count" 0 - in FOR loop (from 0) use lines 38-39 to increase (or not) "count" for each word in the list

Andrei

Level 41

23 November 2020, 10:00

I chose list.get(1) to act like a reference point, to compare each word in the array to that one, to have a point from where to go, so to speak, an anchor around which the comparison of other values can revolve around. The point of count[i] is not to have it increment after each loop. count[i] has the size of the list. The point of count[i] is to hold the values for the future keys of map of the repeating words. For each element of the array, if the element in the list.get(i) is repeating the let's say count[0] becomes 2. So in the map, the first element will have value 2. "- put to "result" map all unique words with "count" 0 " - this is what line 40 does, because the word appears but it is not repeating and instead of 0 is 1, because it has 1 occurence

Misiu

Level 41 , Gdansk, Poland

23 November 2020, 12:39

useful

For each next "i" count[i] = 0, so you can't get "count[i]" bigger then 1 For example: There was one Car, so in the map there is ("Car", 1). You are finding next Car at position i = 4. At the moment count[4] = 0. You are increasing count[4], so count[4] = 1. And that value you are putting into the map ("Car", 1). So repetitions of Car are 1, not 2. See my post here: https://codegym.cc/help/13995

Andrei

Level 41

23 November 2020, 13:28

Thank you! It makes more sense now. :)