?
Solved

wordcount challenge

Posted on 2016-09-30
11
Medium Priority
?
200 Views
Last Modified: 2016-10-04
Hi,

I am working on below challenge.
http://codingbat.com/prob/p117630

The classic word-count algorithm: given an array of strings, return a Map<String, Integer> with a key for each different string, with the value the number of times that string appears in the array.

wordCount(["a", "b", "a", "c", "b"]) → {"b": 2, "c": 1, "a": 2}
wordCount(["c", "b", "a"]) → {"b": 1, "c": 1, "a": 1}
wordCount(["c", "c", "c", "c"]) → {"c": 4}

do i need to worry about sorting like alphabetica in output?
0
Comment
Question by:gudii9
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
11 Comments
 
LVL 21

Accepted Solution

by:
Amitkumar Panchal earned 1000 total points
ID: 41824748
Similar question as mentioned https://www.experts-exchange.com/questions/28973619/wordmultiple-challenge.html?cid=1752.

Answer is the same. :) Example results are not ordered, so need not to worry about.
0
 
LVL 37

Assisted Solution

by:zzynx
zzynx earned 1000 total points
ID: 41826278
>> do i need to worry about sorting like alphabetica in output?
The challenge doesn't mention that, so: No.
Moreover, the're talking about returning a Map.
A Map hasn't any sorting of its keys.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41826807
public Map<String, Integer> wordCount(String[] strings) {
  Map<String, Integer> map=new HashMap();
  for(int i=0;i<strings.length;i++){
  String test=strings[i];
  if(map.containsKey(test)){
  int count=map.get(test);
  map.put(test,count+1);
  }
  
  else{
  map.put(test,1);
  }
}
  return map;
}

Open in new window


above passes all tests. any improvements or alternate approaches.
0
Will your db performance match your db growth?

In Percona’s white paper “Performance at Scale: Keeping Your Database on Its Toes,” we take a high-level approach to what you need to think about when planning for database scalability.

 
LVL 7

Author Comment

by:gudii9
ID: 41826810
public Map < String, Integer > wordCount(String[] strings) {
    Map < String, Integer > map = new SortedMap();
    for (int i = 0; i < strings.length; i++) {
        String test = strings[i];
        if (map.containsKey(test)) {
            int count = map.get(test);
            map.put(test, count + 1);
        } else {
            map.put(test, 1);
        }
    }
    return map;
}

Open in new window


why above gives below compilation error?
Compile problems:


Error:      Map<String, Integer> map=new SortedMap();
                                   ^^^^^^^^^
Cannot instantiate the type SortedMap


see Example Code to help with compile problems
does not codingbat supports SortedMap?
0
 
LVL 32

Expert Comment

by:phoffric
ID: 41826916
Working with SortedMap in this question is off-topic as the challenge questions requires you to use a Map.
You should look up the SortedMap API if you want to ask another question.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41826991
SortedMap is implementation of Map interface similar to HashMap right? I thought codingbat supports that as well but looks no. Only HashMap? I will test in ecipse and see how it works
0
 
LVL 7

Author Comment

by:gudii9
ID: 41828698
looks like no SortedMap class is there. It seems to be interface. TreeMap is only implenetation class.
https://www.tutorialspoint.com/java/java_sortedmap_interface.htm
0
 
LVL 7

Author Comment

by:gudii9
ID: 41828707
package test;

import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;

public class WordCount {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		// wordCount(["a", "b", "a", "c", "b"]) → {"b": 2, "c": 1, "a": 2}
		// Map<String, String> map = new HashMap();
		// map.put("b", null);
		// map.put("a", "Hi");
		String[] arr = { "a", "b", "a", "c", "b" };
		System.out.println("--->" + mapAB(arr));

	}

	/*
	 * private static String mapAB(String[] arr) { // TODO Auto-generated method
	 * stub return null; }
	 */

	public Map<String, Integer> wordCount(String[] strings) {
		Map<String, Integer> map = new TreeMap();
		for (int i = 0; i < strings.length; i++) {
			String test = strings[i];
			if (map.containsKey(test)) {
				int count = map.get(test);
				map.put(test, count + 1);
			}

			else {
				map.put(test, 1);
			}
		}
		return map;
	}

}

Open in new window


above gives below error at line at line 9

The method mapAB(String[]) is undefined for the type WordCount

Not sure why and how to fix it?
0
 
LVL 7

Author Comment

by:gudii9
ID: 41828719
package test;

import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;

public class WordCount {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		// wordCount(["a", "b", "a", "c", "b"]) → {"b": 2, "c": 1, "a": 2}
		// Map<String, String> map = new HashMap();
		// map.put("b", null);
		// map.put("a", "Hi");
		String[] arr = { "a", "b", "a", "c", "b" };
		System.out.println("--->" + wordCount(arr));

	}

	/*
	 * private static String mapAB(String[] arr) { // TODO Auto-generated method
	 * stub return null; }
	 */

	public static Map<String, Integer> wordCount(String[] strings) {
		Map<String, Integer> map = new TreeMap();
		for (int i = 0; i < strings.length; i++) {
			String test = strings[i];
			if (map.containsKey(test)) {
				int count = map.get(test);
				map.put(test, count + 1);
			}

			else {
				map.put(test, 1);
			}
		}
		return map;
	}

}

Open in new window


i fixed it. now i see sorted key output

--->{a=2, b=2, c=1}
0
 
LVL 7

Author Comment

by:gudii9
ID: 41828722
package test;

import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;

public class WordCount {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		// wordCount(["a", "b", "a", "c", "b"]) → {"b": 2, "c": 1, "a": 2}
		// Map<String, String> map = new HashMap();
		// map.put("b", null);
		// map.put("a", "Hi");
		String[] arr = { "a","e","b", "a", "c", "b" };
		System.out.println("--->" + wordCount(arr));

	}

	/*
	 * private static String mapAB(String[] arr) { // TODO Auto-generated method
	 * stub return null; }
	 */

	public static Map<String, Integer> wordCount(String[] strings) {
		Map<String, Integer> map = new HashMap();
		for (int i = 0; i < strings.length; i++) {
			String test = strings[i];
			if (map.containsKey(test)) {
				int count = map.get(test);
				map.put(test, count + 1);
			}

			else {
				map.put(test, 1);
			}
		}
		return map;
	}

}

Open in new window


even after changing TreeMap to HashMap back again why and how it is sorting?

--->{a=2, b=2, c=1, e=1}
0
 
LVL 7

Author Comment

by:gudii9
ID: 41828740
public Map<String, Integer> wordCount(String[] strings) {
  Map<String, Integer> map=new TreeMap();
  for(int i=0;i<strings.length;i++){
  String test=strings[i];
  if(map.containsKey(test)){
  int count=map.get(test);
  map.put(test,count+1);
  }
  
  else{
  map.put(test,1);
  }
}
  return map;
}

Open in new window


codingbat supported TreeMap as well
0

Featured Post

New feature and membership benefit!

New feature! Upgrade and increase expert visibility of your issues with Priority Questions.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

This is about my first experience with programming Arduino.
What do responsible coders do? They don't take detrimental shortcuts. They do take reasonable security precautions, create important automation, implement sufficient logging, fix things they break, and care about users.
The goal of the video will be to teach the user the difference and consequence of passing data by value vs passing data by reference in C++. An example of passing data by value as well as an example of passing data by reference will be be given. Bot…
The viewer will learn how to user default arguments when defining functions. This method of defining functions will be contrasted with the non-default-argument of defining functions.
Suggested Courses

752 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question