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Sorting algortithm with index in java

I have a set of values for srting algorithm. I have successfully sorted them out. But I also want to have the index for each element after sorting. For example like :
Array = [95, 53, 24, 10]
Output after sorting should be like :
10 at index 3, 24 at index 2, 53 at index 1 and 95 at index 0
I have used the following logic for sorting. But unable to get the indexes
for (int p = 0; p < ((list.size()) - 1); p++) {
int min = p;
count++;
for(int q=p+1; q<list.size();q++) {
if(doubleArray[q] < doubleArray[min]) {
min = q;
}
}
double smallNumber = doubleArray[p];
doubleArray[p] = doubleArray[min];
doubleArray[min] = smallNumber;
}

As this is probably homework, just some ideas:
before sorting, create a copy of your initial array
after sorting, iterate the original array, and then find the index of each value in the sorted array, and print that
the tricky part is dealing with values that show up repeatedly. but that is something that depends on your exact requirements.
Alternatively, you could look into introducing a helpful data structure, such as a Pair<Integer, Integer> class. The first entry represents the value, the second one an index. Then you can define your own "sorting" on that class.

As previously suggested, I would also recommend using an additional Item class which stores the item on which you want to sort and the initial index:
public class Item<T extends Comparable<T>> implements Comparable<Item<T>> {
public final T item;
public final int index;
public Item(T item, int index) {
if (item == null)
throw new NullPointerException("the given item is null!");
this.item = item;
this.index = index;
}
#Override
public int compareTo(Item<T> t) {
if (t == null)
return 1;
return item.compareTo(t.item);
}
}
When you need to sort the array of doubles, you first create an ArrayList containing the Items which store the doubles of the input array and the initial index. Since the Item class implements the Comparable interface, you can use Collections.sort for sorting (which will be faster than your bubblesort implementation):
public static void sort(Integer... array) {
List<Item<Integer>> copy = new ArrayList<Item<Integer>>(array.length);
// copy the input array
for (int i = 0; i < array.length; ++i)
copy.add(new Item<Integer>(array[i], i));
Collections.sort(copy);
for (Item<Integer> t : copy)
System.out.println(t.item + " at index " + t.index);
}

Try this:
Create a Pair class like
class Pair {
int val;
int index;
}
sort it by valuez
index will keep the initial index

I would suggest below approach:
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.SortedSet;
import java.util.TreeSet;
public class trial
{
public static void main(String[] args)
{
List<Integer> aList = Arrays.asList(95, 53, 24, 10);
Map<Integer, Integer> aMap = new HashMap<>();
int index = 0;
for( Integer aInteger : aList )
{
aMap.put(aInteger, index);
index++;
}
SortedSet<Integer> keys = new TreeSet<>(aMap.keySet());
for( Integer key : keys )
{
Integer value = aMap.get(key);
System.out.println(key + " at index " + value);
}
}
}

Here you find the old index and shorted value
Map<Integer, Integer> map1 = numbers.stream().collect(Collectors.toMap(i -> i, i -> numbers.indexOf(i))). entrySet().stream().sorted(Map.Entry.comparingByKey()).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
(oldValue, newValue) -> oldValue, LinkedHashMap::new));
//output {10=3, 24=2, 53=1, 95=0}

I have just tried the following and it worked.
int[] index = {0,1,2,3}
for (int p=0;p<((list.size())-1);p++)
{
int min = p;
count++;
for(int q=p+1; q<list.size();q++)
{
if(doubleArray[q]< doubleArray[min])
{
min = q;
}
}
double smallNumber = doubleArray[p];
doubleArray[p] = doubleArray[min];
doubleArray[min] = smallNumber;
store = index[p];
index[p] = index[min];
index[min] = store;
}
}
and it worked.

Returning Elements From ArrayList Alphabetically

I'm working on a program to manipulate chemical formulae, and I'm writing a method which needs to loop through an ArrayList called "terms" and return the first one alphabetically.
e.g. terms = {Term('H',4),Term('C',2),Term('H',4),Term('C',1)} would return Term('C',2)
I've written this code so far but it's not working. I'm a real beginner to the Java language.
public Term nextElement()
{
int i = 0;
for (i = 0; i < terms.size()-1; i++)
{
int j = 1;
while (i + j <= terms.size())
if (terms.get(i).getElement() > terms.get(i+j).getElement())
{
terms.remove(i+j++);
return terms.get(i);
}
}
return null;
}
I'd appreciate any ideas or suggestions to solve this problem. Thanks!

You have two options here:
Let your Term class implement Comparable interface and override its compareTo() method. Then you can use Collections.sort(listOfTerms) to sort them and loop through.
Add class TermComparator which implements Comparator interface, use Collections.sort(listOfTerms, new TermComparator()) and loops through the sorted list.

1- you can implement Comparable and override compareTo()
int compareTo(Object obj){
Term term = (Term)obj;
if(term.getElement < this.getElement())
return 1;
else if (term.getElement == this.getElement())
return 0;
else
return -1;
}
then use
Collection.sort(terms);
2-
public Term nextElement()
{
char minElement = 'Z';
int index = 0;
for (int i = 0; i < terms.size(); i++)
{
if (terms.get(i).getElement() < minElement)
{
minElement = terms.get(i).getElement();
index = i;
}
}
Term temp = terms.get(index);
terms.remove(index)
return temp;
}

use Collections.sort(terms);it will arrange the list alphabetically.

This is what you need to do:
List<Term> terms = //your list
Collections.sort(terms, new Comparator<Term>() {
#Override
public int compare(Term t1, Term t2) {
return t1.getElement().compareTo(t2.getElement());
}
});

CODE:
public class CodeSample {
public static void main(String[] args) {
List<Term> terms=new ArrayList<Term>();
terms.add(new Term('H',4));
terms.add(new Term('C',2));
terms.add(new Term('H',4));
terms.add(new Term('C',1));
System.out.println("Before Sorting");
for(Term term:terms){
System.out.print(term.toString().concat(" "));
}
Collections.sort(terms,new Comparator<Term>() {
#Override
public int compare(Term object1, Term object2) {
if (object1.getElement() != object2.getElement()) {
return object1.getElement() - object2.getElement();
} else {
return object2.getCount() - object1.getCount();
}
}
});
//Sorted terms
System.out.println("After Sorting");
for(Term term:terms){
System.out.print(term.toString().concat(" "));
}
}
public static class Term{
private char element;
private int count;
public Term(char element, int count) {
super();
this.element = element;
this.count = count;
}
public char getElement() {
return element;
}
public int getCount() {
return count;
}
#Override
public String toString() {
return "Term [element=" + element + ", count=" + count + "]";
}
}
}
OUTPUT:
Before Sorting
Term [element=H, count=4] Term [element=C, count=2] Term [element=H, count=4] Term [element=C, count=1]
After Sorting
Term [element=C, count=2] Term [element=C, count=1] Term [element=H, count=4] Term [element=H, count=4]

Algorithm course: Output of int sort and method to sort Strings

My assignment asks me to make a TV show program, where I can input shows, delete, modify and sort them. What I'm stuck on is the sorting part. With the show, it asks for the name, day a new episode premieres, and time. Those are the keys I need to sort it by.
The program prompts the user to input one of those keys, then the program needs to sort (sorting by day will sort alphabetically).
I made a class and used an array. Here is the class:
public class showInfo
{
String name;
String day;
int time;
}
And the method to sort by time in the code:
public static void intSort()
{
int min;
for (int i = 0; i < arr.length; i++)
{
// Assume first element is min
min = i;
for (int j = i+1; j < arr.length; j++)
{
if (arr[j].time < arr[min].time)
{
min = j;
}
}
if (min != i)
{
int temp = arr[i].time;
arr[i].time = arr[min].time;
arr[min].time = temp;
}
}
System.out.println("TV Shows by Time");
for(int i = 0; i < arr.length; i++)
{
System.out.println(arr[i].name + " - " + arr[i].day + " - " + arr[i].time + " hours");
}
}
When I call it and output it in the main, it only shows "TV Shows by Time" and not the list. Why is this?
Also, I need to make ONE method that I will be able to use to sort both the day AND the name (both Strings). How can I do this without using those specific arrays (arr[i].name, arr[i].day) in the method?
Any help would be greatly appreciated! Thanks in advance!

In this part of your code
if (min != i) {
int temp = arr[i].time;
arr[i].time = arr[min].time;
arr[min].time = temp;
}
You're just changing the time when you should move the whole object instead. To fix it, the code must behave like this:
if (min != i) {
//saving the object reference from arr[i] in a temp variable
showInfo temp = arr[i];
//swapping the elements
arr[i] = arr[min];
arr[min] = temp;
}
I̶t̶ ̶w̶o̶u̶l̶d̶ ̶b̶e̶ ̶b̶e̶t̶t̶e̶r̶ ̶t̶o̶ ̶u̶s̶e̶ ̶ Arrays#sort ̶w̶h̶e̶r̶e̶ ̶y̶o̶u̶ ̶p̶r̶o̶v̶i̶d̶e̶ ̶a̶ ̶c̶u̶s̶t̶o̶m̶ ̶̶C̶o̶m̶p̶a̶r̶a̶t̶o̶r̶̶ ̶o̶f̶ ̶t̶h̶e̶ ̶c̶l̶a̶s̶s̶ ̶b̶e̶i̶n̶g̶ ̶s̶o̶r̶t̶e̶d̶ ̶(̶i̶f̶ ̶y̶o̶u̶ ̶a̶r̶e̶ ̶a̶l̶l̶o̶w̶e̶d̶ ̶t̶o̶ ̶u̶s̶e̶ ̶t̶h̶i̶s̶ ̶a̶p̶p̶r̶o̶a̶c̶h̶)̶.̶ ̶S̶h̶o̶r̶t̶ ̶e̶x̶a̶m̶p̶l̶e̶:̶
showInfo[] showInfoArray = ...
//your array declared and filled with data
//sorting the array
Arrays.sort(showInfoArray, new Comparator<showInfo>() {
#Override
public int compare(showInfo showInfo1, showInfo showInfo2) {
//write the comparison logic
//basic implementation
if (showInfo1.getTime() == showInfo2.getTime()) {
return showInfo1.getName().compareTo(showInfo2.getName());
}
return Integer.compare(showInfo1.getTime(), showInfo2.getTime());
}
});
//showInfoArray will be sorted...
Since you have to use a custom made sorting algorithm and support different ways to sort the data, then you just have to change the way you compare your data. This mean, in your current code, change this part
if (arr[j].time < arr[min].time) {
min = j;
}
To something more generic like
if (compare(arr[j], arr[min]) < 0) {
min = j;
}
Where you only need to change the implementation of the compare method by the one you need. Still, it will be too complex to create and maintain a method that can support different ways to compare the data. So the best option seems to be a Comparator<showInfo>, making your code look like this:
if (showInfoComparator.compare(arr[j], arr[min]) < 0) {
min = j;
}
where the showInfoComparator holds the logic to compare the elements. Now your intSort would become into something more generic:
public static void genericSort(Comparator<showInfo> showInfoComparator) {
//your current implementation with few modifications
//...
//using the comparator to find the minimum element
if (showInfoComparator.compare(arr[j], arr[min]) < 0) {
min = j;
}
//...
//swapping the elements directly in the array instead of swapping part of the data
if (min != i) {
int temp = arr[i].time;
arr[i].time = arr[min].time;
arr[min].time = temp;
}
//...
}
Now, you just have to write a set of Comparator<showInfo> implementations that supports your custom criteria. For example, here's one that compares showInfo instances using the time field:
public class ShowInfoTimeComparator implements Comparator<showInfo> {
#Override
public int compare(showInfo showInfo1, showInfo showInfo2) {
//write the comparison logic
return Integer.compare(showInfo1.getTime(), showInfo2.getTime());
}
}
Another comparator that uses the name field:
public class ShowInfoNameComparator implements Comparator<showInfo> {
#Override
public int compare(showInfo showInfo1, showInfo showInfo2) {
//write the comparison logic
return showInfo1.getName().compareTo(showInfo2.getName());
}
}
Now in your code you can call it like this1:
if (*compare by time*) {
genericSort(showInfoArray, new ShowInfoTimeComparator());
}
if (*compare by name*) {
genericSort(showInfoArray, new ShowInfoNameComparator());
}
if (*another custom rule*) {
genericSort(showInfoArray, new ShowInfoAnotherCustomRuleComparator());
}
where now you can implement a custom rule like compare showInfo objects using two or more fields. Taking as example your name and day fields (as stated in the question):
public class ShowInfoNameAndDayComparator implements Comparator<showInfo> {
#Override
public int compare(showInfo showInfo1, showInfo showInfo2) {
//write the comparison logic
int nameComparisonResult = showInfo1.getName().compareTo(showInfo2.getName());
if (nameComparisonResult == 0) {
return showInfo1.getDay().compareTo(showInfo2.getDay());
}
return nameComparisonResult;
}
}
1: There are other ways to solve this instead using lot of if statements, but looks like that's outside the question scope. If not, edit the question and add it to show another ways to solve this.
Other tips for your current code:
Declare the names of the classes using CamelCase, where the first letter of the class name is Upper Case, so your showInfo class must be renamed to ShowInfo.
To access to the fields of a class, use proper getters and setters instead of marking the fields as public or leaving the with default scope. This mean, your ShowInfo class should become into:
public class ShowInfo {
private String name;
private String day;
private int time;
public String getName() {
return this.name;
}
public void setName(String name) {
this.name = name;
}
//similar for other fields in the class
}

Use selection sort algorithm which is easy to implement,
for (int i = 0; i < arr.length; i++)
{
for (int j = i + 1; j < arr.length; j++)
{
if (arr[i].time > arr[j].time) // Here ur code that which should be compare
{
ShowInfo temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
}
no need to check min element. go through this wiki http://en.wikipedia.org/wiki/Selection_sort

Why not you use a Collection for this sort of a thingy to work. Moreover, in your added example, you are simply changing one attribute of a given object, while sorting, though you not changing the position of the object as a whole, inside the given list.
Create a List which will contain the references of all the Shows, now compare each attribute of one Show with another, in the List. Once the algorithm feels like, that swapping needs to be done, simply pick the reference from the List, save it in a temp variable, replace it with a new reference at this location, and set duplicate to the one stored in the temp variable. You are done, List is sorted :-)
Here is one small example for the same, for help :
import java.io.*;
import java.util.*;
public class Sorter {
private BufferedReader input;
private List<ShowInfo> showList;
public Sorter() {
showList = new ArrayList<ShowInfo>();
input = new BufferedReader(
new InputStreamReader((System.in)));
}
private void createList() throws IOException {
for (int i = 0; i < 5; i++) {
System.out.format("Enter Show Name :");
String name = input.readLine();
System.out.format("Enter Time of the Show : ");
int time = Integer.parseInt(input.readLine());
ShowInfo show = new ShowInfo(name, time);
showList.add(show);
}
}
private void performTask() {
try {
createList();
} catch (Exception e) {
e.printStackTrace();
}
sortByTime(showList);
}
private void sortByTime(List<ShowInfo> showList) {
int min;
for (int i = 0; i < showList.size(); i++) {
// Assume first element is min
min = i;
for (int j = i+1; j < showList.size(); j++) {
if (showList.get(j).getTime() <
showList.get(min).getTime()) {
min = j;
}
}
if (min != i) {
ShowInfo temp = showList.get(i);
showList.set(i, showList.get(min));
showList.set(min, temp);
}
}
System.out.println("TV Shows by Time");
for(int i = 0; i < showList.size(); i++) {
System.out.println(showList.get(i).getName() +
" - " + showList.get(i).getTime());
}
}
public static void main(String[] args) {
new Sorter().performTask();
}
}
class ShowInfo {
private String name;
int time;
public ShowInfo(String n, int t) {
name = n;
time = t;
}
public String getName() {
return name;
}
public int getTime() {
return time;
}
}
EDIT 2 :
For sorting By Name you can use this function :
private void sortByName(List<ShowInfo> showList) {
int min;
for (int i = 0; i < showList.size(); i++) {
// Assume first element is min
min = i;
for (int j = i+1; j < showList.size(); j++) {
int value = (showList.get(j).getName()).compareToIgnoreCase(
showList.get(min).getName());
if (value < 0)
min = j;
}
if (min != i) {
ShowInfo temp = showList.get(i);
showList.set(i, showList.get(min));
showList.set(min, temp);
}
}
System.out.println("TV Shows by Time");
for(int i = 0; i < showList.size(); i++) {
System.out.println(showList.get(i).getName() +
" - " + showList.get(i).getTime());
}
}
EDIT 3 :
Added Comparable<?> Interface, to the existing class to perform sorting based on specified input. Though one can improve on the logic, by using Enumeration, though leaving it for the OP to try his/her hands on :-)
import java.io.*;
import java.util.*;
public class Sorter {
private BufferedReader input;
private List<ShowInfo> showList;
private int command;
public Sorter() {
showList = new ArrayList<ShowInfo>();
input = new BufferedReader(
new InputStreamReader((System.in)));
command = -1;
}
private void createList() throws IOException {
for (int i = 0; i < 5; i++) {
System.out.format("Enter Show Name :");
String name = input.readLine();
System.out.format("Enter Time of the Show : ");
int time = Integer.parseInt(input.readLine());
ShowInfo show = new ShowInfo(name, time);
showList.add(show);
}
}
private void performTask() {
try {
createList();
} catch (Exception e) {
e.printStackTrace();
}
System.out.format("How would you like to sort : %n");
System.out.format("Press 0 : By Name%n");
System.out.format("Press 1 : By Time%n");
try {
command = Integer.parseInt(input.readLine());
} catch (Exception e) {
e.printStackTrace();
}
sortList(showList);
}
private void sortList(List<ShowInfo> showList) {
int min;
for (int i = 0; i < showList.size(); i++) {
// Assume first element is min
min = i;
for (int j = i+1; j < showList.size(); j++) {
showList.get(j).setValues(command);
int value = showList.get(j).compareTo(showList.get(min));
if (value < 0) {
min = j;
}
}
if (min != i) {
Collections.swap(showList, i, min);
}
}
System.out.println("TV Shows by Time");
for(int i = 0; i < showList.size(); i++) {
System.out.println(showList.get(i).getName() +
" - " + showList.get(i).getTime());
}
}
public static void main(String[] args) {
new Sorter().performTask();
}
}
class ShowInfo implements Comparable<ShowInfo> {
private String name;
private int time;
private int command;
public ShowInfo(String n, int t) {
name = n;
time = t;
}
public String getName() {
return name;
}
public int getTime() {
return time;
}
public void setValues(int cmd) {
command = cmd;
}
public int compareTo(ShowInfo show) {
int lastCmp = 1;
if (command == 0) {
lastCmp = name.compareTo(show.name);
} else if (command == 1) {
if (time < show.time) {
lastCmp = -1;
} else if (time == show.time) {
lastCmp = 0;
} else if (time > show.time) {
lastCmp = 1;
}
}
return lastCmp;
}
}

java programming & finding the mode of an array

i have a task where i need to find the mode of an array. which means i am looking for the int which is most frequent. i have kinda finished that, but the task also says if there are two modes which is the same, i should return the smallest int e.g {1,1,1,2,2,2} should give 1 (like in my file which i use that array and it gives 2)
public class theMode
{
public theMode()
{
int[] testingArray = new int[] {1,1,1,2,2,2,4};
int mode=findMode(testingArray);
System.out.println(mode);
}
public int findMode(int[] testingArray)
{
int modeWeAreLookingFor = 0;
int frequencyOfMode = 0;
for (int i = 0; i < testingArray.length; i++)
{
int currentIndexOfArray = testingArray[i];
int frequencyOfEachInArray = howMany(testingArray,currentIndexOfArray);
if (frequencyOfEachInArray > frequencyOfMode)
{
modeWeAreLookingFor = currentIndexOfArray;
frequencyOfMode = modeWeAreLookingFor;
}
}
return modeWeAreLookingFor;
}
public int howMany(int[] testingArray, int c)
{
int howManyOfThisInt=0;
for(int i=0; i < testingArray.length;i++)
{
if(testingArray[i]==c){
howManyOfThisInt++;
}
}
return howManyOfThisInt;
}
public static void main(String[] args)
{
new theMode();
}
}
as you see my algorithm returns the last found mode or how i should explain it.

I'd approach it differently. Using a map you could use each unique number as the key and then the count as the value. step through the array and for each number found, check the map to see if there is a key with that value. If one is found increment its value by 1, otherwise create a new entry with the value of 1.
Then you can check the value of each map entry to see which has the highest count. If the current key has a higher count than the previous key, then it is the "current" answer. But you have the possibility of keys with similar counts so you need to store each 'winnning' answer.
One way to approach this is to check each map each entry and remove each entry that is less than the current highest count. What you will be left with is a map of all "highest counts". If you map has only one entry, then it's key is the answer, otherwise you will need to compare the set of keys to determine the lowest.

Hint: You're updating ModeWeAreLookingFor when you find a integer with a strictly higher frequency. What if you find an integer that has the same frequency as ModeWeAreLookingFor ?
Extra exercice: In the first iteration of the main loop execution, you compute the frequency of '1'. On the second iteration (and the third, and the fourth), you re-compute this value. You may save some time if you store the result of the first computation. Could be done with a Map.
Java code convention states that method names and variable name should start with a lower case character. You would have a better syntax coloring and code easier to read if you follow this convention.

this might work with a little modification.
http://www.toves.org/books/java/ch19-array/index.html#fig2
if ((count > maxCount) || (count == maxCount && nums[i] < maxValue)) {
maxValue = nums[i];
maxCount = count;
}

since it seems there are no other way, i did a hashmap after all. i am stuck once again in the logics when it comes to comparing frequencys and and the same time picking lowest integer if equal frequencys.
public void theMode()
{
for (Integer number: intAndFrequencyMap.keySet())
{
int key = number;
int value = intAndFrequencyMap.get(number);
System.out.println("the integer: " +key + " exists " + value + " time(s).");
int lowestIntegerOfArray = 0;
int highestFrequencyOfArray = 0;
int theInteger = 0;
int theModeWanted = 0;
if (value > highestFrequencyOfArray)
{
highestFrequencyOfArray = value;
theInteger = number;
}
else if (value == highestFrequencyOfArray)
{
if (number < theInteger)
{
number = theInteger;
}
else if (number > theInteger)
{
}
else if (number == theInteger)
{
number = theInteger;
}
}
}
}

Completed:
import java.util.Arrays;
public class TheMode
{
//Probably not the most effective solution, but works without hashmap
//or any sorting algorithms
public TheMode()
{
int[] testingArray = new int[] {2,3,5,4,2,3,3,3};
int mode = findMode(testingArray);
System.out.println(Arrays.toString(testingArray));
System.out.println("The lowest mode is: " + mode);
int[] test2 = new int[] {3,3,2,2,1};
int mode2=findMode(test2);
System.out.println(Arrays.toString(test2));
System.out.println("The lowest mode is: " +mode2);
int[] test3 = new int[] {4,4,5,5,1};
int mode3 = findMode(test3);
System.out.println(Arrays.toString(test3));
System.out.println(The lowest mode is: " +mode3);
}
public int findMode(int[] testingArray)
{
int modeWeAreLookingFor = 0;
int frequencyOfMode = 0;
for (int i = 0; i < testingArray.length; i++)
{
int currentIndexOfArray = testingArray[i];
int countIntegerInArray = howMany(testingArray, currentIndexOfArray);
if (countIntegerInArray == frequencyOfMode)
{
if (modeWeAreLookingFor > currentIndexOfArray)
{
modeWeAreLookingFor = currentIndexOfArray;
}
}
else if (countIntegerInArray > frequencyOfMode)
{
modeWeAreLookingFor = currentIndexOfArray;
frequencyOfMode = countIntegerInArray;
}
}
return modeWeAreLookingFor;
}
public int howMany(int[] testingArray, int c)
{
int howManyOfThisInt=0;
for(int i=0; i < testingArray.length;i++)
{
if(testingArray[i]==c){
howManyOfThisInt++;
}
}
return howManyOfThisInt;
}
public static void main(String[] args)
{
new TheMode();
}
}

Glad you managed to solve it. As you will now see, there is more than one way to approach a problem. Here's what I meant by using a map
package util;
import java.util.HashMap;
import java.util.Map;
public class MathUtil {
public static void main(String[] args) {
MathUtil app = new MathUtil();
int[] numbers = {1, 1, 1, 2, 2, 2, 3, 4};
System.out.println(app.getMode(numbers));
}
public int getMode(int[] numbers) {
int mode = 0;
Map<Integer, Integer> numberMap = getFrequencyMap(numbers);
int highestCount = 0;
for (int number : numberMap.keySet()) {
int currentCount = numberMap.get(number);
if (currentCount > highestCount) {
highestCount = currentCount;
mode = number;
} else if (currentCount == highestCount && number < mode) {
mode = number;
}
}
return mode;
}
private Map<Integer,Integer> getFrequencyMap(int[] numbers){
Map<Integer, Integer> numberMap = new HashMap<Integer, Integer>();
for (int number : numbers) {
if (numberMap.containsKey(number)) {
int count = numberMap.get(number);
count++;
numberMap.put(number, count);
} else {
numberMap.put(number, 1);
}
}
return numberMap;
}
}

Implement reading of contact from phone book Using Hashtable in J2ME

I ran into a bind whereby I had to sort the data read from the phones PIM. In doing this I lost the other to which each contact field was referenced to the telephone number because I made use of 2 separate vectors as illustrated below
Before sorting
Nna - +445535533
Ex - +373773737
Ab - +234575757
After sorting.(Which shouldn't be)
Ab - +445535533
Ex - +373773737
Nna - +234575757
This gives an undesired behavior since the sort removes the index to index pointer of the vectors and a selected name (in a Multiple list Box) will get a wrong number.
Alternatively,
I used a hashtable, with the intention of using the names as keys and numbers as the values.
But this pairing means duplicate names being used as keys will not be allowed. Thus I made it a i.e the phone number as keys instead.
I don't want to sound like a cry baby so I stop here for a while and so you the code with a hope u guys would understand it
MY QUESTION
1. Is there a better way/algorithm to implement this?
2. How do I implement the getSelectedItems() in such a ways that it grabs the numbers of the selected indexes of a MULTIPLE CHOICE LIST from a hashTable
import java.util.Enumeration;
import java.util.Vector;
import java.util.Hashtable;
import javax.microedition.lcdui.List;
import javax.microedition.pim.Contact;
import javax.microedition.pim.ContactList;
import javax.microedition.pim.PIM;
import javax.microedition.pim.PIMException;
/**
*
* #author nnanna
*/
public class LoadContacts implements Operation {
private boolean available;
private Vector telNames = new Vector();
Vector telNumbers = new Vector();
Hashtable Listcontact = new Hashtable();
private String[] names;
public Vector getTelNames() {
return telNames;
}
public Hashtable getListcontact() {
return Listcontact;
}
public void execute() {
try {
// go through all the lists
String[] allContactLists = PIM.getInstance().listPIMLists(PIM.CONTACT_LIST);
if (allContactLists.length != 0) {
for (int i = 0; i < allContactLists.length; i++) {
System.out.println(allContactLists[i]);
System.out.println(allContactLists.length);
loadNames(allContactLists[i]);
System.out.println("Execute()");
}
} else {
available = false;
}
} catch (PIMException e) {
available = false;
} catch (SecurityException e) {
available = false;
}
}
private void loadNames(String name) throws PIMException, SecurityException {
ContactList contactList = null;
try {
contactList = (ContactList) PIM.getInstance().openPIMList(PIM.CONTACT_LIST, PIM.READ_ONLY, name);
// First check that the fields we are interested in are supported(MODULARIZE)
if (contactList.isSupportedField(Contact.FORMATTED_NAME) && contactList.isSupportedField(Contact.TEL)) {
Enumeration items = contactList.items();
Hashtable temp = new Hashtable();
while (items.hasMoreElements()) {
Contact contact = (Contact) items.nextElement();
int telCount = contact.countValues(Contact.TEL);
int nameCount = contact.countValues(Contact.FORMATTED_NAME);
if (telCount > 0 && nameCount > 0) {
String contactName = contact.getString(Contact.FORMATTED_NAME, 0);
// go through all the phone availableContacts
for (int i = 0; i < telCount; i++) {
System.out.println("Read Telno");
int telAttributes = contact.getAttributes(Contact.TEL, i);
String telNumber = contact.getString(Contact.TEL, i);
Listcontact.put(telNumber, contactName);
temp.put(contactName, telNumber);
}
names = getSortedList();
// Listcontact = temp;
System.out.println(temp + "-------");
System.out.println(Listcontact + "*******");
shortenName(contactName, 20);
}
available = true;
}
} else {
available = false;
}
} finally {
// always close it
if (contactList != null) {
contactList.close();
}
}
}
private void shortenName(String name, int length) {
if (name.length() > length) {
name = name.substring(0, 17) + "...";
}
}
public Vector getSelectedItems(List lbx) {
boolean[] arrSel = new boolean[lbx.size()];
Vector selectedNumbers = new Vector();
int selected = lbx.getSelectedFlags(arrSel);
String selectedString;
String result = "";
for (int i = 0; i < arrSel.length; i++) {
if (arrSel[i]) {
selectedString = lbx.getString(lbx.getSelectedFlags(arrSel));
result = result + " " + i;
System.out.println(Listcontact.get(selectedString));
// System.out.println(telNumbers.elementAt(i));
}
}
return selectedNumbers;
}
private String[] sortResults(String data[]) {
RecordSorter sorter = new RecordSorter();
boolean changed = true;
while (changed) {
changed = false;
for (int j = 0; j < (data.length - 1); j++) {
String a = data[j], b = data[j + 1];
if (a != null && b != null) {
int order = sorter.compare(a.getBytes(), b.getBytes());
if (order == RecordSorter.FOLLOWS) {
changed = true;
data[j] = b;
data[j + 1] = a;
}
}
}
}
return data;
}
public String[] getNames() {
return names;
}
Vector elements = new Vector();
private String[] getValueArray(Hashtable value) {
System.out.println(Listcontact + " c");
Enumeration e = value.elements();
while (e.hasMoreElements()) {
elements.addElement(e.nextElement());
}
String[] elementsArray = new String[elements.size()];
elements.copyInto(elementsArray);
elements.removeAllElements();
System.out.println(elementsArray + " k");
return elementsArray;
}
public void getDuplicates(Vector realValue) {
Vector duplicate = new Vector();
Enumeration e = realValue.elements();
for (int i = 0; e.hasMoreElements(); i++) {
if (duplicate.isEmpty() || !duplicate.elementAt(i).equals(e.nextElement())) {
break;
} else {
duplicate.addElement(e.nextElement());
}
}
}
public String[] getSortedList() {
return sortResults(getValueArray(Listcontact));
}
}

Let me reiterate you requirement: You want a method that will sort the contacts read from native phonebook, then alphabetically sort them on name.
Following is the approach,
Replace the vectors and hash-tables in your code with a single vector, say contactListVector, containing elements of type ContactItem, don't worry this class is explained below. Fundamentally the contact's name and number(s) are linked together in a ContactItem, hence you do not have to worry about there mappings which reduces the usage of redundant data structures.
class ContactItem {
private String name;
private String tnumber; //this can also be a data structure
//for storing multiple numbers
ContactItem( String name, String tnumber) {
this.name = name;
this.tnumber = tnumber;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getTnumber() {
return tnumber;
}
public void setTnumber(String tnumber) {
this.tnumber = tnumber;
}
}
You can reuse the sorting algorithm on contactListVector by comparing the member variable ContactItem.name of the vector element. Also you can deploy different sorts on member variables numbers and/or names. Also there are lots of libraries for JavaME available that have better sorting algorithm's implemented if need be use them.
I would recommend you to perform the sorting once on the contactListVector elements at the end of your method loadNames(...) maybe in the finally block triggered by some boolean variable. The current sorting call in each iteration on items enumeration is expensive and time consuming.
Also you can serialize / deserialize the ContactItem thus persist your contact list.
Let me know if you need detailed explanation.

What about inserting the contact name and numbers inside a recordStore , so you can later make a sort by creating a class which implements RecordComparator.

This statement in your code makes no sense:
selectedString = lbx.getString(lbx.getSelectedFlags(arrSel))
Per lcdui List API documentation above will return the string located at the index equal to the number of selected elements why would you need that?
If you need to output selected text for debugging purposes, use lbx.getString(i) instead.
To implement the getSelectedItems() in such a ways that it grabs the numbers of the selected indexes of a MULTIPLE CHOICE LIST do about as follows:
public Vector getSelectedItems(List lbx) {
boolean[] arrSel = new boolean[lbx.size()];
Vector selectedNumbers = new Vector();
int selected = lbx.getSelectedFlags(arrSel);
System.out.println("selected: [" + selected + "] elements in list");
String selectedString;
String result = "";
for (int i = 0; i < arrSel.length; i++) {
if (arrSel[i]) {
// here, i is the selected index
selectedNumbers.addElement(new Integer(i)); // add i to result
String selectedString = lbx.getString(i);
System.out.println("selected [" + selectedString
+ "] text at index: [" + i + "]");
}
}
return selectedNumbers;
}
As for sorting needs, just drop the HashTable and use Vector of properly designed objects instead as suggested in another answer - with your own sorting algorithm or one from some 3rd party J2ME library.

I would suggest you to have Contact class with name and Vector of numbers. And instead of sorting names array sort the array of contacts.