Find the modal value of an integear array.

How would you do it with pencil and paper?

You could do it with an iteration after sorting, checking for value changes and counting how often one value appears, and storing that value at the same time. Needs a little fiddling with variables, though.

Alternately, you could just once iterate over the unsorted array, and use a Hashmap, with the numbers as keys: if a key is already inside, increase the counter you store as its value. Otherwise, add the number and a counter with the value of 1.

If you use 1.5, autoboxing should make this quite convenient.

CeciNEstPasUnProgrammeura at 2007-7-15 13:59:36 >

Hi,

1st of all left your 5 dukes for later when You will ask harder questions :-))

2nd-ly:

When you would like to use sort and count the array try something like this

import java.util.Arrays;

public class ArraySample {

public static void main(String[] args) {

int a[] = {1, 17, 6, 4, 5, 5, 6, 4 , 2 , 2 , 2 , 2};

// This will be our result

int maxCount = -1;

int maxNumber = -1;

// We use this to keep track on current group of numbers

int currentCount = -1;

int currentNumber = -1;

// Sorty first -- array given by referce is sorted (nothing returned!)

Arrays.sort(a);

// Let's walk sorted array and count

for(int index = a.length - 1; index >= 0; index--) {

// Let's test if we are still counting same group of number e.g. '3'

if (a[index] != currentNumber) {

// If not let's look if this isn't by any chance the largest group

if (maxCount < currentCount) {

// Yes! Store information about it

maxCount = currentCount;

maxNumber = currentNumber;

}

// Store information about new group of numbers we accountered

currentNumber = a[index];

currentCount = 0;// reset counter

}

currentCount++;

}

// Finished, just print result stored in maxNumber

System.out.println( "Modal value is " + String.valueOf(maxNumber) +

" occured "+String.valueOf(maxCount)+" time(s)" );

}

}

Note that there are also another possibilities, but they involve few different Java classes (collection classes) -- I don't know how proficient with Java are you now, so IMHO this simple solution based on Your idea of sort and count would be enaough.

Message was edited by:

a3cchan

Word on complexity for both principles (sorting vs using collections to count size):

For sorting it's simple:

(1 + log(n) ) * n -- n*log(n) {for sort} + n {for walking the array} (assuming that sort complexity is around n*log(n) )

For collections it may be around:

2 * n * log(n) -- slightly better

I think it's like:

n {for walking array} * ( log(n) {key lookup} + log(n) {key insert} )

a3cchana at 2007-7-15 13:59:36 >

Ok, I assumed only positive numbers. Slighly better (and i think sufficient) solution is using Integer.MIN_VALUE instead of each -1.

Correct example can use flag to determine initial loop (firstLoop)

import java.util.Arrays;

public class ArraySample {

public static void main(String[] args) {

// Targeted array

int a[] = {1, 17, 6, 4, 5, 5, 6, 4 , 2 , 2 , 2 , 2};

// Test array size

if (a == null || a.length == 0) {

// Empty => noting to do

System.out.println("Array is empty.");

System.exit(-1);

}

// This will be our result

int maxCount = Integer.MIN_VALUE;

int maxNumber = Integer.MIN_VALUE;

// We use this to keep track on current group of numbers

int currentCount = Integer.MIN_VALUE;

int currentNumber = Integer.MIN_VALUE;

// This is used to skip first test

boolean firstLoop = true;

// Sorty first -- array given by referce is sorted (nothing returned!)

Arrays.sort(a);

// Let's walk sorted array and count

for(int index = a.length - 1; index >= 0; index--) {

// Let's test if we are still counting same group of number e.g. '3'

if (firstLoop || a[index] != currentNumber) {

// If not let's look if this isn't by any chance the largest group

if (maxCount < currentCount) {

// Yes! Store information about it

maxCount = currentCount;

maxNumber = currentNumber;

}

// Store information about new group of numbers we accounted

currentNumber = a[index];

currentCount = 0;// reset counter

firstLoop = false;

}

currentCount++;

}

// Finished, just print result stored in maxNumber

System.out.println( "Modal value is " + String.valueOf(maxNumber) +

" occured "+String.valueOf(maxCount)+" time(s)" );

}

}

CeciNEstPasUnProgrammeur> do not behave like kid, pls :-P It was just answer on "learning Java" question. Be supportive :-P

a3cchana at 2007-7-15 13:59:36 >

Wo thanks i will just take a while for me to properly understand that because we havent done that muhc programming over all i think my biggest problem is i havent had enough practice with arrays. ah well Thanks i have another question tho, with methods.

This is a small extract of a programme which i am making and i am unsure on how to make the method work this like.

public class Marks

{

public static int[] enterMarks()

{

int[] classMarks = new int[20];

System.out.println("Please enter 20 makes, between 0 and 20: ");

for(int i=0; i<classMarks.length; i++)

{

classMarks[i] = getScannerInput.anInt("Mark " + (i+1) + "= ");

while((classMarks[i] < 0) || (classMarks[i] > 20))//check to make sure the marks are between 0 and 20

{

System.out.println("That is an invalid mark, please try again.");

classMarks[i] = getScannerInput.anInt("Mark " + (i+1) + ": ");

}

}

for(int k=0; k><classMarks.length; k++)

System.out.print(classMarks[k] + ", ");

return classMarks;

}

public static void main(String[] args)

{

int[] Marks = enterMarks();

}

}

I can make it work if i make classMarks a global array but i want to make it work like that.>

RedBraina at 2007-7-15 13:59:36 >

:

:

public static void main(String[] args){

int[] marks = enterMarks();

find(marks);

}

public static void find(int[] marks) {

// Do something fun here

}

kajbja at 2007-7-15 13:59:36 >

> And I was just trying to make sure you didn't make it

> simpler than it is. Using magic numbers as flags if

> you can get random input is no good idea.

Ok, I confess that using -1 was quite a stupid idea -- minus one is dangerous number :-) Maybe just maybe Integer.MIN_VALUE. Third option, equal to boolean flag, is using Integer class and null. But code w/o autoboxing, mentioned by CeciNEstPasUnProgrammeur, looks nasty.

a3cchana at 2007-7-15 13:59:36 >