# Help; algorithm to determine 'range'

• July 29th, 2010, 01:50 PM
b_jones10634
Help; algorithm to determine 'range'
Hello community,

I'm trying to develop a function to determine the 'range' of some integer values. 'Range' is the best way I can describe it, but it is different than the mathematical 'range', which is simply the absolute of the highest value minus the absolute of the lowest value.

It's best to describe with an example.

Lets say I have a list of Integers (pre sorted), containing values [2, 3, 4, 7, 8, 9, 14, 16, 18, 19, 20].

I need to build a string that describes the 'ranges' found within these values. So, for the above list of integers, the corresponding 'range' string would be "2-4, 7-9, 14, 16, 18-20".

I spent the afternoon here trying to create a function that would accomodate this requirement, but I can't seem to develop it in a way that will acount for all possible scenarios.

Following is my (not quite working) code:

Code Java:

Collections.sort( channelNums );
String channelRange = "";
int previousNum = 0;
boolean x = false;

for( int i = 0; i < channelNums.size(); i++ )
{
int currentNum = channelNums.get( i ).intValue();

if( i == 0 )
channelRange = Integer.toString( currentNum );
else
{
if( currentNum == (previousNum + 1) )
{
x = true;
if( !channelRange.endsWith( "-" ) )
channelRange += "-";
}
else
channelRange += ", " + Integer.toString( currentNum );
}

previousNum = currentNum;
}

if( x )
channelRange += Integer.toString( previousNum );

Any input would be highly appreciated. Thanks in advance.
• July 29th, 2010, 02:09 PM
helloworld922
Re: Help; algorithm to determine 'range'
simple algorithm:

Code Pseudo:

startOfCurrentRange = list[0]
lastNumber = startOfCurrentRange - 1
rangeString = startOfCurrentRange + "-"
for every number in the list:
if number != lastNumber + 1:
the next number is not continuous
rangeString += lastNumber
rangeString += ", " + number + "-"
end if
lastNumber = number
end for
remove last character of rangeString (it's an extra "-")

One note: You'll need to determine what the behavior should be if you have two equal numbers in the list (if this input is allowed). For example: {1, 2, 3, 3}
• July 30th, 2010, 01:05 PM
b_jones10634
Re: Help; algorithm to determine 'range'
Thanks for your reply, unfortunately the algorithm you supplied does not account for all scenarios (it assumes that it will always start with a range, but what if the set is [6, 8, 10, 14, 18]).

And, FYI, for the scope of my project, the list of integers will NEVER have duplicate values.
• July 30th, 2010, 01:41 PM
b_jones10634
Re: Help; algorithm to determine 'range'
I think I've developed an algorithm that accounts for all necessary scenarios.

Code Java:

int startOfCurrentRange = channelNums.get( 0 ).intValue();
int lastNumber = startOfCurrentRange - 1;
String channelRange = startOfCurrentRange + "-";
for( int i = 0; i < channelNums.size(); i++ )
{

int currentNumber = channelNums.get( i ).intValue();
if( currentNumber != (lastNumber + 1) )
{
if( channelRange.charAt( channelRange.length() - 1 ) == '-' )
channelRange = channelRange.substring( 0, channelRange.length() - 1 );

if( Integer.parseInt( Character.toString( channelRange.charAt( channelRange.length() - 1 ) ) ) == lastNumber )
channelRange += ", " + currentNumber + "-";
else
channelRange += "-" + lastNumber + ", " + currentNumber + "-";
}

if( i == channelNums.size() -1 )
channelRange += currentNumber;

lastNumber = currentNumber;
}

// Remove trailing '-'
if( channelRange.charAt( channelRange.length() - 1 ) == '-' )
channelRange = channelRange.substring( 0, channelRange.length() - 1 );

// Remove trailing single channel range (ex. replace traling '12-12' with '12')
if( channelRange.substring( (channelRange.length() - (Integer.toString( lastNumber ).length() * 2 + 1)), channelRange.length() )
.compareToIgnoreCase( Integer.toString( lastNumber ) + "-" + Integer.toString( lastNumber ) ) == 0 )
channelRange = channelRange.substring( 0, (channelRange.length() - (Integer.toString( lastNumber ).length() * 2 + 1)) ) +
lastNumber;

• July 30th, 2010, 02:13 PM
mcmillhj
Re: Help; algorithm to determine 'range'
I wrote this one, I think mine is a little simpler than yours. Let me know what you think.

Code :

public class TestRange
{
public static void main(String[] args)
{
int[] intArray = {2,3,4,7,9,10,11,15,16,19};
String rangeString = "";

//first element in the array
int last = intArray[0];
int current = 0;
int rangeStart = -1;

for(int i = 1; i <= intArray.length - 1; i++)
{
//assign the current number
current = intArray[i];

//test for range
if ( current != (last+1) )
{
//if there is a range append it to the rangeString
if(rangeStart != -1)
{
rangeString +=  rangeStart + "-" + last + ", ";

//reset the range start variable
rangeStart = -1;
}
else //if there isnt a range add the individual int to the rangeString
{
rangeString += last + ", ";
}

if( i == (intArray.length-1))
rangeString += current;
}
else
{
//if there are consecutive numbers, store the starting number
if( rangeStart == -1)
rangeStart = last;

//if this is the last element append any remaining ranges
if( i == (intArray.length-1))
rangeString +=rangeStart + "-" + last;
}
//assign the new last variable
last = current;
}

//print rangeString
System.out.println("rangeString: " + rangeString);
}
}

This was my output:
rangeString: 2-4, 7, 9-11, 15-16, 19

Hunter
• July 30th, 2010, 02:33 PM
b_jones10634
Re: Help; algorithm to determine 'range'
Hey Hunter,

Thanks for your input, your solution does seem slightly less convoluted. I wasn't particular happy with my solution based on all the string comparisons and substrings.

I think I'll give yours a go and see how it works out.

Thanks again.
• July 30th, 2010, 05:31 PM
helloworld922
Re: Help; algorithm to determine 'range'
it's easy to find the start of the range: it's the smallest number in the list, and since you're list is already sorted, it's always going to be the very first element of the list (as indicated by the first line of the pseudo-code).
• August 3rd, 2010, 08:49 AM
b_jones10634
Re: Help; algorithm to determine 'range'
HelloWorld,

Thanks for your reply, but I'm not sure you understood the problem at hand.

The task is not to find the start of the range... yes that is very easy to accomplish.
The task is to find the range of values in the given list. Your pseudo code attempts to satisfy this problem, but does not account for all scenarios.

For instance, your pseudo code algorithm assumes that the list of numbers will always start with a range of values (ex. [1, 2, 4, 5, 8]), but one of the most simples cases may incorporate a series that does NOT start with a range (ex. [1, 4, 5, 6, 9]).

A more suitable approach is to utilize the method either Hunter or I developed.

Further more, Hunter, your algorithm is very straight forward, however I found one small problem. The last range in the value is alwas 1 LESS than the actual value. For instance, if the series is [1, 2, 3, 4, 5], your algorithm will produce the range string = "1-4".
• August 3rd, 2010, 09:02 AM
b_jones10634
Re: Help; algorithm to determine 'range'
Hunter,

Changing one line in your algoritm produces the correct result

Change this line:

Code :

rangeString += rangeStart + "-" + previousNum;

to this:

Code :

rangeString += rangeStart + "-" + currentNum;

• August 4th, 2010, 12:58 PM
mcmillhj
Re: Help; algorithm to determine 'range'
When I make the change you suggest my algorithm actually does not work.

with this input and your suggested change: {2,3,4,9,11,12,13,16,19,19}

I get this output:

Code :

rangeString: 2-9, 9, 11-16, 16, 19, 19

With no changes I get this (which is correct):

Code :

rangeString: 2-4, 9, 11-13, 16, 19, 19

What kind of errors were you getting with my solution?

Hunter
• August 4th, 2010, 02:33 PM
hamsterofdeath
Re: Help; algorithm to determine 'range'
just for fun, in scala it would be:
Code :

def main(args: Array[String]) {
val test = List(2,3,4,9,11,12,13,16,19,19,19,19,21,22,23,24)
val result = {
//get the ranges. add an artificial "end marker" that will be dropped after the algorithm finished
val listAndEndMarker = test :+ test.max + 2
val ranges = (firstElement /: listAndEndMarker.sliding(2, 1))((result, currentTuple) => {
//if the difference is > 1, add the current tuple to the result list
if (currentTuple.last - 1 > currentTuple.head) result ++ currentTuple else result
}).dropRight(1)
//format them
}
println(result)
}

this can even be written a bit shorter by inlining the vals. the output is List(2-4, 9, 11-13, 16, 19, 21-24)
• August 4th, 2010, 02:58 PM
b_jones10634
Re: Help; algorithm to determine 'range'
Hunter,

When I implemented your solution, the results would be near perfect, expect the last number in the string would actually be the second last number in the list of integers.

I'm not sure why we are getting differing results, the only difference is that you're using an array, where I'm using a List.

With the change I mentioned, I now get the correct results.

Code java:

// Starting values
String retVal = "";
int previousNum = numberList.get( 0 );
int currentNum = 0;
int rangeStart = -1;

// For each number in the list (excluding the first number which was accessed above)
for(int i = 1; i <= numberList.size() - 1; i++)
{
// Assign the current number
currentNum = numberList.get( i );

// Test for range
if ( currentNum != (previousNum + 1) )
{
// If there is a range append it to the channelRange
if( rangeStart != -1 )
{
retVal +=  rangeStart + "-" + previousNum + ", ";

// Reset the range start variable
rangeStart = -1;
}

// If there isnt a range add the individual int to the channelRange
else
retVal += previousNum + ", ";

if( i == (numberList.size() - 1) )
retVal += currentNum;
}
else
{
// If there are consecutive numbers, store the starting number
if( rangeStart == -1 )
rangeStart = previousNum;

// If this is the last element append any remaining ranges
if( i == (numberList.size() - 1) )
retVal += rangeStart + "-" + currentNum;
}

// Assign the new last number
previousNum = currentNum;
}

• August 5th, 2010, 01:25 PM
mcmillhj
Re: Help; algorithm to determine 'range'
Thanks for pointing that out. I wasn't seeing it in my test because the last two numbers in my data set were the same.

Hunter
• August 24th, 2010, 04:57 PM
Perd1t1on
Re: Help; algorithm to determine 'range'
Code :

byte[]array=new byte[10];
for(byte i=0;i<10;i++){
array[i]=(byte)((i*2)+(2*Math.random()));
System.out.println(array[i]);
}
boolean even=false;
byte next=0,range;
List<Byte>ranges=new ArrayList<Byte>();
for(byte i=0;i<10;i++){
if(i<next){
i=next;
}
else{
if(array[i]%2==0)even=true;
else even=false;
if(even==true){
next=(byte)(i+1);
while(next<10){
if(array[next]%2==0)break;
next++;
}
if(next==10)next=(byte)(i+1);
}
if(even==false){
next=(byte)(i+1);
while(next<10){
if(array[next]%2==1)break;
next++;
}
if(next==10)next=(byte)(i+1);
}
}
}
for(byte i=0;i<ranges.size();i++){
if(i==0)System.out.print(array[ranges.get(0)]);
if(i>0){
if(ranges.get(i)-ranges.get(i-1)>1)System.out.print("-"+array[ranges.get(i)]);
else System.out.print(", "+array[ranges.get(i)]);
}
}
System.out.println();

You need to import java.util.*; The code can go straight into main method. I was too lazy to make a sort method so I just made a series of random numbers that are definitely bigger than the previous one and put it into an array. Everytime it's different. Feel free to ask any questions. I did not add documentation, but to generalize:

-The first loop gets the array data
-The second loop makes a list of all beginnings and ends of odds and evens in the range form that you specified. I don't know how to explain that, but I'm 99% sure this is what you need.
-The last loop just outputs the values in the way that you wanted.

As a side note, I used byte because it's smaller than int, that's why you see (byte) everywhere if you didn't know. If you change it to int you may not have to do that.