help me with array

[Manisekar Chinnasami]

itz regarding the array ...
consider an array,

a[] = { 1,2,4,6,7,2,6};

i need to delete all the repeated values and the output such as ,

a[] = {1,4,7} ...

the output should b stored in the same array ....

help me to slove this one ...

[marc weber]

We're here to help, but we're not going to do it for you. Do you have any ideas about how to accomplish this?

[Rob Spoor]

Originally posted by Manisekar Chinnasami:
the output should b stored in the same array ....

Then what are you going to store in the extra spaces? In your example, the array originally holds 7 elements. Afterwards it should hold 3 elements. But an array cannot be resized; it would contain these 3 elements and 4 other elements - what should these be?

[Manisekar Chinnasami]

i ve tried them in many ways ... but i cudnt sort out ...

here is one algorithm ...

for i = 1 to lenthofarray
for j = 1 to lenthofarray
if(i != j)
if(a[i]==a[j]
{store in b array
break;}

but the value is not stored in b array ... i ve tried in this way ... i need the not equal values to b stored in a array ...

(i m trying this for Reed-Muller's logic)
now guide me ...

[Manisekar Chinnasami]

yup... tat is also a problem ... i even need them to resize ...

(actually, i m storing the binary values ... so zero is also an value)

[Ajith George]

Better store the result in a new array, which will make your life more easier.

[Manisekar Chinnasami]

this is not the whole program ... itz just a part ...

the a[] array is processed in a loop with the resulted array values ... if i use a large loop of 15 or 20, then do i ve to create 15 to 20 arrays ???

so storing in a different array is not possible ...

i need them to be stored in the same array ...

[Ajith George]

Then replace one of the repeated entries with a non occuring number , like 999 or 0. Then while retrieving it handle this case also.

[Jitendra Takalkar]

Hi,

Please find the TestAray class which elemets if any duplicate value ..


This can be optimized ....

public class TestArray {

public static void main(String[] args) {
int[] myarray= {1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8};

int duplicateCount=0;
for(int i=0; i<myarray.length; i++) {
for(int j=i; j<myarray.length; j++) {
if(i!=j) {
if(myarray[i]==myarray[j]) {
duplicateCount++;
}
}
}
}

if(myarray.length!=duplicateCount) {
int[] newarray=new int[myarray.length-duplicateCount];
int k=0;
for(int i=0; i<myarray.length; i++) {
boolean isFound=false;
for(int j=0; j<newarray.length && !isFound; j++) {
if(myarray[i]==newarray[j]) {
isFound=true;
}
}

if(!isFound) {
newarray[k++]=myarray[i];
}
}
/*
* Output elements of new array.
*/
for(int i=0;i<newarray.length; i++) {
System.out.println(newarray[i]);
}

myarray=newarray;
}

/*
* Output of old array
*/
for(int i=0; i<myarray.length; i++) {
System.out.println(myarray[i]);
}
}

}

[vanlalhmangaiha khiangte]

Maybe if you are allowed you can use Sets ...
This will not allow duplicate elements

1.Take each element from the Array
2.Add to a set.
3.Convert it back to Array

[Manisekar Chinnasami]

not only the duplicate values should be deleted ... if a value is repeated , then both of them should be removed from the array ...

a[] = { 1,2,3,5,1,3,2}
then, the answer should be a[] = {5}

only 5 should be printed ...

[Rob Spoor]

Set<Integer> keep = new LinkedHashSet<Integer>(); // keeps insertion order
Set<Integer> unique = new HashSet<Integer>();
for (int i: a)
{
    if (unique.contains(i))
    {
        // already encountered
        keep.remove(i);
    }
    else
    {
        // not yet encountered; add and remove if encountered again
        keep.add(i);
    }
    unique.add(i);
}
// create an int[] again
a = new int[keep.size()];
int index = 0;
for (Integer i: keep)
{
    a[index++] = i;
}

[Manisekar Chinnasami]

HELLO FRIENDS,

FINALLY I GOT THE SOLUTION FOR THIS QUERY ...

THANKS FOR YOUR HELP ...

HERE IS THE PROGRAM ...

class Arr
{
public static void main(String[] args)
{
Arr ao = new Arr();
int a[] = {1,2,3,2,1};
int k=0;


for(int i=0;i<a.length;i++)
{
for(int j=0;j<a.length;j++)
if(i!=j)
if(a[i] == a[j])
{
k++;
break;
}
}

int b[] = new int[k];
k=0;

for(int i=0;i<a.length;i++)
{
for(int j=0;j<a.length;j++)
if(i!=j)
if(a[i] == a[j])
{
b[k] = a[i];
k++;
break;
}
}

int g,m=0;
for(int l=0;l<a.length;l++)
{
g = 0;
for(int j=0;j<b.length;j++)
{

if(a[l] == b[j])
{
break;

} g++;
if(g == b.length)
{ a[m] = a[l];
m++;
}

}
}

for(int i=0;i<m;i++)
System.out.println(a[i]);
}
}