Hardik Vats - 1 year ago 70

Java Question

there is a problem which gives me a random number as a pivot, and i have to sort my array w.r.t to this pivot (closest come first then farthest)

for e.g.

`array =[2,7,4,6,4,4,5,3,6,9,1,1,9] and`

pivot=5

expected output: [5,4,4,6,6,3,7,2,1,1,9,9]

is this a variation of counting sort by any chance? if not ! can anyone give me a clue towards solving this problem?

I am encountering a roadblock in thinking on how to handle the counts and the array indices

Therefore, so far i have been able to do this

`class HelloEclipse{`

public static void main(String[] args) {

Scanner sc=new Scanner(System.in);

int N=sc.nextInt();

int pivot=sc.nextInt();

int[] mainArray=new int[N];

int[] differenceArray=new int[N];

int[] differnceCountArray=new int[Integer.MAX_VALUE];

for(int i=0;i<N;i++){

mainArray[i]=sc.nextInt();

differenceArray[i]=pivot-mainArray[i];

if(differenceArray[i]>0){

differnceCountArray[differenceArray[i]]++;}

else{

differnceCountArray[-differenceArray[i]]++;

}

}

}

}

Any suggestions on how to proceed will be helpful!

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Answer Source

Write a suitable Integer-Comparator and use Arays.sort:

```
public class PivotComparator implements Comparator<Integer> {
private int pivot;
public PivotComparator(int pivot) {
super();
this.pivot = pivot;
}
@Override
public int compare(Integer a, Integer b) {
return Math.abs(a - pivot) - Math.abs(b - pivot);
}
public static void main(String[] args) {
Integer[] toSort = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
Comparator<Integer> comp = new PivotComparator(5);
Arrays.sort(toSort, comp);
for (Integer i : toSort) {
System.out.println(i);
}
}
}
```

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