Question:
An array contains both positive and negative numbers in random order. Rearrange the array elements so that all negative numbers appear before all positive numbers.
Examples:
Input: 12, 11, 13, 5, 6, 7, 5, 3, 6
Output: 12 13 5 7 3 6 11 6 5
Note: Order of elements is not important here.
Logic:

Sorting :
 Take the input as the array
 sort the array. ( Either inbuilt or self implemented )
 all the negatives will be seperated from the positives

0 Reference :
 Take the input as array
 Declare two pointers i and j
 i will move from arr [ 0 ]th location to arr [ arr.length  1 ]th location. in this it will check if arr [ i ] < 0 and i is not equal to j . if yes, then it will swap arr [ i ] and arr [ j ]
 if no, then it will simply increment j
 when both i and j will reach the last position, the loop will stop
 print the array
import java.util.*;
public class Solution{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int sortArr[] = new int[n];
int zeroRef[] = new int[n];
for(int i = 0; i < sortArr.length; i++){
sortArr[i] = sc.nextInt();
zeroRef[i] = sortArr[i];
}
//This is sorting solution
Arrays.sort(sortArr);
System.out.println("Segregation through sorting is: ");
for(int i = 0; i < sortArr.length; i++){
System.out.print(sortArr[i]+" ");
}
System.out.println();
//This is 0 reference solution
int j = 0;
for(int i = 0; i < zeroRef.length; i++){
if(zeroRef[i] < 0){
if(i != j){
int temp = zeroRef[i];
zeroRef[i] = zeroRef[j];
zeroRef[j] = temp;
}
j++;
}
}
System.out.println("Segregation through 0 reference method is: ");
for(int i = 0; i < zeroRef.length; i++){
System.out.print(zeroRef[i]+" ");
}
System.out.println();
}
}
Discussion (2)
Use array filter to compare i to abs(i) and separate into two arrays then spread the two new arrays into a single array. Original order is preserved and cost is 2N +1
Thanks for that solution. I'll definetly try that out!