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Partition negative and positive without comparison with 0
  • Difficulty Level : Easy
  • Last Updated : 14 Mar, 2019

Given an array of n integers, both negative and positive, partition them into two different arrays without comparing any element with 0.

Examples:

Input : arr[] = [1, -2, 6, -7, 8]
Output : a[] = [1, 6, 8] 
         b[] = [-2, -7]

Algorithm:

  1. Initialize two empty vectors. Push the first element of the array in any of the two vectors, suppose the first vector. Let it be denoted by x
  2. For every other element, arr[1] to arr[n-1], check if its sign and the sign of x is same or not. If the signs are the same, then push the element in the same vector. Else, push the element in the other vector
  3. After the traversal of the two vectors has completed, print both the vectors

How to check if their signs are opposite or not?
Let the integers to be checked be denoted by x and y. The sign bit is 1 in negative numbers, and 0 in positive numbers. The XOR of x and y will have the sign bit as 1 if and only if they have opposite signs. In other words, XOR of x and y will be a negative number if and only if x and y have opposite signs.

CPP




// CPP program to rearrange positive and 
// negative numbers without comparison 
// with 0.
#include <bits/stdc++.h>
using namespace std;
  
bool oppositeSigns(int x, int y)
{
    return ((x ^ y) < 0);
}
  
void partitionNegPos(int arr[], int n)
{
    vector<int> a, b;
  
    // Push first element to a. 
    a.push_back(arr[0]);
  
    // Now put all elements of same sign
    // in a[] and opposite sign in b[]
    for (int i = 1; i < n; i++) {
        if (oppositeSigns(a[0], arr[i]))
            b.push_back(arr[i]);
        else
            a.push_back(arr[i]);
    }
  
    // Print a[] and b[]
    for (int i = 0; i < a.size(); i++)
        cout << a[i] << ' ';
    cout << '\n';
    for (int i = 0; i < b.size(); i++)
        cout << b[i] << ' ';
}
  
int main()
{
    int arr[] = { 1, -2, 6, -7, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    partitionNegPos(arr, n);
    return 0;
}


Java




// Java program to rearrange positive and 
// negative numbers without comparison 
// with 0.
import java.util.*;
  
class GFG 
{
      
static boolean oppositeSigns(int x, int y)
{
    return ((x ^ y) < 0);
}
  
static void partitionNegPos(int arr[], int n)
{
    Vector<Integer> a = new Vector<Integer>();
    Vector<Integer> b = new Vector<Integer>();
  
    // Push first element to a. 
    a.add(arr[0]);
  
    // Now put all elements of same sign
    // in a[] and opposite sign in b[]
    for (int i = 1; i < n; i++) 
    {
        if (oppositeSigns(a.get(0), arr[i]))
            b.add(arr[i]);
        else
            a.add(arr[i]);
    }
  
    // Print a[] and b[]
    for (int i = 0; i < a.size(); i++)
        System.out.print(a.get(i) + " ");
    System.out.println("");
    for (int i = 0; i < b.size(); i++)
        System.out.print(b.get(i) + " ");
}
  
public static void main(String[] args) 
{
    int arr[] = { 1, -2, 6, -7, 8 };
    int n = arr.length;
    partitionNegPos(arr, n);
}
}
  
// This code has been contributed by 29AjayKumar


Python3




# Python3 program to rearrange positive
# and negative numbers without comparison 
# with 0.
  
def oppositeSigns(x, y):
      
    return ((x ^ y) < 0)
  
def partitionNegPos(arr, n):
      
    a = []
    b = []
      
    # Push first element to a.
    a = a + [arr[0]]
      
    # Now put all elements of same sign
    # in a[] and opposite sign in b[]
    for i in range(1, n) :
        if (oppositeSigns(a[0], arr[i])):
            b = b + [arr[i]]
        else:
            a = a + [arr[i]]
              
    # Print a[] and b[]
    for i in range(0, len(a)):
        print(a[i], end = ' ')
    print("")
      
    for i in range(0, len(b)):
        print(b[i], end = ' ')
  
# Driver code
arr = [1, -2, 6, -7, 8 ]
n = len(arr)
partitionNegPos(arr, n)
  
# This code is contributed by Smitha


C#




// C# program to rearrange positive and 
// negative numbers without comparison 
// with 0.
using System;
using System.Collections.Generic;
  
class GFG 
{
      
static bool oppositeSigns(int x, int y)
{
    return ((x ^ y) < 0);
}
  
static void partitionNegPos(int []arr, int n)
{
    List<int> a = new List<int> ();
    List<int> b = new List<int> ();
  
    // Push first element to a. 
    a.Add(arr[0]);
  
    // Now put all elements of same sign
    // in a[] and opposite sign in b[]
    for (int i = 1; i < n; i++) 
    {
        if (oppositeSigns(a[0], arr[i]))
            b.Add(arr[i]);
        else
            a.Add(arr[i]);
    }
  
    // Print a[] and b[]
    for (int i = 0; i < a.Count; i++)
        Console.Write(a[i] + " ");
    Console.WriteLine("");
    for (int i = 0; i < b.Count; i++)
        Console.Write(b[i] + " ");
}
  
// Driver code
public static void Main() 
{
    int []arr = { 1, -2, 6, -7, 8 };
    int n = arr.Length;
    partitionNegPos(arr, n);
}
}
  
/* This code contributed by PrinciRaj1992 */



Output:

1 6 8
-2 -7

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