Positive elements at even and negative at odd positions (Relative order not maintained)

You have been given an array and you have to make a program to convert that array such that positive elements occur at even numbered places in the array and negative elements occur at odd numbered places in the array. We have to do it in place.

There can be unequal number of positive and negative values and the extra values have to left as it is.

Examples:

Input : arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}
Output : arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}

Input : arr[] = {-1, 3, -5, 6, 3, 6, -7, -4, -9, 10}
Output : arr[] = {3, -1, 6, -5, 3, -7, 6, -4, 10, -9}

The idea is to use Hoare’s partition process of Quick Sort. 
We take two pointers positive and negative. We set the positive pointer at start of the array and the negative pointer at 1st position of the array. 

We move positive pointer two steps forward till it finds a negative element. Similarly, we move negative pointer forward by two places till it finds a positive value at its position. 

If the positive and negative pointers are in the array then we will swap the values at these indexes otherwise we will stop executing the process.

Implementation:

C++

// C++ program to rearrange positive and negative
// numbers
#include <bits/stdc++.h>
using namespace std;

void rearrange(int a[], int size)
{
    int positive = 0, negative = 1;

    while (true) {

        /* Move forward the positive pointer till
         negative number number not encountered */
        while (positive < size && a[positive] >= 0)
            positive += 2;

        /* Move forward the negative pointer till
         positive number number not encountered   */
        while (negative < size && a[negative] <= 0)
            negative += 2;

        // Swap array elements to fix their position.
        if (positive < size && negative < size)
            swap(a[positive], a[negative]);

        /* Break from the while loop when any index
            exceeds the size of the array */
        else
            break;
    }
}

// Driver code
int main()
{
    int arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
    int n = (sizeof(arr) / sizeof(arr[0]));

    rearrange(arr, n);
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";

    return 0;
}

Java

// Java program to rearrange positive
// and negative numbers
import java.io.*;

class GFG {

static void rearrange(int a[], int size)
{
  int positive = 0, negative = 1, temp;

    while (true) {

    /* Move forward the positive pointer till
    negative number number not encountered */
    while (positive < size && a[positive] >= 0)
        positive += 2;

    /* Move forward the negative pointer till
        positive number number not encountered */
    while (negative < size && a[negative] <= 0)
        negative += 2;

    // Swap array elements to fix their position.
    if (positive < size && negative < size) {
        temp = a[positive];
        a[positive] = a[negative];
        a[negative] = temp;
    }

    /* Break from the while loop when any index
    exceeds the size of the array */
    else
        break;
    }
}

// Driver code
public static void main(String args[]) {
    int arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10};
    int n = arr.length;

    rearrange(arr, n);
    for (int i = 0; i < n; i++)
    System.out.print(arr[i] + " ");
}
}

/*This code is contributed by Nikita Tiwari.*/

Python3

# Python 3 program to rearrange
# positive and negative numbers

def rearrange(a, size) :

    positive = 0
    negative = 1

    while (True) :

        # Move forward the positive
        # pointer till negative number
        # number not encountered
        while (positive < size and a[positive] >= 0) :
            positive = positive + 2

        # Move forward the negative
        # pointer till positive number
        # number not encountered
        while (negative < size and a[negative] <= 0) :
            negative = negative + 2

        # Swap array elements to fix
        # their position.
        if (positive < size and negative < size) :
            temp = a[positive]
            a[positive] = a[negative]
            a[negative] = temp

        # Break from the while loop when
        # any index exceeds the size of
        # the array
        else :
            break

# Driver code
arr =[ 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 ]
n = len(arr)

rearrange(arr, n)
for i in range(0, n) :
    print(arr[i], end = " ")

# This code is contributed by Nikita Tiwari.

C#

// C# program to rearrange positive
// and negative numbers
using System;

class GFG {

// Function to rearrange
static void rearrange(int []a, int size)
{
int positive = 0, negative = 1, temp;

    while (true) {

    // Move forward the positive pointer till
    // negative number number not encountered
    while (positive < size && a[positive] >= 0)
        positive += 2;

    // Move forward the negative pointer till
    // positive number number not encountered
    while (negative < size && a[negative] <= 0)
        negative += 2;

    // Swap array elements to fix their position.
    if (positive < size && negative < size) {
        temp = a[positive];
        a[positive] = a[negative];
        a[negative] = temp;
    }

    // Break from the while loop when any
    // index exceeds the size of the array
    else
        break;
    }
}

// Driver Code
public static void Main(String []args) {
    int []arr = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10};
    int n = arr.Length;

    rearrange(arr, n);
    for (int i = 0; i < n; i++)
    Console.Write(arr[i] + " ");
}
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP program to rearrange positive
// and negative numbers

function rearrange(&$a, $size)
{
    $positive = 0;
    $negative = 1;

    while (true)
    {

        /* Move forward the positive
        pointer till negative number
        number not encountered */
        while ($positive < $size &&
               $a[$positive] >= 0)
            $positive += 2;

        /* Move forward the negative
        pointer till positive number
        number not encountered */
        while ($negative < $size &&
               $a[$negative] <= 0)
            $negative += 2;

        // Swap array elements to fix
        // their position.
        if ($positive < $size &&
            $negative < $size)
        {
            $temp = $a[$positive];
            $a[$positive] = $a[$negative];
            $a[$negative] = $temp;
        }


        /* Break from the while loop
        when any index exceeds the
        size of the array */
        else
            break;
    }
}

// Driver code
$arr = array( 1, -3, 5, 6, -3,
              6, 7, -4, 9, 10 );
$n = sizeof($arr);

rearrange($arr, $n);
for ($i = 0; $i < $n; $i++)
    echo $arr[$i] ." ";

// This code is contributed by ChitraNayal
?>

Javascript

<script>

// Javascript  program to rearrange positive
// and negative numbers
function rearrange(a, size)
{
    let  positive = 0;
    let negative = 1;
    let temp;

    while (true)
    {

        // Move forward the positive
        // pointer till negative number
        // number not encountered
        while (positive < size && a[positive] >= 0)
            positive += 2;

        // Move forward the negative pointer
        // till positive number number not encountered
        while (negative < size && a[negative] <= 0)
            negative += 2;

        // Swap array elements to fix their position.
        if (positive < size && negative < size)
        {
            temp = a[positive];
            a[positive] = a[negative];
            a[negative] = temp;
        }

        // Break from the while loop when any index
        // exceeds the size of the array
        else
            break;
    }
}

// Driver code
let arr = [ 1, -3, 5, 6, -3,
            6, 7, -4, 9, 10 ];
let n = arr.length;

rearrange(arr, n);

for(let i = 0; i < n; i++)
    document.write(arr[i] + " ");

// This code is contributed by sravan kumar G

</script>

Output

1 -3 5 -3 6 6 7 -4 9 10 

Time Complexity: O(n²), 
Auxiliary Space: O(1)

Lets explain the working of the code on the first example 
arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10} 
We declare two variables positive and negative positive points to zeroth position and negative points to first position 
positive = 0 negative = 1 
In the first iteration positive will move 4 places to fifth position as a[4] is less than zero and positive = 4. 
Negative will move 2 places and will point to fourth position as a[3]>0 
we will swap positive and negative position values as they are less than size of array. 
After first iteration the array becomes arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}
Now positive points at fourth position and negative points at third position 
In second iteration the positive value will move 6 places and its value will 
more than the size of the array.
The negative pointer will move two steps forward and it will point to 5th position 
As the positive pointer value becomes greater than the array size
we will not perform any swap operation and break out of the while loop. 
The final output will be 
arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}

An example where relative order is not maintained:
{ -1, -2, -3, -4, -5, 6, 7, 8 }; 

Another Approach :-
The idea is to find a positive/negative element which is in incorrect place(i.e. positive at odd and negative at even place) and the then find the element of opposite sign which is also in incorrect position in the remaining array and then swap these two elements. 

Here is the implementation of the above idea.  

C++

// C++ program to rearrange positive
// and negative numbers
#include<iostream>
using namespace std;

// Swap function
void swap(int* a, int i , int j)
{
    int temp = a[i];
    a[i] = a[j];
    a[j] = temp;
    return ;
}

// Print array function
void printArray(int* a, int n)
{
    for(int i = 0; i < n; i++)
        cout << a[i] << " ";
    cout << endl;
    return ;
}

// Driver code
int main()
{
    int arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
    int n = sizeof(arr)/sizeof(arr[0]);

    //before modification
    printArray(arr, n);

    for(int i = 0; i < n; i++)
    {
        if(arr[i] >= 0 && i % 2 == 1)
        {
            // out of order positive element
            for(int j = i + 1; j < n; j++)
            {
                if(arr[j] < 0 && j % 2 == 0)
                {
                    // find out of order negative
                    // element in remaining array
                    swap(arr, i, j);
                    break ;
                }
            }
        }
        else if(arr[i] < 0 && i % 2 == 0)
        {
            // out of order negative element
            for(int j = i + 1; j < n; j++)
            {
                if(arr[j] >= 0 && j % 2 == 1)
                {
                    // find out of order positive
                    // element in remaining array
                    swap(arr, i, j);
                    break;
                }
            }
        }
    }

    //after modification
    printArray(arr, n);
    return 0;
}

// This code is contributed by AnitAggarwal

Java

// Java program to rearrange positive
// and negative numbers
import java.io.*;
import java.util.*;

class GFG
{

    // Swap function
    static void swap(int[] a, int i, int j)
    {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }

    // Print array function
    static void printArray(int[] a, int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }

    // Driver code
    public static void main(String args[])
    {
        int[] arr = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
        int n = arr.length;

        //before modification
        printArray(arr, n);

        for (int i = 0; i < n; i++)
        {
            if (arr[i] >= 0 && i % 2 == 1)
            {
                // out of order positive element
                for (int j = i + 1; j < n; j++)
                {
                    if (arr[j] < 0 && j % 2 == 0)
                    {
                        // find out of order negative
                        // element in remaining array
                        swap(arr, i, j);
                        break ;
                    }
                }
            }

            else if (arr[i] < 0 && i % 2 == 0)
            {
                // out of order negative element
                for (int j = i + 1; j < n; j++)
                {
                    if (arr[j] >= 0 && j % 2 == 1)
                    {
                        // find out of order positive
                        // element in remaining array
                        swap(arr, i, j);
                        break;
                    }
                }
            }
        }

        //after modification
        printArray(arr, n);
    }
}

// This code is contributed by rachana soma

Python3

# Python3 program to rearrange positive
# and negative numbers

# Print array function
def printArray(a, n):
    for i in a:
        print(i, end = " ")
    print()


# Driver code
arr = [1, -3, 5, 6, -3, 6, 7, -4, 9, 10]
n = len(arr)

# before modification
printArray(arr, n)

for i in range(n):

    if(arr[i] >= 0 and i % 2 == 1):

        # out of order positive element
        for j in range(i + 1, n):

            if(arr[j] < 0 and j % 2 == 0):

                # find out of order negative
                # element in remaining array
                arr[i], arr[j] = arr[j], arr[i]
                break

    elif (arr[i] < 0 and i % 2 == 0):

        # out of order negative element
        for j in range(i + 1, n):

            if(arr[j] >= 0 and j % 2 == 1):

                # find out of order positive
                # element in remaining array
                arr[i], arr[j] = arr[j], arr[i]
                break

# after modification
printArray(arr, n);

# This code is contributed
# by mohit kumar

C#

// C# program to rearrange positive
// and negative numbers
using System;

class GFG
{

    // Swap function
    static void swap(int[] a, int i, int j)
    {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }

    // Print array function
    static void printArray(int[] a, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(a[i] + " ");
        Console.WriteLine();
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
        int n = arr.Length;

        //before modification
        printArray(arr, n);

        for (int i = 0; i < n; i++)
        {
            if (arr[i] >= 0 && i % 2 == 1)
            {
                // out of order positive element
                for (int j = i + 1; j < n; j++)
                {
                    if (arr[j] < 0 && j % 2 == 0)
                    {
                        // find out of order negative
                        // element in remaining array
                        swap(arr, i, j);
                        break ;
                    }
                }
            }

            else if (arr[i] < 0 && i % 2 == 0)
            {
                // out of order negative element
                for (int j = i + 1; j < n; j++)
                {
                    if (arr[j] >= 0 && j % 2 == 1)
                    {
                        // find out of order positive
                        // element in remaining array
                        swap(arr, i, j);
                        break;
                    }
                }
            }
        }

        // after modification
        printArray(arr, n);
    }
}

// This code is contributed by Akanksha Rai

Javascript

<script>

// Javascript program to rearrange positive
// and negative numbers

    // Swap function
    function  swap(a,i,j)
    {
        let temp = a[i];
        a[i] = a[j];
        a[j] = temp;

    }

    // Print array function
    function printArray(a,n)
    {
        for (let i = 0; i < n; i++)
            document.write(a[i] + " ");
        document.write("<br>");
    }

     // Driver code
    let arr=[1, -3, 5, 6, -3, 6, 7, -4, 9, 10];
    let n = arr.length;

        //before modification
        printArray(arr, n);

        for (let i = 0; i < n; i++)
        {
            if (arr[i] >= 0 && i % 2 == 1)
            {
                // out of order positive element
                for (let j = i + 1; j < n; j++)
                {
                    if (arr[j] < 0 && j % 2 == 0)
                    {
                        // find out of order negative
                        // element in remaining array
                        swap(arr, i, j);
                        break ;
                    }
                }
            }

            else if (arr[i] < 0 && i % 2 == 0)
            {
                // out of order negative element
                for (let j = i + 1; j < n; j++)
                {
                    if (arr[j] >= 0 && j % 2 == 1)
                    {
                        // find out of order positive
                        // element in remaining array
                        swap(arr, i, j);
                        break;
                    }
                }
            }
        }

        //after modification
        printArray(arr, n);


    // This code is contributed by avanitrachhadiya2155

</script>

Output

1 -3 5 6 -3 6 7 -4 9 10 
1 -3 5 -3 6 6 7 -4 9 10 

Time Complexity: O(n²),
Auxiliary Space: O(1)

This article is contributed by Ashish Madaan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.