Rearrange positive and negative numbers in O(n) time and O(1) extra space

An array contains both positive and negative numbers in random order. Rearrange the array elements so that positive and negative numbers are placed alternatively. A number of positive and negative numbers need not be equal. If there are more positive numbers they appear at the end of the array. If there are more negative numbers, they too appear at the end of the array.
For example, if the input array is [-1, 2, -3, 4, 5, 6, -7, 8, 9], then the output should be [9, -7, 8, -3, 5, -1, 2, 4, 6]
Note: The partition process changes the relative order of elements. I.e. the order of the appearance of elements is not maintained with this approach. See this for maintaining the order of appearance of elements in this problem.
The solution is to first separate positive and negative numbers using the partition process of QuickSort. In the partition process, consider 0 as the value of the pivot element so that all negative numbers are placed before positive numbers. Once negative and positive numbers are separated, we start from the first negative number and first positive number and swap every alternate negative number with the next positive number. 

Flowchart

Flowchart of below code

C++

// A C++ program to put positive
// numbers at even indexes (0, 2, 4,..)
// and negative numbers at odd
// indexes (1, 3, 5, ..)
#include <iostream>
using namespace std;

class GFG
{
    public:
    void rearrange(int [],int);
    void swap(int *,int *);
    void printArray(int [],int);
};

// The main function that rearranges
// elements of given array. It puts
// positive elements at even indexes
// (0, 2, ..) and negative numbers
// at odd indexes (1, 3, ..).
void GFG :: rearrange(int arr[], int n)
{
    // The following few lines are
    // similar to partition process
    // of QuickSort. The idea is to
    // consider 0 as pivot and
    // divide the array around it.
    int i = -1;
    for (int j = 0; j < n; j++)
    {
        if (arr[j] < 0)
        {
            i++;
            swap(&arr[i], &arr[j]);
        }
    }

    // Now all positive numbers are at
    // end and negative numbers at the
    // beginning of array. Initialize
    // indexes for starting point of
    // positive and negative numbers
    // to be swapped
    int pos = i + 1, neg = 0;

    // Increment the negative index by
    // 2 and positive index by 1,
    // i.e., swap every alternate negative
    // number with next positive number
    while (pos < n && neg < pos &&
                     arr[neg] < 0)
    {
        swap(&arr[neg], &arr[pos]);
        pos++;
        neg += 2;
    }
}

// A utility function
// to swap two elements
void GFG :: swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

// A utility function to print an array
void GFG :: printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Driver Code
int main()
{
    int arr[] = {-1, 2, -3, 4,
                  5, 6, -7, 8, 9};
    int n = sizeof(arr) / sizeof(arr[0]);
    GFG test;
    test.rearrange(arr, n);
    test.printArray(arr, n);
    return 0;
}

// This code is contributed
// by vt_Yogesh Shukla 1

C

// A C++ program to put positive numbers at even indexes (0,
// 2, 4,..) and negative numbers at odd indexes (1, 3, 5, ..)
#include <stdio.h>

// prototype for swap
void swap(int *a, int *b);

// The main function that rearranges elements of given array.
// It puts  positive elements at even indexes (0, 2, ..) and
// negative numbers at odd indexes (1, 3, ..).
void rearrange(int arr[], int n)
{
    // The following few lines are similar to partition process
    // of QuickSort.  The idea is to consider 0 as pivot and
    // divide the array around it.
    int i = -1;
    for (int j = 0; j < n; j++)
    {
        if (arr[j] < 0)
        {
            i++;
            swap(&arr[i], &arr[j]);
        }
    }

    // Now all positive numbers are at end and negative numbers
    // at the beginning of array. Initialize indexes for starting
    // point of positive and negative numbers to be swapped
    int pos = i+1, neg = 0;

    // Increment the negative index by 2 and positive index by 1,
    // i.e., swap every alternate negative number with next
    // positive number
    while (pos < n && neg < pos && arr[neg] < 0)
    {
        swap(&arr[neg], &arr[pos]);
        pos++;
        neg += 2;
    }
}

// A utility function to swap two elements
void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

// A utility function to print an array
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        printf("%4d ", arr[i]);
}

// Driver program to test above functions
int main()
{
    int arr[] = {-1, 2, -3, 4, 5, 6, -7, 8, 9};
    int n = sizeof(arr)/sizeof(arr[0]);
    rearrange(arr, n);
    printArray(arr, n);
    return 0;
}

Java

// A JAVA program to put positive numbers at even indexes
// (0, 2, 4,..) and negative numbers at odd indexes (1, 3,
// 5, ..)
import java.io.*;

class Alternate {

    // The main function that rearranges elements of given
    // array.  It puts positive elements at even indexes (0,
    // 2, ..) and negative numbers at odd indexes (1, 3, ..).
    static void rearrange(int arr[], int n)
    {
        // The following few lines are similar to partition
        // process of QuickSort.  The idea is to consider 0
        // as pivot and divide the array around it.
        int i = -1, temp = 0;
        for (int j = 0; j < n; j++)
        {
            if (arr[j] < 0)
            {
                i++;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }

        // Now all positive numbers are at end and negative numbers at
        // the beginning of array. Initialize indexes for starting point
        // of positive and negative numbers to be swapped
        int pos = i+1, neg = 0;

        // Increment the negative index by 2 and positive index by 1, i.e.,
        // swap every alternate negative number with next positive number
        while (pos < n && neg < pos && arr[neg] < 0)
        {
            temp = arr[neg];
            arr[neg] = arr[pos];
            arr[pos] = temp;
            pos++;
            neg += 2;
        }
    }

    // A utility function to print an array
    static void printArray(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + "   ");
    }

    /*Driver function to check for above functions*/
    public static void main (String[] args)
    {
        int arr[] = {-1, 2, -3, 4, 5, 6, -7, 8, 9};
        int n = arr.length;
        rearrange(arr,n);
        System.out.println("Array after rearranging: ");
        printArray(arr,n);
    }
}
/*This code is contributed by Devesh Agrawal*/

Python3

#  Python program to put positive numbers at even indexes (0,  // 2, 4,..) and
#  negative numbers at odd indexes (1, 3, 5, ..)

# The main function that rearranges elements of given array.
# It puts  positive elements at even indexes (0, 2, ..) and
# negative numbers at odd indexes (1, 3, ..).
def rearrange(arr, n):
    # The following few lines are similar to partition process
    # of QuickSort.  The idea is to consider 0 as pivot and
    # divide the array around it.
    i = -1
    for j in range(n):
        if (arr[j] < 0):
            i += 1
            # swapping of arr
            arr[i], arr[j] = arr[j], arr[i]

    # Now all positive numbers are at end and negative numbers
    # at the beginning of array. Initialize indexes for starting
    # point of positive and negative numbers to be swapped
    pos, neg = i+1, 0

    # Increment the negative index by 2 and positive index by 1,
    # i.e., swap every alternate negative number with next
    # positive number
    while (pos < n and neg < pos and arr[neg] < 0):

        # swapping of arr
        arr[neg], arr[pos] = arr[pos], arr[neg]
        pos += 1
        neg += 2

# A utility function to print an array
def printArray(arr, n):

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

# Driver program to test above functions
arr = [-1, 2, -3, 4, 5, 6, -7, 8, 9]
n = len(arr)
rearrange(arr, n)
printArray(arr, n)

# Contributed by Afzal

C#

// A C# program to put positive numbers
// at even indexes (0, 2, 4, ..) and
// negative numbers at odd indexes (1, 3, 5, ..)
using System;

class Alternate {

    // The main function that rearranges elements
    // of given array. It puts positive elements
    // at even indexes (0, 2, ..) and negative
    // numbers at odd indexes (1, 3, ..).
    static void rearrange(int[] arr, int n)
    {
        // The following few lines are similar to partition
        // process of QuickSort. The idea is to consider 0
        // as pivot and divide the array around it.
        int i = -1, temp = 0;
        for (int j = 0; j < n; j++) {
            if (arr[j] < 0) {
                i++;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }

        // Now all positive numbers are at end
        // and negative numbers at the beginning of
        // array. Initialize indexes for starting point
        // of positive and negative numbers to be swapped
        int pos = i + 1, neg = 0;

        // Increment the negative index by 2 and
        // positive index by 1, i.e., swap every
        // alternate negative number with next positive number
        while (pos < n && neg < pos && arr[neg] < 0) {
            temp = arr[neg];
            arr[neg] = arr[pos];
            arr[pos] = temp;
            pos++;
            neg += 2;
        }
    }

    // A utility function to print an array
    static void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }

    /*Driver function to check for above functions*/
    public static void Main()
    {
        int[] arr = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };
        int n = arr.Length;
        rearrange(arr, n);
        printArray(arr, n);
    }
}

// This code is contributed by vt_m.

PHP

<?php
// A PHP program to put positive numbers
// at even indexes (0, 2, 4,..) and negative
// numbers at odd indexes (1, 3, 5, ..)

// The main function that rearranges elements
// of given array. It puts positive elements
// at even indexes (0, 2, ..) and negative
// numbers at odd indexes (1, 3, ..).
function rearrange(&$arr, $n)
{
    // The following few lines are similar
    // to partition process of QuickSort.
    // The idea is to consider 0 as pivot
    // and divide the array around it.
    $i = -1;
    for ($j = 0; $j < $n; $j++)
    {
        if ($arr[$j] < 0)
        {
            $i++;
            swap($arr[$i], $arr[$j]);
        }
    }

    // Now all positive numbers are at end and
    // negative numbers at the beginning of array.
    // Initialize indexes for starting point of
    // positive and negative numbers to be swapped
    $pos = $i + 1;
    $neg = 0;

    // Increment the negative index by 2 and
    // positive index by 1, i.e., swap every
    // alternate negative number with next
    // positive number
    while ($pos < $n && $neg < $pos &&
                        $arr[$neg] < 0)
    {
        swap($arr[$neg], $arr[$pos]);
        $pos++;
        $neg += 2;
    }
}

// A utility function to swap two elements
function swap(&$a, &$b)
{
    $temp = $a;
    $a = $b;
    $b = $temp;
}

// A utility function to print an array
function printArray(&$arr, $n)
{
    for ($i = 0; $i < $n; $i++)
        echo " " . $arr[$i] . " ";
}

// Driver Code
$arr = array(-1, 2, -3, 4, 5, 6, -7, 8, 9);
$n = count($arr);
rearrange($arr, $n);
printArray($arr, $n);

// This code is contributed
// by rathbhupendra
?>

Javascript

<script>

// Javascript program to put positive
// numbers at even indexes
// (0, 2, 4,..) and negative
// numbers at odd indexes (1, 3,
// 5, ..)


    // The main function that
    // rearranges elements of given
    // array. It puts positive elements
    // at even indexes (0,
    // 2, ..) and negative numbers
    // at odd indexes (1, 3, ..).

    function rearrange(arr,n)
    {
        // The following few lines are similar to partition
        // process of QuickSort. The idea is to consider 0
        // as pivot and divide the array around it.
        let i = -1, temp = 0;
        for (let j = 0; j < n; j++)
        {
            if (arr[j] < 0)
            {
                i++;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }

        // Now all positive numbers are
        // at end and negative numbers at
        // the beginning of array.
        // Initialize indexes for starting point
        // of positive and negative numbers
        // to be swapped
        let pos = i+1, neg = 0;

        // Increment the negative index
        // by 2 and positive index by 1, i.e.,
        // swap every alternate negative number
        // with next positive number
        while (pos < n && neg < pos && arr[neg] < 0)
        {
            temp = arr[neg];
            arr[neg] = arr[pos];
            arr[pos] = temp;
            pos++;
            neg += 2;
        }
    }

    // A utility function to print an array
    function printArray(arr,n)
    {
        for (let i = 0; i < n; i++)
            document.write(arr[i] + "   ");
    }

    /*Driver function to check for above functions*/

        let arr = [-1, 2, -3, 4, 5, 6, -7, 8, 9];
        let n = arr.length;
        rearrange(arr,n);
        printArray(arr,n);


// This code is contributed by sravan kumar

</script>

Output: 

    4   -3    5   -1    6   -7    2    8    9

Time Complexity: O(n) where n is number of elements in given array. As, we are using a loop to traverse N times so it will cost us O(N) time 
Auxiliary Space: O(1), as we are not using any extra space.

New Approach:- This approach uses two pointers, i and j, that start at opposite ends of the array. The pointer i moves forward to find the next negative element, while the pointer j moves backward to find the next positive element. When both i and j have found an element that is out of place, they swap positions. This process continues until i and j cross, at which point all elements have been rearranged.

This approach only requires O(1) extra space because it rearranges the elements in place, without creating a separate copy of the array.

 Steps:-

1. The code defines a class called Alternate.
2. The rearrange() method takes an integer array and its length as input parameters.
3. Two variables i and j are initialized to -1 and n, respectively.
4. The while loop executes infinitely until it returns from inside the loop.
5. In the first inner while loop, the value of i is incremented until a negative element is encountered.
6. In the second inner while loop, the value of j is decremented until a positive element is encountered.
7. If i and j cross each other, the rearrangement of all elements is complete and the function returns.
8. If i and j have not crossed each other, the positive and negative elements are swapped.
9. The printArray() method takes an integer array and its length as input parameters.
10. A for loop is used to iterate through the array elements and print them.
11. If the current iteration index modulo 3 is equal to 0, a newline character is printed to break the line.
12. The main() method creates an integer array with some initial values, calls the rearrange() method to rearrange its elements, and then calls the printArray() method to print the rearranged array with every third element on a new line.

Below is the implementation of the above approach: 

C++

#include <iostream>
using namespace std;

void rearrange(int arr[], int n)
{
    int i = -1, j = n;
    while (true) {
        // Move i to the next negative element
        while (i < n - 1 && arr[++i] >= 0) {
        }
        // Move j to the next positive element
        while (j > 0 && arr[--j] < 0) {
        }

        // If i and j cross, we have rearranged all elements
        if (i >= j) {
            return;
        }

        // Swap the positive and negative elements
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
}

void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
        if ((i + 1) % 3 == 0) {
            cout << endl;
        }
    }
}

int main()
{
    int arr[] = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    rearrange(arr, n);
    cout << "Array after rearranging: " << endl;
    printArray(arr, n);
    return 0;
}

// This code is contributed by sarojmcy2e

Java

public class Alternate {
    static void rearrange(int arr[], int n)
    {
        int i = -1, j = n;
        while (true) {
            // Move i to the next negative element
            while (i < n - 1 && arr[++i] >= 0) {
            }

            // Move j to the next positive element
            while (j > 0 && arr[--j] < 0) {
            }

            // If i and j cross, we have rearranged all
            // elements
            if (i >= j) {
                return;
            }

            // Swap the positive and negative elements
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    static void printArray(int arr[], int n)
    {
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
            if ((i + 1) % 3 == 0) {
                System.out.println();
            }
        }
    }

    public static void main(String[] args)
    {
        int arr[] = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };
        int n = arr.length;
        rearrange(arr, n);
        System.out.println("Array after rearranging: ");
        printArray(arr, n);
    }
}

Python3

# Python implementation of above mentioned approach
def rearrange(arr, n):
    i = -1
    j = n
    while True:
        # Move i to the next negative element
        i += 1
        while i < n - 1 and arr[i] >= 0:
            i += 1

        # Move j to the next positive element
        j -= 1
        while j > 0 and arr[j] < 0:
            j -= 1

        # If i and j cross, we have rearranged all elements
        if i >= j:
            return

        # Swap the positive and negative elements
        temp = arr[i]
        arr[i] = arr[j]
        arr[j] = temp


def printArray(arr, n):
    for i in range(n):
        print(arr[i], end=" ")
        if (i + 1) % 3 == 0:
            print()


arr = [-1, 2, -3, 4, 5, 6, -7, 8, 9]
n = len(arr)
rearrange(arr, n)
print("Array after rearranging: ")
printArray(arr, n)

# This code is contributed by Tapesh(tapeshdua420)

C#

using System;

public class Program
{
    public static void Main()
    {
        int[] arr = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };
        int n = arr.Length;

        Rearrange(arr, n);

        Console.WriteLine("Array after rearranging: ");
        PrintArray(arr, n);
    }

    public static void Rearrange(int[] arr, int n)
    {
        int i = -1, j = n;
        while (true)
        {
            // Move i to the next negative element
            while (i < n - 1 && arr[++i] >= 0) {}

            // Move j to the next positive element
            while (j > 0 && arr[--j] < 0) {}

            // If i and j cross, we have rearranged all elements
            if (i >= j) {
                return;
            }

            // Swap the positive and negative elements
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    public static void PrintArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
        {
            Console.Write(arr[i] + " ");

            if ((i + 1) % 3 == 0)
            {
                Console.WriteLine();
            }
        }
    }
}

Output:-

Array after rearranging: 
9 2 8 
4 5 6 
-7 -3 -1 

Time Complexity:-he time complexity of the rearrange function is O(n), since it uses two while loops to iterate over the array once, swapping positive and negative elements as it goes.

Auxiliary Space:-The space complexity of both the rearrange and printArray functions is O(1), since they only use a fixed amount of extra space to store variables (i, j, temp, and n) and don’t create any additional data structures. Therefore, the overall space complexity of the program is also O(1).

This article is compiled by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.