Rearrange array in alternating positive & negative items with O(1) extra space | Set 2

Given an array of positive and negative numbers, arrange them in an alternate fashion such that every positive number is followed by negative and vice-versa. Order of elements in output doesn’t matter. Extra positive or negative elements should be moved to end.

Examples:

Input :
arr[] = {-2, 3, 4, -1}
Output :
arr[] = {-2, 3, -1, 4} OR {-1, 3, -2, 4} OR ..

Input :
arr[] = {-2, 3, 1}
Output :
arr[] = {-2, 3, 1} OR {-2, 1, 3} 

Input : 
arr[] = {-5, 3, 4, 5, -6, -2, 8, 9, -1, -4}
Output :
arr[] = {-5, 3, -2, 5, -6, 4, -4, 9, -1, 8} 
        OR ..

Approach 1:

1. First, sort the array in non-increasing order. Then we will count the number of positive and negative integers.
2. Then swap the one negative and one positive number in the odd positions till we reach our condition.
3. This will rearrange the array elements because we are sorting the array and accessing the element from left to right according to our need.

Below is the implementation of the above approach:

C++

// Below is the implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function which works in the condition when number of
// negative numbers are lesser or equal than positive
// numbers
void fill1(int a[], int neg, int pos)
{
    if (neg % 2 == 1) {
        for (int i = 1; i < neg; i += 2) {
            int c = a[i];
            int d = a[i + neg];
            int temp = c;
            a[i] = d;
            a[i + neg] = temp;
        }
    }
    else {
        for (int i = 1; i <= neg; i += 2) {
            int c = a[i];
            int d = a[i + neg - 1];
            int temp = c;
            a[i] = d;
            a[i + neg - 1] = temp;
        }
    }
}

// Function which works in the condition when number of
// negative numbers are greater than positive numbers
void fill2(int a[], int neg, int pos)
{
    if (pos % 2 == 1) {
        for (int i = 1; i < pos; i += 2) {
            int c = a[i];
            int d = a[i + pos];
            int temp = c;
            a[i] = d;
            a[i + pos] = temp;
        }
    }
    else {
        for (int i = 1; i <= pos; i += 2) {
            int c = a[i];
            int d = a[i + pos - 1];
            int temp = c;
            a[i] = d;
            a[i + pos - 1] = temp;
        }
    }
}

// Reverse the array
void reverse(int a[], int n)
{
    int i, k, t;
    for (i = 0; i < n / 2; i++) {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}

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

// Driver code
int main()
{
    int arr[] = { 2, 3, -4, -1, 6, -9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Given array is ";
    print(arr, n);

    int neg = 0, pos = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] < 0)
            neg++;
        else
            pos++;
    }
    // Sort the array
    sort(arr, arr + n);

    if (neg <= pos)
        fill1(arr, neg, pos);
    else {
        // Reverse the array in this condition
        reverse(arr, n);
        fill2(arr, neg, pos);
    }
    cout << "Rearranged array is  ";
    print(arr, n);
}

// This code is contributed by Potta Lokesh

C

// Below is the implementation of the above approach
#include <stdio.h>
#include <stdlib.h>

// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}

// Function which works in the condition when number of
// negative numbers are lesser or equal than positive
// numbers
void fill1(int a[], int neg, int pos)
{
    if (neg % 2 == 1) {
        for (int i = 1; i < neg; i += 2) {
            int c = a[i];
            int d = a[i + neg];
            int temp = c;
            a[i] = d;
            a[i + neg] = temp;
        }
    }
    else {
        for (int i = 1; i <= neg; i += 2) {
            int c = a[i];
            int d = a[i + neg - 1];
            int temp = c;
            a[i] = d;
            a[i + neg - 1] = temp;
        }
    }
}

// Function which works in the condition when number of
// negative numbers are greater than positive numbers
void fill2(int a[], int neg, int pos)
{
    if (pos % 2 == 1) {
        for (int i = 1; i < pos; i += 2) {
            int c = a[i];
            int d = a[i + pos];
            int temp = c;
            a[i] = d;
            a[i + pos] = temp;
        }
    }
    else {
        for (int i = 1; i <= pos; i += 2) {
            int c = a[i];
            int d = a[i + pos - 1];
            int temp = c;
            a[i] = d;
            a[i + pos - 1] = temp;
        }
    }
}

// Reverse the array
void reverse(int a[], int n)
{
    int i, k, t;
    for (i = 0; i < n / 2; i++) {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}

// Print the array
void print(int a[], int n)
{
    for (int i = 0; i < n; i++)
        printf("%d ", a[i]);
    printf("\n");
}

// Driver code
int main()
{
    int arr[] = { 2, 3, -4, -1, 6, -9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Given array is ");
    print(arr, n);

    int neg = 0, pos = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] < 0)
            neg++;
        else
            pos++;
    }
    // Sort the array
    qsort(arr, n, sizeof(int), cmpfunc);

    if (neg <= pos)
        fill1(arr, neg, pos);
    else {
        // Reverse the array in this condition
        reverse(arr, n);
        fill2(arr, neg, pos);
    }
    printf("Rearranged array is  ");
    print(arr, n);
}

// This code is contributed by Sania Kumari Gupta

Java

// Below is the implementation of the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class Main {

    // function which works in the condition when number of
    // negative numbers are lesser or equal than positive
    // numbers
    static void fill1(int a[], int neg, int pos)
    {
        if (neg % 2 == 1) {
            for (int i = 1; i < neg; i += 2) {
                int c = a[i];
                int d = a[i + neg];
                int temp = c;
                a[i] = d;
                a[i + neg] = temp;
            }
        }
        else {
            for (int i = 1; i <= neg; i += 2) {
                int c = a[i];
                int d = a[i + neg - 1];
                int temp = c;
                a[i] = d;
                a[i + neg - 1] = temp;
            }
        }
    }

    // Function which works in the condition when number of
    // negative numbers are greater than positive numbers
    static void fill2(int a[], int neg, int pos)
    {
        if (pos % 2 == 1) {
            for (int i = 1; i < pos; i += 2) {
                int c = a[i];
                int d = a[i + pos];
                int temp = c;
                a[i] = d;
                a[i + pos] = temp;
            }
        }
        else {
            for (int i = 1; i <= pos; i += 2) {
                int c = a[i];
                int d = a[i + pos - 1];
                int temp = c;
                a[i] = d;
                a[i + pos - 1] = temp;
            }
        }
    }

    // Reverse the array
    static void reverse(int a[], int n)
    {
        int i, k, t;
        for (i = 0; i < n / 2; i++) {
            t = a[i];
            a[i] = a[n - i - 1];
            a[n - i - 1] = t;
        }
    }

    // Print the array
    static void print(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)
        throws java.lang.Exception
    {
        // Given array
        int[] arr = { 2, 3, -4, -1, 6, -9 };
        int n = arr.length;
        System.out.println("Given array is  ");
        print(arr, n);
        int neg = 0, pos = 0;
        for (int i = 0; i < n; i++) {
            if (arr[i] < 0)
                neg++;
            else
                pos++;
        }
        // Sort the array
        Arrays.sort(arr);

        if (neg <= pos) {
            fill1(arr, neg, pos);
        }
        else {
            // reverse the array in this condition
            reverse(arr, n);
            fill2(arr, neg, pos);
        }
        System.out.println("Rearranged array is  ");
        print(arr, n);
    }
}

Python3

# Python3 program for the above approach

# Function which works in the condition
# when number of negative numbers are
# lesser or equal than positive numbers
def fill1(a, neg, pos) :

    if (neg % 2 == 1)  :
        for i in range(1, neg, 2):
            c = a[i]
            d = a[i + neg]
            temp = c
            a[i] = d
            a[i + neg] = temp


    else   :
         for i in range(1, neg+1, 2):
            c = a[i]
            d = a[i + neg - 1]
            temp = c
            a[i] = d
            a[i + neg - 1] = temp

# Function which works in the condition
# when number of negative numbers are
# greater than positive numbers
def fill2(a, neg, pos):
    if (pos % 2 == 1) :
         for i in range(1, pos, 2):
            c = a[i]
            d = a[i + pos]
            temp = c
            a[i] = d
            a[i + pos] = temp


    else  :
        for i in range(1, pos+1, 2):
            c = a[i]
            d = a[i + pos - 1]
            temp = c
            a[i] = d
            a[i + pos - 1] = temp

# Reverse the array
def reverse(a, n) :

    for i in range(n / 2):
        t = a[i]
        a[i] = a[n - i - 1]
        a[n - i - 1] = t

# Print the array
def printt(a, n):
    for i in range(n):
        print(a[i], end = " ")

    print()

# Driver code
if __name__ == "__main__":

    arr = [ 2, 3, -4, -1, 6, -9 ]
    n = len(arr)
    print("Given array is ")
    printt(arr, n)

    neg = 0
    pos = 0
    for i in range(0, n):
        if (arr[i] < 0):
            neg += 1
        else:
            pos += 1

    # Sort the array
    arr.sort()

    if (neg <= pos) :
        fill1(arr, neg, pos)

    else  :

        # Reverse the array in this condition
        reverse(arr, n)
        fill2(arr, neg, pos)

    print("Rearranged array is  ")
    printt(arr, n)

# This code is contributed by sanjoy_62.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;

public class GFG {

    // function which works in the condition when number of
    // negative numbers are lesser or equal than positive
    // numbers
    static void fill1(int[] a, int neg, int pos)
    {
        if (neg % 2 == 1) {
            for (int i = 1; i < neg; i += 2) {
                int c = a[i];
                int d = a[i + neg];
                int temp = c;
                a[i] = d;
                a[i + neg] = temp;
            }
        }
        else {
            for (int i = 1; i <= neg; i += 2) {
                int c = a[i];
                int d = a[i + neg - 1];
                int temp = c;
                a[i] = d;
                a[i + neg - 1] = temp;
            }
        }
    }

    // Function which works in the condition when number of
    // negative numbers are greater than positive numbers
    static void fill2(int[] a, int neg, int pos)
    {
        if (pos % 2 == 1) {
            for (int i = 1; i < pos; i += 2) {
                int c = a[i];
                int d = a[i + pos];
                int temp = c;
                a[i] = d;
                a[i + pos] = temp;
            }
        }
        else {
            for (int i = 1; i <= pos; i += 2) {
                int c = a[i];
                int d = a[i + pos - 1];
                int temp = c;
                a[i] = d;
                a[i + pos - 1] = temp;
            }
        }
    }

    // Reverse the array
    static void reverse(int[] a, int n)
    {
        int i, k, t;
        for (i = 0; i < n / 2; i++) {
            t = a[i];
            a[i] = a[n - i - 1];
            a[n - i - 1] = t;
        }
    }

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

// Driver Code
public static void Main (string[] args) {

     // Given array
        int[] arr = { 2, 3, -4, -1, 6, -9 };
        int n = arr.Length;
        Console.WriteLine("Given array is  ");
        print(arr, n);
        int neg = 0, pos = 0;
        for (int i = 0; i < n; i++) {
            if (arr[i] < 0)
                neg++;
            else
                pos++;
        }

        // Sort the array
        Array.Sort(arr);

        if (neg <= pos) {
            fill1(arr, neg, pos);
        }
        else {
            // reverse the array in this condition
            reverse(arr, n);
            fill2(arr, neg, pos);
        }
        Console.WriteLine("Rearranged array is  ");
        print(arr, n);
}
}

// This code is contributed by splevel62.

Javascript

<script>

// Below is the implementation of the above approach

// Function which works in the condition
// when number of negative numbers are
// lesser or equal than positive numbers
function fill1(a, neg, pos)
{
    if (neg % 2 == 1)
    {
        for (let i = 1; i < neg; i += 2)
        {
            let c = a[i];
            let d = a[i + neg];
            let temp = c;
            a[i] = d;
            a[i + neg] = temp;
        }
    }
    else
    {
        for(let i = 1; i <= neg; i += 2)
        {
            let c = a[i];
            let d = a[i + neg - 1];
            let temp = c;
            a[i] = d;
            a[i + neg - 1] = temp;
        }
    }
}

// Function which works in the condition
// when number of negative numbers are
// greater than positive numbers
function fill2(a, neg, pos)
{
    if (pos % 2 == 1)
    {
        for (let i = 1; i < pos; i += 2)
        {
            let c = a[i];
            let d = a[i + pos];
            let temp = c;
            a[i] = d;
            a[i + pos] = temp;
        }
    }
    else
    {
        for(let i = 1; i <= pos; i += 2)
        {
            let c = a[i];
            let d = a[i + pos - 1];
            let temp = c;
            a[i] = d;
            a[i + pos - 1] = temp;
        }
    }
}

// Reverse the array
function reverse(a, n)
{
    let i, k, t;
    for(i = 0; i < parseInt(n / 2); i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}

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

    document.write('<br>');
}

// Driver code
var arr = [ 2, 3, -4, -1, 6, -9 ];
let n = arr.length;
document.write("Given array is ");
print(arr, n);
let neg = 0, pos = 0;

for(let i = 0; i < n; i++)
{
    if (arr[i] < 0)
        neg++;
    else
        pos++;
}

// Sort the array
arr.sort(function (a, b){return a - b;});

if (neg <= pos)
{
    fill1(arr, neg, pos);
}
else
{

    // Reverse the array in this condition
    reverse(arr, n);
    fill2(arr, neg, pos);
}
document.write("Rearranged array is  ");
print(arr, n);

// This code is contributed by Potta Lokesh

</script>

Output

Given array is 2 3 -4 -1 6 -9 
Rearranged array is  -9 3 -1 2 -4 6 

Time Complexity: O(N*logN)
Space Complexity: O(1)

Efficient Approach: We have already discussed a O(n²) solution that maintains the order of appearance in the array here. If we are allowed to change order of appearance, we can solve this problem in O(n) time and O(1) space.
The idea is to process the array and shift all negative values to the end in O(n) time. After all negative values are shifted to the end, we can easily rearrange array in alternating positive & negative items. We basically swap next positive element at even position from next negative element in this step. 

Following is the implementation of above idea.

C++

// C++ program to rearrange
// array in alternating
// positive & negative items
// with O(1) extra space
#include <bits/stdc++.h>
using namespace std;

// Function to rearrange positive and negative
// integers in alternate fashion. The below
// solution doesn't maintain original order of
// elements
void rearrange(int arr[], int n)
{
    int i = 0, j = n-1;

    // shift all negative values to the end
    while (i < j) {
        while (i <= n - 1 and arr[i] > 0)
            i += 1;
        while (j >= 0 and arr[j] < 0)
            j -= 1;
        if (i < j )
            swap(arr[i], arr[j]);
    }

    // i has index of leftmost
    // negative element
    if (i == 0 || i == n)
        return;

    // start with first positive
    // element at index 0

    // Rearrange array in alternating
    // positive &
    // negative items
    int k = 0;
    while (k < n && i < n ) {
        // swap next positive
        // element at even position
        // from next negative element.
        swap(arr[k], arr[i]);
        i = i + 1;
        k = k + 2;
    }
}

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

// Driver code
int main()
{
    int arr[] = { 2, 3, -4, -1, 6, -9 };

    int n = sizeof(arr) / sizeof(arr[0]);

    cout << "Given array is \n";
    printArray(arr, n);

    rearrange(arr, n);

    cout << "Rearranged array is \n";
    printArray(arr, n);

    return 0;
}

Java

// Java program to rearrange
// array in alternating
// positive & negative
// items with O(1) extra space
class GFG {

    // Function to rearrange positive and negative
    // integers in alternate fashion. The below
    // solution doesn't maintain original order of
    // elements
    static void rearrange(int arr[], int n)
    {
        int i = 0, j = n - 1;

        // shift all negative values to the end
        while (i < j) {
            while (i <= n - 1 && arr[i] > 0)
                i += 1;
            while (j >= 0 && arr[j] < 0)
                j -= 1;
            if (i < j)
                swap(arr, i, j);
        }

        // i has index of leftmost negative element
        if (i == 0 || i == n)
            return;

        // start with first positive
        // element at index 0

        // Rearrange array in alternating positive &
        // negative items
        int k = 0;
        while (k < n && i < n) {
            // swap next positive element
            // at even position
            // from next negative element.
            swap(arr, k, i);
            i = i + 1;
            k = k + 2;
        }
    }

    // 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] + " ");
        System.out.println("");
    }

    static void swap(int arr[], int index1, int index2)
    {
        int c = arr[index1];
        arr[index1] = arr[index2];
        arr[index2] = c;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, -4, -1, 6, -9 };

        int n = arr.length;

        System.out.println("Given array is ");
        printArray(arr, n);

        rearrange(arr, n);

        System.out.println("Rearranged array is ");
        printArray(arr, n);
    }
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to rearrange array
# in alternating positive & negative
# items with O(1) extra space

# Function to rearrange positive and
# negative integers in alternate fashion.
# The below solution does not maintain
# original order of elements


def rearrange(arr, n):
    i = 0
    j = n - 1

    # shift all negative values
    # to the end
    while (i < j):

        while (i <= n - 1 and arr[i] > 0):
            i += 1
        while (j >= 0 and arr[j] < 0):
            j -= 1

        if (i < j):
            temp = arr[i]
            arr[i] = arr[j]
            arr[j] = temp

    # i has index of leftmost
    # negative element
    if (i == 0 or i == n):
        return 0

    # start with first positive element
    # at index 0

    # Rearrange array in alternating
    # positive & negative items
    k = 0
    while (k < n and i < n):

        # swap next positive element at even
        # position from next negative element.
        temp = arr[k]
        arr[k] = arr[i]
        arr[i] = temp
        i = i + 1
        k = k + 2

# Utility function to print an array


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


# Driver code
arr = [2, 3]

n = len(arr)

print("Given array is")
printArray(arr, n)

rearrange(arr, n)

print("Rearranged array is")
printArray(arr, n)

# This code is contributed
# Princi Singh

C#

// C# program to rearrange array
// in alternating positive & negative
// items with O(1) extra space
using System;

class GFG {

    // Function to rearrange positive and
    // negative integers in alternate fashion.
    // The below solution doesn't maintain
    // original order of elements
    static void rearrange(int[] arr, int n)
    {
        int i = 0, j = n - 1;

        // shift all negative values
        // to the end
        while (i < j) {
            while (i <= n - 1 && arr[i] > 0)
                i += 1;
            while (j >= 0 && arr[j] < 0)
                j -= 1;
            if (i < j)
                swap(arr, i, j);
        }

        // i has index of leftmost
        // negative element
        if (i == 0 || i == n)
            return;

        // start with first positive
        // element at index 0

        // Rearrange array in alternating
        // positive & negative items
        int k = 0;
        while (k < n && i < n) {
            // swap next positive element
            // at even position from next
            // negative element.
            swap(arr, k, i);
            i = i + 1;
            k = k + 2;
        }
    }

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

    static void swap(int[] arr, int index1, int index2)
    {
        int c = arr[index1];
        arr[index1] = arr[index2];
        arr[index2] = c;
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 2, 3, -4, -1, 6, -9 };

        int n = arr.Length;

        Console.WriteLine("Given array is ");
        printArray(arr, n);

        rearrange(arr, n);

        Console.WriteLine("Rearranged array is ");
        printArray(arr, n);
    }
}

// This code is contributed
// by 29AjayKumar

PHP

<?php
// PHP program to rearrange array
// in alternating positive & negative
// items with O(1) extra space

// Function to rearrange positive and
// negative integers in alternate fashion.
// The below solution doesn't maintain
// original order of elements
function rearrange(&$arr, $n)
{
    $i = 0;
    $j = $n - 1;

    // shift all negative values
    // to the end
    while ($i < $j)
    {

        while($i <= $n - 1 and $arr[$i] > 0)
            ++$i;
        while ($j >= 0 and $arr[$j] < 0)
            --$j;

        if ($i < $j)
        {
            $temp = $arr[$i];
            $arr[$i] = $arr[$j];
            $arr[$j] = $temp;
        }
    }

    // i has index of leftmost
    // negative element
    if ($i == 0 || $i == $n)
        return;

    // start with first positive element
    // at index 0

    // Rearrange array in alternating
    // positive & negative items
    $k = 0;
    while ($k < $n && $i < $n)
    {
        // swap next positive element at even
        // position from next negative element.
        $temp = $arr[$k];
        $arr[$k] = $arr[$i];
        $arr[$i] = $temp;
        $i = $i + 1;
        $k = $k + 2;
    }
}

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

// Driver code
$arr = array(2, 3, -4, -1, 6, -9);

$n = sizeof($arr);

echo "Given array is \n";
printArray($arr, $n);

rearrange($arr, $n);

echo "Rearranged array is \n";
printArray($arr, $n);

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

Javascript

<script>

// Javascript program to rearrange
// array in alternating
// positive & negative
// items with O(1) extra space

// Function to rearrange positive and negative
// integers in alternate fashion. The below
// solution doesn't maintain original order of
// elements
function rearrange(arr,n)
{
    let i = 0, j = n - 1;

    // Shift all negative values to the end
    while (i < j)
    {
        while(i <= n - 1 && arr[i] > 0)
            i += 1;
        while (j >= 0 && arr[j] < 0)
            j -= 1;

        if (i < j)
            swap(arr, i,j);
    }

    // i has index of leftmost negative element
    if (i == 0 || i == n)
        return;

    // Start with first positive
    // element at index 0

    // Rearrange array in alternating
    // positive & negative items
    let k = 0;
    while (k < n && i < n)
    {

        // Swap next positive element
        // at even position
        // from next negative element.
        swap(arr, k, i);
        i = i + 1;
        k = k + 2;
    }
}

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

    document.write("<br>");
}

function swap(arr, index1, index2)
{
    let c = arr[index1];
    arr[index1] = arr[index2];
    arr[index2] = c;
}

// Driver code
let arr = [ 2, 3, -4, -1, 6, -9 ];
let n = arr.length;

document.write("Given array is <br>");

printArray(arr, n);
rearrange(arr, n);

document.write("Rearranged array is <br>");
printArray(arr, n);

// This code is contributed by rag2127

</script>

Output

Given array is 
2 3 -4 -1 6 -9 
Rearranged array is 
-1 3 -4 2 -9 6 

Time Complexity : O(N)
Space Complexity : O(1)

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