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Rearrange array in alternating positive & negative items with O(1) extra space | Set 2
  • Difficulty Level : Medium
  • Last Updated : 12 Jun, 2021

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:

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);
    }
}
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 = -1, j = n;
 
    // 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)

This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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