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

• Difficulty Level : Medium
• Last Updated : 30 Jun, 2022

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 ` `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`

## 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

 ``

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 ` `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

 ` 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

 ``

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)