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

Given an array having positive and negative numbers, our task is to arrange them in an alternate fashion such that every positive number is followed by a negative number and vice-versa maintaining the order of appearance. The number of positive and negative numbers need not to be equal. If there are more positive numbers then they have to appear at the end of the array , same condition for negative numbers also .

Examples:

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

Input:  arr[] = {-5, -2, 5, 2, 4, 7, 1, 8, 0, -8}
Output: arr[] = {-5, 5, -2, 2, -8, 4, 7, 1, 8, 0}

This question has been asked in many places (See this and this)

NaÃ¯ve Approach: To solve the problem follow the below idea:

• The problem can be easily solved if O(n) extra space is allowed.
• We can store the positive values and negative values in two separate data structures.
• We will start filling the original array with alternating negative and positive values in the same order
in which it appears in the original array.

It becomes interesting due to the limitations that O(1) extra space and order of appearances.

Optimal Approach: To solve the problem in O(1) Auxiliary space follow the below idea:

The idea is to process the array from left to right. While processing, find the first out-of-place element in the remaining unprocessed array. An element is out of place if it is negative and at odd index (0-based index), or if it is positive and at even index (0-based index). Once we find an out-of-place element, we find the first element after it with an opposite sign. We right rotate the subarray between these two elements (including these two).

Illustration:

Let the array be arr[] = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 }

First iteration:

• { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 } , -2 appears on odd index position and is out of place.
We will look for the first element that appears with an opposite sign
• { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 } => perform rotation of subarray between this two elements
• { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 }

Second iteration:

• { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 } , 4 is out of place.
• { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 } => -8 is of different sign
• { -5, 5, -2, 2, -8, 4, 7, 1, 8, 0 } => after performing right rotation between 4 to -8

resultant array arr[] = { -5, 5, -2, 2, -8, 4, 7, 1, 8, 0 }

Follow the steps to solve the problem:

• Maintain a variable to mark if the element is in its correct position or not. Let the variable be outofplace. Initially, it is -1.
• We will iterate over the array.
• If outofplace is -1, we will check if the current index is out of place.
• If the current index is out of place we will update the outofplace with the index value.
• Else we will keep the value as it is.
• If the outofplace is not -1, we will search for the next index which has a different sign than that of the value that is present in outofplace position.
• Now we will pass this two positions to right rotate our array.
• Update the value of outofplace by 2 unit (becuase previously valid elements will now become the out-of-place elements).

Following is the implementation of the above idea.

## C++

 `/*  C++ program to rearrange``   ``positive and negative integers``   ``in alternate fashion while keeping``   ``the order of positive and negative numbers. */``#include ``using` `namespace` `std;` `// Utility function to right rotate all elements between``// [outofplace, cur]``void` `rightrotate(``int` `arr[], ``int` `n, ``int` `outofplace, ``int` `cur)``{``    ``char` `tmp = arr[cur];``    ``for` `(``int` `i = cur; i > outofplace; i--)``        ``arr[i] = arr[i - 1];``    ``arr[outofplace] = tmp;``}` `void` `rearrange(``int` `arr[], ``int` `n)``{``    ``int` `outofplace = -1;` `    ``for` `(``int` `index = 0; index < n; index++) {``        ``if` `(outofplace >= 0) {``            ``// find the item which must be moved into the``            ``// out-of-place entry if out-of-place entry is``            ``// positive and current entry is negative OR if``            ``// out-of-place entry is negative and current``            ``// entry is positive then right rotate``            ``//``            ``// [...-3, -4, -5, 6...] -->   [...6, -3, -4,``            ``// -5...]``            ``//      ^                          ^``            ``//      |                          |``            ``//     outofplace      -->      outofplace``            ``//``            ``if` `(((arr[index] >= 0) && (arr[outofplace] < 0))``                ``|| ((arr[index] < 0)``                    ``&& (arr[outofplace] >= 0))) {``                ``rightrotate(arr, n, outofplace, index);` `                ``// the new out-of-place entry is now 2 steps``                ``// ahead``                ``if` `(index - outofplace >= 2)``                    ``outofplace = outofplace + 2;``                ``else``                    ``outofplace = -1;``            ``}``        ``}` `        ``// if no entry has been flagged out-of-place``        ``if` `(outofplace == -1) {``            ``// check if current entry is out-of-place``            ``if` `(((arr[index] >= 0) && (!(index & 0x01)))``                ``|| ((arr[index] < 0) && (index & 0x01))) {``                ``outofplace = index;``            ``}``        ``}``    ``}``}` `// A utility function to print an array 'arr[]' of size 'n'``void` `printArray(``int` `arr[], ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++)``        ``cout << arr[i] << ``" "``;``    ``cout << endl;``}` `// Driver code``int` `main()``{` `    ``int` `arr[] = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << ``"Given array is \n"``;``    ``printArray(arr, n);``    ` `  ``// Function Call``    ``rearrange(arr, n);` `    ``cout << ``"Rearranged array is \n"``;``    ``printArray(arr, n);` `    ``return` `0;``}`

## Java

 `import` `java.io.*;``class` `RearrangeArray {``    ``// Utility function to right rotate all elements``    ``// between [outofplace, cur]``    ``void` `rightrotate(``int` `arr[], ``int` `n, ``int` `outofplace,``                     ``int` `cur)``    ``{``        ``int` `tmp = arr[cur];``        ``for` `(``int` `i = cur; i > outofplace; i--)``            ``arr[i] = arr[i - ``1``];``        ``arr[outofplace] = tmp;``    ``}` `    ``void` `rearrange(``int` `arr[], ``int` `n)``    ``{``        ``int` `outofplace = -``1``;` `        ``for` `(``int` `index = ``0``; index < n; index++) {``            ``if` `(outofplace >= ``0``) {``                ``// find the item which must be moved into``                ``// the out-of-place entry if out-of-place``                ``// entry is positive and current entry is``                ``// negative OR if out-of-place entry is``                ``// negative and current entry is negative``                ``// then right rotate``                ``//``                ``// [...-3, -4, -5, 6...] -->   [...6, -3,``                ``// -4, -5...]``                ``//      ^                          ^``                ``//      |                          |``                ``//     outofplace      -->      outofplace``                ``//``                ``if` `(((arr[index] >= ``0``)``                     ``&& (arr[outofplace] < ``0``))``                    ``|| ((arr[index] < ``0``)``                        ``&& (arr[outofplace] >= ``0``))) {``                    ``rightrotate(arr, n, outofplace, index);` `                    ``// the new out-of-place entry is now 2``                    ``// steps ahead``                    ``if` `(index - outofplace >= ``2``)``                        ``outofplace = outofplace + ``2``;``                    ``else``                        ``outofplace = -``1``;``                ``}``            ``}` `            ``// if no entry has been flagged out-of-place``            ``if` `(outofplace == -``1``) {``                ``// check if current entry is out-of-place``                ``if` `(((arr[index] >= ``0``)``                     ``&& ((index & ``0x01``) == ``0``))``                    ``|| ((arr[index] < ``0``)``                        ``&& (index & ``0x01``) == ``1``))``                    ``outofplace = index;``            ``}``        ``}``    ``}` `    ``// A utility function to print``    ``// an array 'arr[]' of size 'n'``    ``void` `printArray(``int` `arr[], ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);``        ``System.out.println(``""``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``RearrangeArray rearrange = ``new` `RearrangeArray();``        ``/* int arr[n] = {-5, 3, 4, 5, -6,``                         ``-2, 8, 9, -1, -4};``        ``int arr[] = {-5, -3, -4, -5, -6,``                     ``2 , 8, 9, 1 , 4};``        ``int arr[] = {5, 3, 4, 2, 1,``                     ``-2 , -8, -9, -1 , -4};``        ``int arr[] = {-5, 3, -4, -7,``                     ``-1, -2 , -8, -9, 1 , -4};*/``        ``int` `arr[] = { -``5``, -``2``, ``5``, ``2``, ``4``, ``7``, ``1``, ``8``, ``0``, -``8` `};``        ``int` `n = arr.length;` `        ``System.out.println(``"Given array is "``);``        ``rearrange.printArray(arr, n);` `        ``rearrange.rearrange(arr, n);` `        ``System.out.println(``"RearrangeD array is "``);``        ``rearrange.printArray(arr, n);``    ``}``}` `// This code has been contributed by Mayank Jaiswal`

## Python

 `# Python3 program to rearrange``# positive and negative integers``# in alternate fashion and``# maintaining the order of positive``# and negative numbers` `# rotates the array to right by once``# from index 'outOfPlace to cur'`  `def` `rightRotate(arr, n, outOfPlace, cur):``    ``temp ``=` `arr[cur]``    ``for` `i ``in` `range``(cur, outOfPlace, ``-``1``):``        ``arr[i] ``=` `arr[i ``-` `1``]``    ``arr[outOfPlace] ``=` `temp``    ``return` `arr`  `def` `rearrange(arr, n):``    ``outOfPlace ``=` `-``1``    ``for` `index ``in` `range``(n):``        ``if``(outOfPlace >``=` `0``):` `            ``# if element at outOfPlace place in``            ``# negative and if element at index``            ``# is positive we can rotate the``            ``# array to right or if element``            ``# at outOfPlace place in positive and``            ``# if element at index is negative we``            ``# can rotate the array to right``            ``if``((arr[index] >``=` `0` `and` `arr[outOfPlace] < ``0``) ``or``               ``(arr[index] < ``0` `and` `arr[outOfPlace] >``=` `0``)):``                ``arr ``=` `rightRotate(arr, n, outOfPlace, index)``                ``if``(index``-``outOfPlace > ``2``):``                    ``outOfPlace ``+``=` `2``                ``else``:``                    ``outOfPlace ``=` `-` `1` `        ``if``(outOfPlace ``=``=` `-``1``):` `            ``# conditions for A[index] to``            ``# be in out of place``            ``if``((arr[index] >``=` `0` `and` `index ``%` `2` `=``=` `0``) ``or``               ``(arr[index] < ``0` `and` `index ``%` `2` `=``=` `1``)):``                ``outOfPlace ``=` `index``    ``return` `arr`  `# Driver Code``arr ``=` `[``-``5``, ``-``2``, ``5``, ``2``, ``4``,``       ``7``, ``1``, ``8``, ``0``, ``-``8``]` `print``(``"Given Array is:"``)``print``(arr)` `print``(``"\nRearranged array is:"``)``print``(rearrange(arr, ``len``(arr)))` `# This code is contributed``# by Charan Sai`

## C#

 `// Rearrange array in alternating positive``// & negative items with O(1) extra space``using` `System;` `class` `GFG {``    ``// Utility function to right rotate``    ``// all elements between [outofplace, cur]``    ``static` `void` `rightrotate(``int``[] arr, ``int` `n,``                            ``int` `outofplace, ``int` `cur)``    ``{``        ``int` `tmp = arr[cur];``        ``for` `(``int` `i = cur; i > outofplace; i--)``            ``arr[i] = arr[i - 1];``        ``arr[outofplace] = tmp;``    ``}` `    ``static` `void` `rearrange(``int``[] arr, ``int` `n)``    ``{``        ``int` `outofplace = -1;` `        ``for` `(``int` `index = 0; index < n; index++) {``            ``if` `(outofplace >= 0) {``                ``// find the item which must be moved``                ``// into the out-of-place entry if out-of-``                ``// place entry is positive and current``                ``// entry is negative OR if out-of-place``                ``// entry is negative and current entry``                ``// is negative then right rotate``                ``// [...-3, -4, -5, 6...] --> [...6, -3, -4,``                ``// -5...]``                ``//     ^                         ^``                ``//     |                         |``                ``//     outofplace     -->     outofplace``                ``//``                ``if` `(((arr[index] >= 0)``                     ``&& (arr[outofplace] < 0))``                    ``|| ((arr[index] < 0)``                        ``&& (arr[outofplace] >= 0))) {``                    ``rightrotate(arr, n, outofplace, index);` `                    ``// the new out-of-place entry``                    ``// is now 2 steps ahead``                    ``if` `(index - outofplace > 2)``                        ``outofplace = outofplace + 2;``                    ``else``                        ``outofplace = -1;``                ``}``            ``}` `            ``// if no entry has been flagged out-of-place``            ``if` `(outofplace == -1) {``                ``// check if current entry is out-of-place``                ``if` `(((arr[index] >= 0)``                     ``&& ((index & 0x01) == 0))``                    ``|| ((arr[index] < 0)``                        ``&& (index & 0x01) == 1))``                    ``outofplace = index;``            ``}``        ``}``    ``}` `    ``// A utility function to print an``    ``// array 'arr[]' of size 'n'``    ``static` `void` `printArray(``int``[] arr, ``int` `n)``    ``{``        ``for` `(``int` `i = 0; i < n; i++)``            ``Console.Write(arr[i] + ``" "``);``        ``Console.WriteLine(``""``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 };``        ``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 Sam007`

## Javascript

 ``

## PHP

 ` ``\$outofplace``; ``\$i``--)``        ``\$arr``[``\$i``] = ``\$arr``[``\$i` `- 1];``    ``\$arr``[``\$outofplace``] = ``\$tmp``;``}` `function` `rearrange(&``\$arr``, ``\$n``)``{``    ``\$outofplace` `= -1;` `    ``for` `(``\$index` `= 0; ``\$index` `< ``\$n``; ``\$index` `++)``    ``{``        ``if` `(``\$outofplace` `>= 0)``        ``{``            ``// find the item which must be moved ``            ``// into the out-of-place entry if ``            ``// out-of-place entry is positive and``            ``// current entry is negative OR if ``            ``// out-of-place entry is negative``            ``// and current entry is negative then ``            ``// right rotate``            ``// [...-3, -4, -5, 6...] --> [...6, -3, -4, -5...]``            ``//     ^                         ^``            ``//     |                         |``            ``//     outofplace     -->     outofplace``            ``//``            ``if` `(((``\$arr``[``\$index``] >= 0) && (``\$arr``[``\$outofplace``] < 0)) || ``                ``((``\$arr``[``\$index``] < 0) && (``\$arr``[``\$outofplace``] >= 0)))``            ``{``                ``rightrotate(``\$arr``, ``\$n``, ``\$outofplace``, ``\$index``);` `                ``// the new out-of-place entry is ``                ``// now 2 steps ahead``                ``if` `(``\$index` `- ``\$outofplace` `> 2)``                    ``\$outofplace` `= ``\$outofplace` `+ 2;``                ``else``                    ``\$outofplace` `= -1;``            ``}``        ``}` `        ``// if no entry has been flagged out-of-place``        ``if` `(``\$outofplace` `== -1)``        ``{``            ``// check if current entry is out-of-place``            ``if` `(((``\$arr``[``\$index``] >= 0) && (!(``\$index` `& 0x01)))``                ``|| ((``\$arr``[``\$index``] < 0) && (``\$index` `& 0x01)))``            ``{``                ``\$outofplace` `= ``\$index``;``            ``}``        ``}``    ``}``}` `// A utility function to print an``// array 'arr[]' of size 'n'``function` `printArray(&``\$arr``, ``\$n``)``{``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``    ``echo` `\$arr``[``\$i``].``" "``;``    ``echo` `"\n"``;``}` `// Driver Code` `// arr = array(-5, 3, 4, 5, -6, -2, 8, 9, -1, -4);``// arr = array(-5, -3, -4, -5, -6, 2 , 8, 9, 1 , 4);``// arr = array(5, 3, 4, 2, 1, -2 , -8, -9, -1 , -4);``// arr = array(-5, 3, -4, -7, -1, -2 , -8, -9, 1 , -4);``\$arr` `= ``array``(-5, -2, 5, 2, 4, 7, 1, 8, 0, -8);``\$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``?>`

Output
```Given array is
-5 -2 5 2 4 7 1 8 0 -8
Rearranged array is
-5 5 -2 2 -8 4 7 1 8 0

```

Time Complexity: O(N^2), as we are using a loop to traverse N times and calling function to right rotate each time which will cost O (N).
Auxiliary Space: O(1).

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