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

• Difficulty Level : Medium
• Last Updated : 29 May, 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 maintaining the order of appearance.
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 in the end of the array.

Examples :

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```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 at many places (See this and this)

Naive Approach :
The above problem can be easily solved if O(n) extra space is allowed. It becomes interesting due to the limitations that O(1) extra space and order of appearances.
The idea is to process 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, or it is positive and at even index. Once we find an out of place element, we find the first element after it with opposite sign. We right rotate the subarray between these two elements (including these two).

Following is the implementation of above idea.

## C++

 `/*  C++ program to rearrange``   ``positive and negative integers``   ``in alternate fashion while keeping``   ``the order of positive and negative numbers. */``#include ``#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 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'``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);` `    ``cout << ``"Given array is \n"``;``    ``printArray(arr, n);` `    ``rearrange(arr, n);` `    ``cout << ``"Rearranged array is \n"``;``    ``printArray(arr, n);` `    ``return` `0;``}`

## Java

 `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`

## 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)

Space Complexity : O(1)

Efficient Approach :

We first sort the array in non-increasing order.Then we will count the number of poitive and negative integers. Then we will swap the one negative and one positive

number till we reach our condition. This will rearrange the array elements because we are sorting the array and accessing the element from left to right according to our need.

## 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;``            ``}``        ``}``    ``}``    ``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;``        ``}``    ``}``    ``static` `void` `print(``int` `a[], ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(a[i] + ``" "``);``        ``System.out.println();``    ``}``    ``public` `static` `void` `main(String[] args)``        ``throws` `java.lang.Exception``    ``{``        ``// Given array``        ``int``[] arr = { -``5``, -``2``, ``5``, ``2``, ``4``, ``7``, ``1``, ``8``, ``0``, -``8` `};``        ``int` `n = arr.length;` `        ``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);``        ``}``        ``print(arr, n);``    ``}``}`

## Python3

 `# Below is the implementation of the above approach` `def` `rearrange(arr, n):``    ``# sort the array``    ``arr.sort()` `    ``# initialize two pointers``    ``# one pointing to the negative number``    ``# one pointing to the positive number``    ``i, j ``=` `1``, ``1``    ``while` `j < n:``        ``if` `arr[j] > ``0``:``            ``break``        ``j ``+``=` `1` `    ``# swap the numbers until the given condition gets satisfied``    ``while` `(arr[i] < ``0``) ``and` `(j < n):``        ``# swaping``        ``arr[i], arr[j] ``=` `arr[j], arr[i]  ``        ` `        ``# increment i by 2``        ``# because a negative number is followed by a positive number``        ``i ``+``=` `2`     `        ``j ``+``=` `1``    ` `    ``return``(arr)``  `   `# Driver Code``# Given array``arr ``=` `[``-``5``, ``-``2``, ``5``, ``2``, ``4``, ``7``, ``1``, ``8``, ``0``, ``-``8``]` `ans ``=` `rearrange(arr, ``len``(arr))``for` `num ``in` `ans:``    ``print``(num, end ``=` `" "``)` `# This code is contributed by Tharun Reddy`
Output
`-8 1 -2 0 -5 2 4 5 7 8 `

Time Complexity: O(N*logN)

Space Complexity: O(1)

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