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

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.

Example:

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)

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 C++ implementation of above idea.

/*  C++ program to rearrange positive and negative integers in alternate     fashion while keeping the order of positive and negative numbers. */ #include <iostream> #include <assert.h> 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 program to test abive function int main() {     //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 = 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; }

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