Rearrange positive and negative numbers in O(n) time and O(1) extra space | GeeksforGeeks

Rearrange positive and negative numbers in O(n) time and O(1) extra space | GeeksforGeeks
An array contains both positive and negative numbers in random order. Rearrange the array elements so that positive and negative numbers are placed alternatively. 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.

For example, if the input array is [-1, 2, -3, 4, 5, 6, -7, 8, 9], then the output should be [9, -7, 8, -3, 5, -1, 2, 4, 6]

X1. The solution is to first separate positive and negative numbers using partition process of QuickSort. In the partition process, consider 0 as value of pivot element so that all negative numbers are placed before positive numbers. Once negative and positive numbers are separated, we start from the first negative number and first positive number, and swap every alternate negative number with next positive number.

    static void rearrange(int arr[], int n)

    {

        // The following few lines are similar to partition

        // process of QuickSort.  The idea is to consider 0

        // as pivot and divide the array around it.

        int i = -1, temp = 0;

        for (int j = 0; j < n; j++)

        {

            if (arr[j] < 0)

            {

                i++;

                temp = arr[i];

                arr[i] = arr[j];

                arr[j] = temp;

            }

        }

        // Now all positive numbers are at end and negative numbers at

        // the beginning of array. Initialize indexes for starting point

        // of positive and negative numbers to be swapped

        int pos = i+1, neg = 0;

        // Increment the negative index by 2 and positive index by 1, i.e.,

        // swap every alternate negative number with next positive number

        while (pos < n && neg < pos && arr[neg] < 0)

        {

            temp = arr[neg];

            arr[neg] = arr[pos];

            arr[pos] = temp;

            pos++;

            neg += 2;

        }

    }

http://algorithms.tutorialhorizon.com/rearrange-positive-and-negative-elements-at-alternate-positions-in-an-array-in-o1-extra-space/
//no need to do this

        int high = 0;

        while (arrA[high] < 0)

            high++;

http://www.ritambhara.in/rearrange-array-into-alternate-positive-and-negative-numbers/
DIfferent partition code: less swap.

void arrange(int *arr, int n)

{

    // Partition logic of QuickSort.

    // Bring all -ve number at start.

    int low = 0, high = n-1;

    while(low<high)

    {

        while(arr[low] < 0) low++;

        while(arr[high] > 0) high--;

        if(low<high)

            swap(arr, low, high);

    }

    // At this point low will be pointint to the first positive number in the array.

    // If there are no jpositive numbers in array, then low will be out of bond

    if(low>=n)

       return;

    // First indexes of positive and negative numbers will be as below

    int positive = low, negative = 0;

    // swap every alternate negative number with next positive number

    while (positive < n && negative < positive && arr[negative] < 0)

    {

        swap(arr, positive, negative);

        positive++;

        negative += 2;

    }

}

http://www.geeksforgeeks.org/rearrange-array-alternating-positive-negative-items-o1-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} 

 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).

// 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;

            }

        }

    }

}

TODO: https://ideone.com/VgIvYZ
http://comproguide.blogspot.com/2014/05/separating-positive-and-negative.html
Performance is not best, but concise.

void seggregate(int arr[], int len)


{


if( len <= 1 )


return;


 


int nIndex  = 0;


int i;


for( i = 0; i < len; i++ )


{


if( arr[i] <= 0 )


{


swap(arr[i],arr[nIndex]);


nIndex++;


}


}


}

http://www.cnblogs.com/yuzhangcmu/p/4175620.html
http://blog.csdn.net/nicaishibiantai/article/details/43448341
Method-1: Using shift and Swap:
The good thing about above code is that it maintains the order in which element are present in the original array. But it is an O(n2) time algorithm.
http://www.ritambhara.in/rearrange-array-into-alternate-positive-and-negative-numbers/

Also check: Rearrange Positive and Negative Numbers of Array On Each Side in O(nlogn) | Algorithms
Objec­tive: Rearrange Pos­i­tive and Neg­a­tive Num­bers of an Array so that one side you have pos­i­tive num­bers and other side with neg­a­tive Inte­gers with­out chang­ing their respec­tive order.
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