Positive elements at even and negative at odd positions

3

You have been given an array and you have to make a program to convert that array such that positive elements occur at even numbered places in the array and negative elements occur at odd numbered places in the array. We have to do it in place and the relative order of the elements must be maintained.
There can be unequal number of positive and negative values and the extra values have to left as it is.

Examples:

Input : arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}
Output : arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}

Input : arr[] = {-1, 3, -5, 6, 3, 6, -7, -4, -9, 10}
Output : arr[] = {3, -1, 6, -5, 3, -7, 6, -4, 10, -9}

The idea is to use Hoare’s partition process of Quick Sort.
We take two pointers positive and negative. We set the positive pointer at start of the array and the negative pointer at 1st position of the array.
We move positive pointer two steps forward till it finds a negative element. Similarly we move negative pointer forward by two places till it finds a positive value at its position.
If the positive and negative pointers are in the array then we will swap the values at these indexes otherwise we will stop executing the process.

// C++ program to rearrange positive and negative
// numbers
#include <bits/stdc++.h>
using namespace std;

void rearrange(int a[], int size)
{
    int positive = 0, negative = 1;

    while (true) {

        /* Move forward the positive pointer till 
         negative number number not encountered */
        while (positive < size && a[positive] >= 0)
            positive += 2;

        /* Move forward the negative pointer till 
         positive number number not encountered   */
        while (negative < size && a[negative] <= 0)
            negative += 2;

        // Swap array elements to fix their position.
        if (positive < size && negative < size)
            swap(&a[positive], &a[negative]);

        /* Break from the while loop when any index 
            exceeds the size of the array */
        else
            break;
    }
}

// Driver code
int main()
{
    int arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
    int n = (sizeof(arr) / sizeof(arr[0]));

    rearrange(arr, n); 
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";

    return 0;
}

Output:

1 -3 5 -3 6 6 7 -4 9 10 

Lets explain the working of the code on the first example
arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}
We declare two variables positive and negative positive points to zeroth position and negative points to first position
positive = 0 negative = 1
In the first iteration positive will move 4 places to fifth position as a[4] is less than zero and positive = 4.
Negative will move 2 places and will point to fourth position as a[3]>0
we will swap positive and negative position values as they are less than size of array.
After first iteration the array becomes arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}

Now positive points at fourth position and negative points at third position
In second iteration the positive value will move 6 places and its value will
more than the size of the array.
The negative pointer will move two steps forward and it will point to 5th position
As the positive pointer value becomes greater than the array size we will not perform any swap operation and break out of the while loop.
The final output will be
arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}

This article is contributed by Ashish Madaan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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