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

• Difficulty Level : Medium
• Last Updated : 29 Apr, 2021

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]
Note: The partition process changes relative order of elements. I.e. the order of the appearance of elements is not maintained with this approach. See this for maintaining order of appearance of elements in this problem.
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. 
 

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution. 
 

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// A C++ program to put positive

// numbers at even indexes (0, 2, 4,..)

// and negative numbers at odd

// indexes (1, 3, 5, ..)

#include <iostream>

using namespace std;

class GFG

{

    public:

    void rearrange(int [],int);

    void swap(int *,int *);

    void printArray(int [],int);

};

// The main function that rearranges

// elements of given array. It puts

// positive elements at even indexes

// (0, 2, ..) and negative numbers

// at odd indexes (1, 3, ..).

void GFG :: 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;

    for (int j = 0; j < n; j++)

    {

        if (arr[j] < 0)

        {

            i++;

            swap(&arr[i], &arr[j]);

        }

    }

    // 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)

    {

        swap(&arr[neg], &arr[pos]);

        pos++;

        neg += 2;

    }

}

// A utility function

// to swap two elements

void GFG :: swap(int *a, int *b)

{

    int temp = *a;

    *a = *b;

    *b = temp;

}

// A utility function to print an array

void GFG :: printArray(int arr[], int n)

{

    for (int i = 0; i < n; i++)

        cout << arr[i] << " ";

}

// Driver Code

int main()

{

    int arr[] = {-1, 2, -3, 4,

                  5, 6, -7, 8, 9};

    int n = sizeof(arr) / sizeof(arr[0]);

    GFG test;

    test.rearrange(arr, n);

    test.printArray(arr, n);

    return 0;

}

// This code is contributed

// by vt_Yogesh Shukla 1

Output:  

4   -3    5   -1    6   -7    2    8    9

Time Complexity: O(n) where n is number of elements in given array. 
Auxiliary Space: O(1)