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

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  • Difficulty Level : Medium
  • Last Updated : 30 Sep, 2022
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Given an array of positive and negative numbers, arrange them in an alternate fashion such that every positive number is followed by a negative and vice-versa maintaining the order of appearance. The 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 at the end of the array.

Examples: 

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 in many places (See this and this)

Complete Interview Preparation - GFG

Naive Approach: 

The above problem can be easily solved if O(n) extra space is allowed.

  • We can store the positive values and negative values in two separate data structures.
  • We will start filling the original array with alternating negative and positive values in the same order 
    in which it appears in the original array.

It becomes interesting due to the limitations that O(1) extra space and order of appearances. 

Optimal Approach:

The idea is to process the 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 (0-based index), or if it is positive and at even index (0-based index). Once we find an out-of-place element, we find the first element after it with an opposite sign. We right rotate the subarray between these two elements (including these two).

Illustration:

Let the array be arr[] = { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 }

First iteration: 

  • { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 } , -2 appears on odd index position and is out of place.
    We will look for the first element that appears with an opposite sign
  • { -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 } => perform rotation of subarray between this two elements
  • { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 }

Second iteration: 

  • { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 } , 4 is out of place.
  • { -5, 5, -2, 2, 4, 7, 1, 8, 0, -8 } =>-8 is of diffenrent sign 
  • { -5, 5, -2, 2, -8, 4, 7, 1, 8, 0 } => after performing right rotation between 4 to -8

Algorithm:
We will maintain a variable to mark if the element is in its correct position or not. Let the variable be outofplace. Initially, it is -1.

  • We will iterate over the array
  • If outofplace is -1, we will check if the current index is out of place.
    • If the current index is out of place we will update the outofplace with the index value.
    • Else we will keep the value as it is.
  • If the outofplace is not -1, we will search for the next index which has a different sign than that of the value that is present in outofplace position.

Following is the implementation of the above idea.  

C++




/*  C++ program to rearrange
   positive and negative integers
   in alternate fashion while keeping
   the order of positive and negative numbers. */
#include <assert.h>
#include <iostream>
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[0]);
 
    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




<?php
// PHP program to rearrange positive and
// negative integers in alternate fashion
// while keeping the order of positive
// and negative numbers.
 
// Utility function to right rotate all
// elements between [outofplace, cur]
function rightrotate(&$arr, $n,
                      $outofplace, $cur)
{
    $tmp = $arr[$cur];
    for ($i = $cur; $i > $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
?>

Javascript




<script>
    // Utility function to right rotate all elements
    // between [outofplace, cur]
    function rightrotate(arr , n , outofplace , cur) {
        var tmp = arr[cur];
        for (i = cur; i > outofplace; i--)
            arr[i] = arr[i - 1];
        arr[outofplace] = tmp;
    }
 
    function rearrange(arr , n) {
        var outofplace = -1;
 
        for (var 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'
    function printArray(arr , n) {
        for (i = 0; i < n; i++)
            document.write(arr[i] + " ");
        document.write("");
    }
 
    // Driver Code
     
    /*
         * var arr[n] = [-5, 3, 4, 5, -6, -2, 8, 9, -1, -4]; var arr = [-5, -3, -4,
         * -5, -6, 2 , 8, 9, 1 , 4]; var arr = [5, 3, 4, 2, 1, -2 , -8, -9, -1 , -4];
         * var arr = [-5, 3, -4, -7, -1, -2 , -8, -9, 1 , -4];
         */
        var arr = [ -5, -2, 5, 2, 4, 7, 1, 8, 0, -8 ];
        var n = arr.length;
 
        document.write("Given array is ");
        printArray(arr, n);
 
        rearrange(arr, n);
 
        document.write("<br/>RearrangeD array is ");
        printArray(arr, n);
         
// This code is contributed by gauravrajput1
</script>

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*N), as we are using a loop to traverse N times and calling function to right rotate each time which will cost O (N).
Auxiliary Space: O(1).

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