Rearrange array in alternating positive & negative items with O(1) extra space | Set 2

30 Jun, 2022
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. Order of elements in output doesn’t matter. Extra positive or negative elements should be moved to end.
Examples: 
Input :
arr[] = {-2, 3, 4, -1}
Output :
arr[] = {-2, 3, -1, 4} OR {-1, 3, -2, 4} OR ..

Input :
arr[] = {-2, 3, 1}
Output :
arr[] = {-2, 3, 1} OR {-2, 1, 3} 

Input : 
arr[] = {-5, 3, 4, 5, -6, -2, 8, 9, -1, -4}
Output :
arr[] = {-5, 3, -2, 5, -6, 4, -4, 9, -1, 8} 
        OR ..
Approach 1:
First, sort the array in non-increasing order. Then we will count the number of positive and negative integers.Then swap the one negative and one positive number in the odd positions till we reach our condition.This will rearrange the array elements because we are sorting the array and accessing the element from left to right according to our need.
First, sort the array in non-increasing order. Then we will count the number of positive and negative integers.
Then swap the one negative and one positive number in the odd positions till we reach our condition.
This will rearrange the array elements because we are sorting the array and accessing the element from left to right according to our need.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// Below is the implementation of the above approach#include <bits/stdc++.h>using namespace std; // Function which works in the condition// when number of negative numbers are// lesser or equal than positive numbersvoid fill1(int a[], int neg, int pos){    if (neg % 2 == 1)    {        for(int i = 1; i < neg; i += 2)        {            int c = a[i];            int d = a[i + neg];            int temp = c;            a[i] = d;            a[i + neg] = temp;        }    }    else    {        for(int i = 1; i <= neg; i += 2)        {            int c = a[i];            int d = a[i + neg - 1];            int temp = c;            a[i] = d;            a[i + neg - 1] = temp;        }    }} // Function which works in the condition// when number of negative numbers are// greater than positive numbersvoid fill2(int a[], int neg, int pos){    if (pos % 2 == 1)    {        for(int i = 1; i < pos; i += 2)        {            int c = a[i];            int d = a[i + pos];            int temp = c;            a[i] = d;            a[i + pos] = temp;        }    }    else    {        for(int i = 1; i <= pos; i += 2)        {            int c = a[i];            int d = a[i + pos - 1];            int temp = c;            a[i] = d;            a[i + pos - 1] = temp;        }    }} // Reverse the arrayvoid reverse(int a[], int n){    int i, k, t;    for(i = 0; i < n / 2; i++)    {        t = a[i];        a[i] = a[n - i - 1];        a[n - i - 1] = t;    }} // Print the arrayvoid print(int a[], int n){    for(int i = 0; i < n; i++)        cout << a[i] << " ";             cout << endl;} // Driver codeint main(){    int arr[] = { 2, 3, -4, -1, 6, -9 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << "Given array is ";    print(arr, n);         int neg = 0, pos = 0;    for(int i = 0; i < n; i++)    {        if (arr[i] < 0)            neg++;        else            pos++;    }         // Sort the array    sort(arr, arr + n);     if (neg <= pos)    {        fill1(arr, neg, pos);    }    else    {                 // Reverse the array in this condition        reverse(arr, n);        fill2(arr, neg, pos);    }    cout << "Rearranged array is  ";    print(arr, n);} // This code is contributed by Potta Lokesh
// Below is the implementation of the above approachimport java.io.*;import java.lang.*;import java.util.*;public class Main {       // function which works in the condition when number of    // negative numbers are lesser or equal than positive    // numbers    static void fill1(int a[], int neg, int pos)    {        if (neg % 2 == 1) {            for (int i = 1; i < neg; i += 2) {                int c = a[i];                int d = a[i + neg];                int temp = c;                a[i] = d;                a[i + neg] = temp;            }        }        else {            for (int i = 1; i <= neg; i += 2) {                int c = a[i];                int d = a[i + neg - 1];                int temp = c;                a[i] = d;                a[i + neg - 1] = temp;            }        }    }       // Function which works in the condition when number of    // negative numbers are greater than positive numbers    static void fill2(int a[], int neg, int pos)    {        if (pos % 2 == 1) {            for (int i = 1; i < pos; i += 2) {                int c = a[i];                int d = a[i + pos];                int temp = c;                a[i] = d;                a[i + pos] = temp;            }        }        else {            for (int i = 1; i <= pos; i += 2) {                int c = a[i];                int d = a[i + pos - 1];                int temp = c;                a[i] = d;                a[i + pos - 1] = temp;            }        }    }       // Reverse the array    static void reverse(int a[], int n)    {        int i, k, t;        for (i = 0; i < n / 2; i++) {            t = a[i];            a[i] = a[n - i - 1];            a[n - i - 1] = t;        }    }       // Print the array    static void print(int a[], int n)    {        for (int i = 0; i < n; i++)            System.out.print(a[i] + " ");        System.out.println();    }       // Driver Code    public static void main(String[] args)        throws java.lang.Exception    {        // Given array        int[] arr = { 2, 3, -4, -1, 6, -9 };        int n = arr.length;        System.out.println("Given array is  ");        print(arr, n);        int neg = 0, pos = 0;        for (int i = 0; i < n; i++) {            if (arr[i] < 0)                neg++;            else                pos++;        }        // Sort the array        Arrays.sort(arr);         if (neg <= pos) {            fill1(arr, neg, pos);        }        else {            // reverse the array in this condition            reverse(arr, n);            fill2(arr, neg, pos);        }        System.out.println("Rearranged array is  ");        print(arr, n);    }}
# Python3 program for the above approach # Function which works in the condition# when number of negative numbers are# lesser or equal than positive numbersdef fill1(a, neg, pos) :         if (neg % 2 == 1)  :        for i in range(1, neg, 2):            c = a[i]            d = a[i + neg]            temp = c            a[i] = d            a[i + neg] = temp                  else   :         for i in range(1, neg+1, 2):            c = a[i]            d = a[i + neg - 1]            temp = c            a[i] = d            a[i + neg - 1] = temp         # Function which works in the condition# when number of negative numbers are# greater than positive numbersdef fill2(a, neg, pos):    if (pos % 2 == 1) :         for i in range(1, pos, 2):            c = a[i]            d = a[i + pos]            temp = c            a[i] = d            a[i + pos] = temp                  else  :        for i in range(1, pos+1, 2):            c = a[i]            d = a[i + pos - 1]            temp = c            a[i] = d            a[i + pos - 1] = temp         # Reverse the arraydef reverse(a, n) :         for i in range(n / 2):        t = a[i]        a[i] = a[n - i - 1]        a[n - i - 1] = t     # Print the arraydef printt(a, n):    for i in range(n):        print(a[i], end = " ")             print() # Driver codeif __name__ == "__main__":         arr = [ 2, 3, -4, -1, 6, -9 ]    n = len(arr)    print("Given array is ")    printt(arr, n)         neg = 0    pos = 0    for i in range(0, n):        if (arr[i] < 0):            neg += 1        else:            pos += 1           # Sort the array    arr.sort()     if (neg <= pos) :        fill1(arr, neg, pos)         else  :                 # Reverse the array in this condition        reverse(arr, n)        fill2(arr, neg, pos)         print("Rearranged array is  ")    printt(arr, n) # This code is contributed by sanjoy_62.
// C# program for the above approachusing System;using System.Collections.Generic;using System.Linq; public class GFG {         // function which works in the condition when number of    // negative numbers are lesser or equal than positive    // numbers    static void fill1(int[] a, int neg, int pos)    {        if (neg % 2 == 1) {            for (int i = 1; i < neg; i += 2) {                int c = a[i];                int d = a[i + neg];                int temp = c;                a[i] = d;                a[i + neg] = temp;            }        }        else {            for (int i = 1; i <= neg; i += 2) {                int c = a[i];                int d = a[i + neg - 1];                int temp = c;                a[i] = d;                a[i + neg - 1] = temp;            }        }    }        // Function which works in the condition when number of    // negative numbers are greater than positive numbers    static void fill2(int[] a, int neg, int pos)    {        if (pos % 2 == 1) {            for (int i = 1; i < pos; i += 2) {                int c = a[i];                int d = a[i + pos];                int temp = c;                a[i] = d;                a[i + pos] = temp;            }        }        else {            for (int i = 1; i <= pos; i += 2) {                int c = a[i];                int d = a[i + pos - 1];                int temp = c;                a[i] = d;                a[i + pos - 1] = temp;            }        }    }        // Reverse the array    static void reverse(int[] a, int n)    {        int i, k, t;        for (i = 0; i < n / 2; i++) {            t = a[i];            a[i] = a[n - i - 1];            a[n - i - 1] = t;        }    }        // Print the array    static void print(int[] a, int n)    {        for (int i = 0; i < n; i++)            Console.Write(a[i] + " ");        Console.WriteLine();    } // Driver Codepublic static void Main (string[] args) {          // Given array        int[] arr = { 2, 3, -4, -1, 6, -9 };        int n = arr.Length;        Console.WriteLine("Given array is  ");        print(arr, n);        int neg = 0, pos = 0;        for (int i = 0; i < n; i++) {            if (arr[i] < 0)                neg++;            else                pos++;        }           // Sort the array        Array.Sort(arr);          if (neg <= pos) {            fill1(arr, neg, pos);        }        else {            // reverse the array in this condition            reverse(arr, n);            fill2(arr, neg, pos);        }        Console.WriteLine("Rearranged array is  ");        print(arr, n);}} // This code is contributed by splevel62.
<script> // Below is the implementation of the above approach // Function which works in the condition// when number of negative numbers are// lesser or equal than positive numbersfunction fill1(a, neg, pos){    if (neg % 2 == 1)    {        for (let i = 1; i < neg; i += 2)        {            let c = a[i];            let d = a[i + neg];            let temp = c;            a[i] = d;            a[i + neg] = temp;        }    }    else    {        for(let i = 1; i <= neg; i += 2)        {            let c = a[i];            let d = a[i + neg - 1];            let temp = c;            a[i] = d;            a[i + neg - 1] = temp;        }    }} // Function which works in the condition// when number of negative numbers are// greater than positive numbersfunction fill2(a, neg, pos){    if (pos % 2 == 1)    {        for (let i = 1; i < pos; i += 2)        {            let c = a[i];            let d = a[i + pos];            let temp = c;            a[i] = d;            a[i + pos] = temp;        }    }    else    {        for(let i = 1; i <= pos; i += 2)        {            let c = a[i];            let d = a[i + pos - 1];            let temp = c;            a[i] = d;            a[i + pos - 1] = temp;        }    }} // Reverse the arrayfunction reverse(a, n){    let i, k, t;    for(i = 0; i < parseInt(n / 2); i++)    {        t = a[i];        a[i] = a[n - i - 1];        a[n - i - 1] = t;    }} // Print the arrayfunction print(a, n){    for(let i = 0; i < n; i++)        document.write(a[i] + " ");             document.write('<br>');} // Driver codevar arr = [ 2, 3, -4, -1, 6, -9 ];let n = arr.length;document.write("Given array is ");print(arr, n);let neg = 0, pos = 0; for(let i = 0; i < n; i++){    if (arr[i] < 0)        neg++;    else        pos++;} // Sort the arrayarr.sort(function (a, b){return a - b;}); if (neg <= pos){    fill1(arr, neg, pos);}else{         // Reverse the array in this condition    reverse(arr, n);    fill2(arr, neg, pos);}document.write("Rearranged array is  ");print(arr, n); // This code is contributed by Potta Lokesh </script>
Given array is 2 3 -4 -1 6 -9 
Rearranged array is  -9 3 -1 2 -4 6 
Time Complexity: O(N*logN)Space Complexity: O(1)
Efficient Approach: We have already discussed a O(n2) solution that maintains the order of appearance in the array here. If we are allowed to change order of appearance, we can solve this problem in O(n) time and O(1) space.The idea is to process the array and shift all negative values to the end in O(n) time. After all negative values are shifted to the end, we can easily rearrange array in alternating positive & negative items. We basically swap next positive element at even position from next negative element in this step. 
Following is the implementation of above idea.
C++
Java
Python3
C#
PHP
Javascript
// C++ program to rearrange// array in alternating// positive & negative items// with O(1) extra space#include <bits/stdc++.h>using namespace std; // Function to rearrange positive and negative// integers in alternate fashion. The below// solution doesn't maintain original order of// elementsvoid rearrange(int arr[], int n){    int i = 0, j = n-1;     // shift all negative values to the end    while (i < j) {        while (i <= n - 1 and arr[i] > 0)            i += 1;        while (j >= 0 and arr[j] < 0)            j -= 1;        if (i < j )            swap(arr[i], arr[j]);    }     // i has index of leftmost    // negative element    if (i == 0 || i == n)        return;     // start with first positive    // element at index 0     // Rearrange array in alternating    // positive &    // negative items    int k = 0;    while (k < n && i < n ) {        // swap next positive        // element at even position        // from next negative element.        swap(arr[k], arr[i]);        i = i + 1;        k = k + 2;    }} // Utility function to print an arrayvoid printArray(int arr[], int n){    for (int i = 0; i < n; i++)        cout << arr[i] << " ";    cout << endl;} // Driver codeint main(){    int arr[] = { 2, 3, -4, -1, 6, -9 };     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 program to rearrange// array in alternating// positive & negative// items with O(1) extra spaceclass GFG {     // Function to rearrange positive and negative    // integers in alternate fashion. The below    // solution doesn't maintain original order of    // elements    static void rearrange(int arr[], int n)    {        int i = 0, j = n - 1;         // shift all negative values to the end        while (i < j) {            while (i <= n - 1 && arr[i] > 0)                i += 1;            while (j >= 0 && arr[j] < 0)                j -= 1;            if (i < j)                swap(arr, i, j);        }         // i has index of leftmost negative element        if (i == 0 || i == n)            return;         // start with first positive        // element at index 0         // Rearrange array in alternating positive &        // negative items        int k = 0;        while (k < n && i < n) {            // swap next positive element            // at even position            // from next negative element.            swap(arr, k, i);            i = i + 1;            k = k + 2;        }    }     // Utility function to print an array    static void printArray(int arr[], int n)    {        for (int i = 0; i < n; i++)            System.out.print(arr[i] + " ");        System.out.println("");    }     static void swap(int arr[], int index1, int index2)    {        int c = arr[index1];        arr[index1] = arr[index2];        arr[index2] = c;    }     // Driver code    public static void main(String[] args)    {        int arr[] = { 2, 3, -4, -1, 6, -9 };         int n = arr.length;         System.out.println("Given array is ");        printArray(arr, n);         rearrange(arr, n);         System.out.println("Rearranged array is ");        printArray(arr, n);    }} // This code is contributed by 29AjayKumar
# Python3 program to rearrange array# in alternating positive & negative# items with O(1) extra space # Function to rearrange positive and# negative integers in alternate fashion.# The below solution does not maintain# original order of elements  def rearrange(arr, n):    i = 0    j = n - 1     # shift all negative values    # to the end    while (i < j):         while (i <= n - 1 and arr[i] > 0):            i += 1        while (j >= 0 and arr[j] < 0):            j -= 1         if (i < j):            temp = arr[i]            arr[i] = arr[j]            arr[j] = temp     # i has index of leftmost    # negative element    if (i == 0 or i == n):        return 0     # start with first positive element    # at index 0     # Rearrange array in alternating    # positive & negative items    k = 0    while (k < n and i < n):         # swap next positive element at even        # position from next negative element.        temp = arr[k]        arr[k] = arr[i]        arr[i] = temp        i = i + 1        k = k + 2 # Utility function to print an array  def printArray(arr, n):    for i in range(n):        print(arr[i], end=" ")    print("\n")  # Driver codearr = [2, 3] n = len(arr) print("Given array is")printArray(arr, n) rearrange(arr, n) print("Rearranged array is")printArray(arr, n) # This code is contributed# Princi Singh
// C# program to rearrange array// in alternating positive & negative// items with O(1) extra spaceusing System; class GFG {     // Function to rearrange positive and    // negative integers in alternate fashion.    // The below solution doesn't maintain    // original order of elements    static void rearrange(int[] arr, int n)    {        int i = 0, j = n - 1;         // shift all negative values        // to the end        while (i < j) {            while (i <= n - 1 && arr[i] > 0)                i += 1;            while (j >= 0 && arr[j] < 0)                j -= 1;            if (i < j)                swap(arr, i, j);        }         // i has index of leftmost        // negative element        if (i == 0 || i == n)            return;         // start with first positive        // element at index 0         // Rearrange array in alternating        // positive & negative items        int k = 0;        while (k < n && i < n) {            // swap next positive element            // at even position from next            // negative element.            swap(arr, k, i);            i = i + 1;            k = k + 2;        }    }     // Utility function to print an array    static void printArray(int[] arr, int n)    {        for (int i = 0; i < n; i++)            Console.Write(arr[i] + " ");        Console.WriteLine("");    }     static void swap(int[] arr, int index1, int index2)    {        int c = arr[index1];        arr[index1] = arr[index2];        arr[index2] = c;    }     // Driver code    public static void Main()    {        int[] arr = { 2, 3, -4, -1, 6, -9 };         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 29AjayKumar
<?php// PHP program to rearrange array// in alternating positive & negative// items with O(1) extra space // Function to rearrange positive and// negative integers in alternate fashion.// The below solution doesn't maintain// original order of elementsfunction rearrange(&$arr, $n){    $i = 0;    $j = $n - 1;     // shift all negative values    // to the end    while ($i < $j)    {          while($i <= $n - 1 and $arr[$i] > 0)            ++$i;        while ($j >= 0 and $arr[$j] < 0)            --$j;                 if ($i < $j)        {            $temp = $arr[$i];            $arr[$i] = $arr[$j];            $arr[$j] = $temp;        }            }     // i has index of leftmost    // negative element    if ($i == 0 || $i == $n)        return;     // start with first positive element    // at index 0     // Rearrange array in alternating    // positive & negative items    $k = 0;    while ($k < $n && $i < $n)    {        // swap next positive element at even        // position from next negative element.        $temp = $arr[$k];        $arr[$k] = $arr[$i];        $arr[$i] = $temp;        $i = $i + 1;        $k = $k + 2;    }} // Utility function to print an arrayfunction printArray(&$arr, $n){    for ($i = 0; $i < $n; $i++)    echo $arr[$i] . " ";    echo "\n";} // Driver code$arr = array(2, 3, -4, -1, 6, -9); $n = sizeof($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?>
<script> // Javascript program to rearrange// array in alternating// positive & negative// items with O(1) extra space // Function to rearrange positive and negative// integers in alternate fashion. The below// solution doesn't maintain original order of// elementsfunction rearrange(arr,n){    let i = 0, j = n - 1;      // Shift all negative values to the end    while (i < j)    {        while(i <= n - 1 && arr[i] > 0)            i += 1;        while (j >= 0 && arr[j] < 0)            j -= 1;                     if (i < j)            swap(arr, i,j);    }      // i has index of leftmost negative element    if (i == 0 || i == n)        return;      // Start with first positive    // element at index 0      // Rearrange array in alternating    // positive & negative items    let k = 0;    while (k < n && i < n)    {                 // Swap next positive element        // at even position        // from next negative element.        swap(arr, k, i);        i = i + 1;        k = k + 2;    }} // Utility function to print an arrayfunction printArray(arr, n){    for(let i = 0; i < n; i++)        document.write(arr[i] + " ");             document.write("<br>");} function swap(arr, index1, index2){    let c = arr[index1];    arr[index1] = arr[index2];    arr[index2] = c;} // Driver codelet arr = [ 2, 3, -4, -1, 6, -9 ];let n = arr.length; document.write("Given array is <br>"); printArray(arr, n);rearrange(arr, n); document.write("Rearranged array is <br>");printArray(arr, n); // This code is contributed by rag2127 </script>
Given array is 
2 3 -4 -1 6 -9 
Rearranged array is 
-1 3 -4 2 -9 6 
Time Complexity : O(N)Space Complexity : O(1)
This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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