Place array in alternating positive & negative items with O(1) extra space

Software Development
#1

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:

  1. First, sort the array in non-increasing order. Then we will count the number of positive and negative integers.
  2. Then swap the one negative and one positive number in the odd positions till we reach our condition.
  3. 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:

  • Java

// Below is the implementation of the above approach

import 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);

}

}

Output

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)