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.
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)