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LogoutHello Everyone,
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 maintaining the order of appearance.
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.
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}
Naive Approach :
The above problem can be easily solved if O(n) extra space is allowed. It becomes interesting due to the limitations that O(1) extra space and order of appearances.
The idea is to process 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, or it is positive and at even index. Once we find an out of place element, we find the first element after it with opposite sign. We right rotate the subarray between these two elements (including these two).
Following is the implementation of 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;
}
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^2)
Space Complexity : O(1)
Efficient Approach :
We first sort the array in non-increasing order.Then we will count the number of poitive and negative integers. Then we will swap the one negative and one positive
number 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.
- 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;
}
}
}
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;
}
}
static void print( int a[], int n)
{
for ( int i = 0 ; i < n; i++)
System.out.print(a[i] + " " );
System.out.println();
}
public static void main(String[] args)
throws java.lang.Exception
{
// Given array
int [] arr = { - 5 , - 2 , 5 , 2 , 4 , 7 , 1 , 8 , 0 , - 8 };
int n = arr.length;
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);
}
print(arr, n);
}
}
Output
-8 1 -2 0 -5 2 4 5 7 8
Time Complexity: O(N*logN)
Space Complexity: O(1)