You have been given an array and you have to make a program to convert that array such that positive elements occur at even numbered places in the array and negative elements occur at odd numbered places in the array. We have to do it in place.
There can be unequal number of positive and negative values and the extra values have to left as it is.
Examples:
Input : arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10} Output : arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10} Input : arr[] = {-1, 3, -5, 6, 3, 6, -7, -4, -9, 10} Output : arr[] = {3, -1, 6, -5, 3, -7, 6, -4, 10, -9}
The idea is to use Hoare’s partition process of Quick Sort.
We take two pointers positive and negative. We set the positive pointer at start of the array and the negative pointer at 1st position of the array.
We move positive pointer two steps forward till it finds a negative element. Similarly, we move negative pointer forward by two places till it finds a positive value at its position.
If the positive and negative pointers are in the array then we will swap the values at these indexes otherwise we will stop executing the process.
Implementation:
// C++ program to rearrange positive and negative // numbers #include <bits/stdc++.h> using namespace std;
void rearrange( int a[], int size)
{ int positive = 0, negative = 1;
while ( true ) {
/* Move forward the positive pointer till
negative number number not encountered */
while (positive < size && a[positive] >= 0)
positive += 2;
/* Move forward the negative pointer till
positive number number not encountered */
while (negative < size && a[negative] <= 0)
negative += 2;
// Swap array elements to fix their position.
if (positive < size && negative < size)
swap(a[positive], a[negative]);
/* Break from the while loop when any index
exceeds the size of the array */
else
break ;
}
} // Driver code int main()
{ int arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
int n = ( sizeof (arr) / sizeof (arr[0]));
rearrange(arr, n);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
} |
// Java program to rearrange positive // and negative numbers import java.io.*;
class GFG {
static void rearrange( int a[], int size)
{ int positive = 0 , negative = 1 , temp;
while ( true ) {
/* Move forward the positive pointer till
negative number number not encountered */
while (positive < size && a[positive] >= 0 )
positive += 2 ;
/* Move forward the negative pointer till
positive number number not encountered */
while (negative < size && a[negative] <= 0 )
negative += 2 ;
// Swap array elements to fix their position.
if (positive < size && negative < size) {
temp = a[positive];
a[positive] = a[negative];
a[negative] = temp;
}
/* Break from the while loop when any index
exceeds the size of the array */
else
break ;
}
} // Driver code public static void main(String args[]) {
int arr[] = { 1 , - 3 , 5 , 6 , - 3 , 6 , 7 , - 4 , 9 , 10 };
int n = arr.length;
rearrange(arr, n);
for ( int i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
} } /*This code is contributed by Nikita Tiwari.*/ |
// C# program to rearrange positive // and negative numbers using System;
class GFG {
// Function to rearrange static void rearrange( int []a, int size)
{ int positive = 0, negative = 1, temp;
while ( true ) {
// Move forward the positive pointer till
// negative number number not encountered
while (positive < size && a[positive] >= 0)
positive += 2;
// Move forward the negative pointer till
// positive number number not encountered
while (negative < size && a[negative] <= 0)
negative += 2;
// Swap array elements to fix their position.
if (positive < size && negative < size) {
temp = a[positive];
a[positive] = a[negative];
a[negative] = temp;
}
// Break from the while loop when any
// index exceeds the size of the array
else
break ;
}
} // Driver Code public static void Main(String []args) {
int []arr = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10};
int n = arr.Length;
rearrange(arr, n);
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
} } // This code is contributed by Nitin Mittal. |
<script> // Javascript program to rearrange positive // and negative numbers function rearrange(a, size)
{ let positive = 0;
let negative = 1;
let temp;
while ( true )
{
// Move forward the positive
// pointer till negative number
// number not encountered
while (positive < size && a[positive] >= 0)
positive += 2;
// Move forward the negative pointer
// till positive number number not encountered
while (negative < size && a[negative] <= 0)
negative += 2;
// Swap array elements to fix their position.
if (positive < size && negative < size)
{
temp = a[positive];
a[positive] = a[negative];
a[negative] = temp;
}
// Break from the while loop when any index
// exceeds the size of the array
else
break ;
}
} // Driver code let arr = [ 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 ];
let n = arr.length; rearrange(arr, n); for (let i = 0; i < n; i++)
document.write(arr[i] + " " );
// This code is contributed by sravan kumar G </script> |
1 -3 5 -3 6 6 7 -4 9 10
Time Complexity: O(n2),
Auxiliary Space: O(1)
Lets explain the working of the code on the first example
arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}
We declare two variables positive and negative positive points to zeroth position and negative points to first position
positive = 0 negative = 1
In the first iteration positive will move 4 places to fifth position as a[4] is less than zero and positive = 4.
Negative will move 2 places and will point to fourth position as a[3]>0
we will swap positive and negative position values as they are less than size of array.
After first iteration the array becomes arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}
Now positive points at fourth position and negative points at third position
In second iteration the positive value will move 6 places and its value will
more than the size of the array.
The negative pointer will move two steps forward and it will point to 5th position
As the positive pointer value becomes greater than the array size
we will not perform any swap operation and break out of the while loop.
The final output will be
arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}
An example where relative order is not maintained:
{ -1, -2, -3, -4, -5, 6, 7, 8 };
Another Approach :-
The idea is to find a positive/negative element which is in incorrect place(i.e. positive at odd and negative at even place) and the then find the element of opposite sign which is also in incorrect position in the remaining array and then swap these two elements.
Here is the implementation of the above idea.
// C++ program to rearrange positive // and negative numbers #include<iostream> using namespace std;
// Swap function void swap( int * a, int i , int j)
{ int temp = a[i];
a[i] = a[j];
a[j] = temp;
return ;
} // Print array function void printArray( int * a, int n)
{ for ( int i = 0; i < n; i++)
cout << a[i] << " " ;
cout << endl;
return ;
} // Driver code int main()
{ int arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
int n = sizeof (arr)/ sizeof (arr[0]);
//before modification
printArray(arr, n);
for ( int i = 0; i < n; i++)
{
if (arr[i] >= 0 && i % 2 == 1)
{
// out of order positive element
for ( int j = i + 1; j < n; j++)
{
if (arr[j] < 0 && j % 2 == 0)
{
// find out of order negative
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
else if (arr[i] < 0 && i % 2 == 0)
{
// out of order negative element
for ( int j = i + 1; j < n; j++)
{
if (arr[j] >= 0 && j % 2 == 1)
{
// find out of order positive
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
}
//after modification
printArray(arr, n);
return 0;
} // This code is contributed by AnitAggarwal |
// Java program to rearrange positive // and negative numbers import java.io.*;
import java.util.*;
class GFG
{ // Swap function
static void swap( int [] a, int i, int j)
{
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
// Print array function
static void printArray( 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[])
{
int [] arr = { 1 , - 3 , 5 , 6 , - 3 , 6 , 7 , - 4 , 9 , 10 };
int n = arr.length;
//before modification
printArray(arr, n);
for ( int i = 0 ; i < n; i++)
{
if (arr[i] >= 0 && i % 2 == 1 )
{
// out of order positive element
for ( int j = i + 1 ; j < n; j++)
{
if (arr[j] < 0 && j % 2 == 0 )
{
// find out of order negative
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
else if (arr[i] < 0 && i % 2 == 0 )
{
// out of order negative element
for ( int j = i + 1 ; j < n; j++)
{
if (arr[j] >= 0 && j % 2 == 1 )
{
// find out of order positive
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
}
//after modification
printArray(arr, n);
}
} // This code is contributed by rachana soma |
# Python3 program to rearrange positive # and negative numbers # Print array function def printArray(a, n):
for i in a:
print (i, end = " " )
print ()
# Driver code arr = [ 1 , - 3 , 5 , 6 , - 3 , 6 , 7 , - 4 , 9 , 10 ]
n = len (arr)
# before modification printArray(arr, n) for i in range (n):
if (arr[i] > = 0 and i % 2 = = 1 ):
# out of order positive element
for j in range (i + 1 , n):
if (arr[j] < 0 and j % 2 = = 0 ):
# find out of order negative
# element in remaining array
arr[i], arr[j] = arr[j], arr[i]
break
elif (arr[i] < 0 and i % 2 = = 0 ):
# out of order negative element
for j in range (i + 1 , n):
if (arr[j] > = 0 and j % 2 = = 1 ):
# find out of order positive
# element in remaining array
arr[i], arr[j] = arr[j], arr[i]
break
# after modification printArray(arr, n); # This code is contributed # by mohit kumar |
// C# program to rearrange positive // and negative numbers using System;
class GFG
{ // Swap function
static void swap( int [] a, int i, int j)
{
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
// Print array function
static void printArray( int [] a, int n)
{
for ( int i = 0; i < n; i++)
Console.Write(a[i] + " " );
Console.WriteLine();
}
// Driver code
public static void Main()
{
int [] arr = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 };
int n = arr.Length;
//before modification
printArray(arr, n);
for ( int i = 0; i < n; i++)
{
if (arr[i] >= 0 && i % 2 == 1)
{
// out of order positive element
for ( int j = i + 1; j < n; j++)
{
if (arr[j] < 0 && j % 2 == 0)
{
// find out of order negative
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
else if (arr[i] < 0 && i % 2 == 0)
{
// out of order negative element
for ( int j = i + 1; j < n; j++)
{
if (arr[j] >= 0 && j % 2 == 1)
{
// find out of order positive
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
}
// after modification
printArray(arr, n);
}
} // This code is contributed by Akanksha Rai |
<script> // Javascript program to rearrange positive // and negative numbers // Swap function
function swap(a,i,j)
{
let temp = a[i];
a[i] = a[j];
a[j] = temp;
}
// Print array function
function printArray(a,n)
{
for (let i = 0; i < n; i++)
document.write(a[i] + " " );
document.write( "<br>" );
}
// Driver code
let arr=[1, -3, 5, 6, -3, 6, 7, -4, 9, 10];
let n = arr.length;
//before modification
printArray(arr, n);
for (let i = 0; i < n; i++)
{
if (arr[i] >= 0 && i % 2 == 1)
{
// out of order positive element
for (let j = i + 1; j < n; j++)
{
if (arr[j] < 0 && j % 2 == 0)
{
// find out of order negative
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
else if (arr[i] < 0 && i % 2 == 0)
{
// out of order negative element
for (let j = i + 1; j < n; j++)
{
if (arr[j] >= 0 && j % 2 == 1)
{
// find out of order positive
// element in remaining array
swap(arr, i, j);
break ;
}
}
}
}
//after modification
printArray(arr, n);
// This code is contributed by avanitrachhadiya2155
</script> |
1 -3 5 6 -3 6 7 -4 9 10 1 -3 5 -3 6 6 7 -4 9 10
Time Complexity: O(n2),
Auxiliary Space: O(1)