# Sort array after converting elements to their squares

Given a array of both positive and negative integers ‘arr[]’ which are sorted. Task is to sort square of the numbers of the Array.
Examples:

```Input  : arr[] =  {-6, -3, -1, 2, 4, 5}
Output : 1, 4, 9, 16, 25, 36

Input  : arr[] = {-5, -4, -2, 0, 1}
Output : 0, 1, 4, 16, 25
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple solution is to first convert each array elements into its square and than apply any “O(nlogn)” sorting algorithm to sort the array elements.

Below is the implementation of above idea

## C/C++

```// C++ program to Sort square of the numbers
// of the array
#include<bits/stdc++.h>
using namespace std;

// Function to sort an square array
void sortSquares(int arr[], int n)
{
// First convert each array elements
// into its square
for (int i = 0 ; i < n ; i++)
arr[i] = arr[i] * arr[i];

// Sort an array using "sort STL function "
sort(arr, arr+n);
}

// Driver program to test above function
int main()
{
int arr[] = { -6 , -3 , -1 , 2 , 4 , 5 };
int n = sizeof(arr)/sizeof(arr[0]);

cout << "Before sort " << endl;
for (int i = 0; i < n; i++)
cout << arr[i] << " " ;
sortSquares(arr, n);

cout << "\nAfter Sort " << endl;
for (int i = 0 ; i < n ; i++)
cout << arr[i] << " " ;

return 0;
}
```

## Java

```// Java program to Sort square of the numbers
// of the array
import java.util.*;
import java.io.*;

class GFG
{
// Function to sort an square array
public static void sortSquares(int arr[])
{
int n = arr.length;

// First convert each array elements
// into its square
for (int i = 0 ; i < n ; i++)
arr[i] = arr[i] * arr[i];

// Sort an array using "inbuild sort function"
// in Arrays class.
Arrays.sort(arr);
}

// Driver program to test above function
public static void main (String[] args)
{
int arr[] = { -6 , -3 , -1 , 2 , 4 , 5 };
int n = arr.length;

System.out.println("Before sort ");
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");

sortSquares(arr);
System.out.println("");
System.out.println("After Sort ");
for (int i = 0 ; i < n ; i++)
System.out.print(arr[i] + " ");

}
}
```

Output:

```Before sort
-6 -3 -1 2 4 5
After Sort
1 4 9 16 25 36
```

Time complexity : O(n log n)

Efficient solution is based on the fact that given array is already sorted. We do following two steps.

1. Divide the array into two part “Negative and positive “.
2. Use merge function to merge two sorted arrays into a single sorted array.

Below is the implementation of above idea.

## C/C++

```// C++ program to Sort square of the numbers of the array
#include<bits/stdc++.h>
using namespace std;

// function to sort array after doing squares of elements
void sortSquares(int arr[], int n)
{
// first dived array into part negative and positive
int K = 0;
for (K = 0 ; K < n; K++)
if (arr[K] >= 0 )
break;

// Now do the same process that we learn
// in merge sort to merge to two sorted array
// here both two half are sorted and we traverse
// first half in reverse meaner because
// first half contain negative element
int i = K-1; // Initial index of first half
int j = K; // Initial index of second half
int ind = 0; // Initial index of temp array

// store sorted array
int temp[n];
while (i >= 0 && j < n)
{
if (arr[i] * arr[i] < arr[j] * arr[j])
{
temp[ind] = arr[i] * arr[i];
i--;
}
else
{
temp[ind] = arr[j] * arr[j];
j++;
}
ind++;
}

/* Copy the remaining elements of first half */
while (i >= 0)
{
temp[ind] = arr[i] * arr[i];
i--;
ind++;
}

/* Copy the remaining elements of second half */
while (j < n)
{
temp[ind] = arr[j] * arr[j];
j++;
ind++;
}

// copy 'temp' array into original array
for (int i = 0 ; i < n; i++)
arr[i] = temp[i];
}

// Driver program to test above function
int main()
{
int arr[] = { -6 , -3 , -1 , 2 , 4 , 5 };
int n = sizeof(arr)/sizeof(arr[0]);

cout << "Before sort " << endl;
for (int i = 0; i < n; i++)
cout << arr[i] << " " ;
sortSquares(arr, n);

cout << "\nAfter Sort " << endl;
for (int i = 0 ; i < n ; i++)
cout << arr[i] << " " ;

return 0;
}
```

## Java

```// Java program to Sort square of the numbers
// of the array
import java.util.*;
import java.io.*;

class GFG
{
// Function to sort an square array
public static void sortSquares(int arr[])
{
int n = arr.length;
// first dived array into part negative and positive
int k;
for(k = 0; k < n; k++)
{
if(arr[k] >= 0)
break;
}

// Now do the same process that we learn
// in merge sort to merge to two sorted array
// here both two half are sorted and we traverse
// first half in reverse meaner because
// first half contain negative element
int i = k-1; // Initial index of first half
int j = k; // Initial index of second half
int ind = 0; // Initial index of temp array

int[] temp = new int[n];
while(i >= 0 && j < n)
{
if(arr[i] * arr[i] < arr[j] * arr[j])
{
temp[ind] = arr[i] * arr[i];
i--;
}
else{

temp[ind] = arr[j] * arr[j];
j++;

}
ind++;
}

while(i >= 0)
{
temp[ind++] = arr[i] * arr[i];
i--;
}
while(j < n)
{
temp[ind++] = arr[j] * arr[j];
j++;
}

// copy 'temp' array into original array
for (int x = 0 ; x < n; x++)
arr[x] = temp[x];
}

// Driver program to test above function
public static void main (String[] args)
{
int arr[] = { -6 , -3 , -1 , 2 , 4 , 5 };
int n = arr.length;

System.out.println("Before sort ");
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");

sortSquares(arr);
System.out.println("");
System.out.println("After Sort ");
for (int i = 0 ; i < n ; i++)
System.out.print(arr[i] + " ");

}
}
```
```Before sort
-6 -3 -1 2 4 5
After Sort
1 4 9 16 25 36
```

Time complexity: O(n)
space complexity: O(n)

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