# Place array in alternating positive & negative items with O(1) extra space - Part 2

Efficient Approach :
We have already discussed a O(n2) solution that maintains the order of appearance in the array. If we are allowed to change order of appearance, we can solve this problem in O(n) time and O(1) space.
The idea is to process the array and shift all negative values to the end in O(n) time. After all negative values are shifted to the end, we can easily rearrange array in alternating positive & negative items. We basically swap next positive element at even position from next negative element in this step.

Following is the implementation of above idea.

`// Java program to rearrange`

`// array in alternating`

`// positive & negative`

`// items with O(1) extra space`

`class` `GFG {`

`// Function to rearrange positive and negative`

`// integers in alternate fashion. The below`

`// solution doesn't maintain original order of`

`// elements`

`static` `void` `rearrange(` `int` `arr[], ` `int` `n)`

`{`

` ` `int` `i = ` `0` `, j = n - ` `1` `;`

` ` `// shift all negative values to the end`

` ` `while` `(i < j)`

` ` `{`

` ` `while` `(i <= n - ` `1` `&& arr[i] > ` `0` `)`

` ` `i += ` `1` `;`

` ` `while` `(j >= ` `0` `&& arr[j] < ` `0` `)`

` ` `j -= ` `1` `;`

` ` `if` `(i < j)`

` ` `swap(arr, i,j);`

` ` `}`

` ` `// i has index of leftmost negative element`

` ` `if` `(i == ` `0` `|| i == n)`

` ` `return` `;`

` ` `// start with first positive`

` ` `// element at index 0`

` ` `// Rearrange array in alternating positive &`

` ` `// negative items`

` ` `int` `k = ` `0` `;`

` ` `while` `(k < n && i < n)`

` ` `{`

` ` `// swap next positive element`

` ` `// at even position`

` ` `// from next negative element.`

` ` `swap(arr,k,i);`

` ` `i = i + ` `1` `;`

` ` `k = k + ` `2` `;`

` ` `}`

`}`

`// Utility function to print an array`

`static` `void` `printArray(` `int` `arr[], ` `int` `n)`

`{`

` ` `for` `(` `int` `i = ` `0` `; i < n; i++)`

` ` `System.out.print(arr[i] + ` `" "` `);`

` ` `System.out.println(` `""` `);`

`}`

`static` `void` `swap(` `int` `arr[], ` `int` `index1, ` `int` `index2)`

`{`

` ` `int` `c = arr[index1];`

` ` `arr[index1]=arr[index2];`

` ` `arr[index2]=c;`

`}`

` ` `// Driver code`

` ` `public` `static` `void` `main(String[] args)`

` ` `{`

` ` `int` `arr[] = {` `2` `, ` `3` `, -` `4` `, -` `1` `, ` `6` `, -` `9` `};`

` ` `int` `n = arr.length;`

` ` `System.out.println(` `"Given array is "` `);`

` ` `printArray(arr, n);`

` ` `rearrange(arr, n);`

` ` `System.out.println(` `"Rearranged array is "` `);`

` ` `printArray(arr, n);`

` ` `}`

`}`

Output:

Given array is 2 3 -4 -1 6 -9 Rearranged array is -1 3 -4 2 -9 6

Time Complexity : O(N)

Space Complexity : O(1)