Java Program For Converting Array Into Zig-Zag Fashion

Given an array of DISTINCT elements, rearrange the elements of array in zig-zag fashion in O(n) time. The converted array should be in form a c e . 

Example:

Input: arr[] = {4, 3, 7, 8, 6, 2, 1} 
Output: arr[] = {3, 7, 4, 8, 2, 6, 1}

Input: arr[] = {1, 4, 3, 2} 
Output: arr[] = {1, 4, 2, 3}

A Simple Solution is to first sort the array. After sorting, exclude the first element, swap the remaining elements in pairs. (i.e. keep arr[0] as it is, swap arr[1] and arr[2], swap arr[3] and arr[4], and so on). 
Time complexity: O(N log N) since we need to sort the array first.

We can convert in O(n) time using an efficient approach. The idea is to use a modified one pass of bubble sort.

• Maintain a flag for representing which order(i.e. ) currently we need.
• If the current two elements are not in that order then swap those elements otherwise not.

Let us see the main logic using three consecutive elements A, B, C.

Suppose we are processing B and C currently and the current relation is ‘ C. Since current relation is ‘‘ i.e., A must be greater than B. So, the relation is A > B and B > C. We can deduce A > C. So if we swap B and C then the relation is A > C and C A C B 
Refer this for more explanation.

Below image is a dry run of the above approach:

Below is the implementation of above approach:

Java

// Java program to sort an array
// in Zig-Zag form
import java.util.Arrays;

class Test
{
    static int arr[] = new int[]{4, 3, 7,
                                 8, 6, 2, 1};

    // Method for zig-zag conversion
    // of array
    static void zigZag()
    {
        // Flag true indicates relation "<"
        // is expected, else ">" is expected.
        // The first expected relation is "<"
        boolean flag = true;

        int temp =0;

        for (int i=0; i<=arr.length-2; i++)
        {
            // "<" relation expected
            if (flag)
            {
                /* If we have a situation like
                   A > B > C, we get A > B < C
                   by swapping B and C */
                if (arr[i] > arr[i+1])
                {
                    // swap
                    temp = arr[i];
                    arr[i] = arr[i+1];
                    arr[i+1] = temp;
                }

            }

            // ">" relation expected
            else
            {
                /* If we have a situation like
                   A < B < C, we get A < C > B
                   by swapping B and C */
                if (arr[i] < arr[i+1])
                {
                    // swap
                    temp = arr[i];
                    arr[i] = arr[i+1];
                    arr[i+1] = temp;
                }
            }

            // flip flag
            flag = !flag;
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        zigZag();
        System.out.println(Arrays.toString(arr));
    }
}

Output:

3  7  4  8  2  6  1 

Time complexity: O(n) 
Auxiliary Space: O(1) 

Please refer complete article on Convert array into Zig-Zag fashion for more details!