Question

The zigzag sorting problem takes an array data of size n and outputs a per- mutation where data[1] <data[2] > data[3] = data[4] > data[5] S..., all the way to the end of the array. (In general data[i] = data[i+1] if i is odd and data[i] > data[i+1] if i is even.) Answer the following questions about developing algorithms for the zigzag sorting problem. 1. Give pseudocode for a brute force algorithm for zigzag sorting. Your pseudocode just needs to give the main idea for how to solve this problem using brute force; you do not need to provide code. 2. Give pseudocode for a divide-and-conquer algorithm for zigzag sorting. Hint: if you solve n = 2 as a base case, you can always divide n approxi- mately in half (n/2+1) so that the first half is always even. For example, 3 = 2 + 1, 4 = 2 + 2,5 = 2 + 3, 6 = 4 + 2,7 = 4 + 3,8 = 4 + 4, 9 = 4 + 5, 10 = 6 + 4, etc. You can do this by dividing by 2 and rounding down, then adding 1 if the result is odd (or rounding up and subtracting 1 if odd). 3. Analyze the complexity of your divide-and-conquer algorithm. You may assume that the #1 in the recursive call does not change the complexity of the algorithm.

Answer 1

Brute force approach solution:

// Program for zig-zag conversion of array

void zigZag(int arr[], int n) {

// Flag true indicates relation "<" is expected,

// else ">" is expected. The first expected relation

// is "<"

bool flag = true;

for (int i=0; i<=n-2; i++)

    {

        if (flag) /* "<" relation expected */

        {

            /* 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(arr[i], arr[i+1]);

        }

        else /* ">" relation expected */

        {

            /* 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(arr[i], arr[i+1]);

        }

        flag = !flag; /* flip flag */

    }

}

Complexity of above solution O(N) where N=total number of elements in array.

if we solve above problem using divinde and conquer still complexity would be O(N) as we need to traverse every elements of array.