Convert 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 < b > c < d > e < f
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. < or >) 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 ‘<‘, but we have B > C. Since current relation is ‘<‘ previous relation must be ‘>’ 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 < B. Finally we get the desired order 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:

C++

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// C++ program to sort an array in Zig-Zag form 
#include <iostream> 
using namespace std; 
  
// 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 */
    
  
// Driver program 
int main() 
    int arr[] = {4, 3, 7, 8, 6, 2, 1}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    zigZag(arr, n); 
    for (int i=0; i<n; i++) 
        cout << arr[i] << " "
    return 0; 

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Java

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// 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++) 
        
            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 
                    temp = arr[i]; 
                    arr[i] = arr[i+1]; 
                    arr[i+1] = temp; 
                
                  
            
            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 
                    temp = arr[i]; 
                    arr[i] = arr[i+1]; 
                    arr[i+1] = temp; 
                
            
            flag = !flag; /* flip flag */
        
    
      
    // Driver method to test the above function 
    public static void main(String[] args) 
    
        zigZag(); 
        System.out.println(Arrays.toString(arr)); 
    

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Python

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# Python program to sort an array in Zig-Zag form 
  
# Program for zig-zag conversion of array 
def zigZag(arr, n): 
    # Flag true indicates relation "<" is expected, 
    # else ">" is expected. The first expected relation 
    # is "<" 
    flag = True
    for i in range(n-1): 
        # "<" relation expected 
        if flag is True
            # 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]: 
                arr[i],arr[i+1] = arr[i+1],arr[i] 
            # ">" 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]: 
                arr[i],arr[i+1] = arr[i+1],arr[i] 
        flag = bool(1 - flag) 
    print(arr) 
  
# Driver program 
arr = [4, 3, 7, 8, 6, 2, 1
n = len(arr) 
zigZag(arr, n) 
  
# This code is contributed by Pratik Chhajer 

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C#

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// C# program to sort an array in Zig-Zag form
using System;
  
class GFG{
      
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 "<"
    bool 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();
    foreach(int i in arr)
        Console.Write(i + " ");
}
}
  
// This code is contributed by amal kumar choubey

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Output: 

3  7  4  8  2  6  1 

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

This article is contributed by Siva Krishna Aleti. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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Improved By : Amal Kumar Choubey

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