Sort an array in wave form

2.6

Given an unsorted array of integers, sort the array into a wave like array. An array ‘arr[0..n-1]’ is sorted in wave form if arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4] >= …..

Examples:

 Input:  arr[] = {10, 5, 6, 3, 2, 20, 100, 80}
 Output: arr[] = {10, 5, 6, 2, 20, 3, 100, 80} OR
                 {20, 5, 10, 2, 80, 6, 100, 3} OR
                 any other array that is in wave form

 Input:  arr[] = {20, 10, 8, 6, 4, 2}
 Output: arr[] = {20, 8, 10, 4, 6, 2} OR
                 {10, 8, 20, 2, 6, 4} OR
                 any other array that is in wave form

 Input:  arr[] = {2, 4, 6, 8, 10, 20}
 Output: arr[] = {4, 2, 8, 6, 20, 10} OR
                 any other array that is in wave form

 Input:  arr[] = {3, 6, 5, 10, 7, 20}
 Output: arr[] = {6, 3, 10, 5, 20, 7} OR
                 any other array that is in wave form

A Simple Solution is to use sorting. First sort the input array, then swap all adjacent elements.

For example, let the input array be {3, 6, 5, 10, 7, 20}. After sorting, we get {3, 5, 6, 7, 10, 20}. After swapping adjacent elements, we get {5, 3, 7, 6, 20, 10}.

Below are implementations of this simple approach.

C++

// A C++ program to sort an array in wave form using
// a sorting function
#include<iostream>
#include<algorithm>
using namespace std;

// A utility method to swap two numbers.
void swap(int *x, int *y)
{
    int temp = *x;
    *x = *y;
    *y = temp;
}

// This function sorts arr[0..n-1] in wave form, i.e., 
// arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4] >= arr[5]..
void sortInWave(int arr[], int n)
{
    // Sort the input array
    sort(arr, arr+n);

    // Swap adjacent elements
    for (int i=0; i<n-1; i += 2)
        swap(&arr[i], &arr[i+1]);
}

// Driver program to test above function
int main()
{
    int arr[] = {10, 90, 49, 2, 1, 5, 23};
    int n = sizeof(arr)/sizeof(arr[0]);
    sortInWave(arr, n);
    for (int i=0; i<n; i++)
       cout << arr[i] << " ";
    return 0;
}

Python

# Python function to sort the array arr[0..n-1] in wave form,
# i.e., arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4] >= arr[5]
def sortInWave(arr, n):
    
    #sort the array
    arr.sort()
   
    # Swap adjacent elements
    for i in range(0,n-1,2):
        arr[i], arr[i+1] = arr[i+1], arr[i]

# Driver progrM
arr = [10, 90, 49, 2, 1, 5, 23]
sortInWave(arr, len(arr))
for i in range(0,len(arr)):
    print arr[i],
    
# This code is contributed by __Devesh Agrawal__

Java

// Java implementation of naive method for sorting
// an array in wave form.
import java.util.*;

class SortWave
{
    // A utility method to swap two numbers.
    void swap(int arr[], int a, int b)
    {
        int temp = arr[a];
        arr[a] = arr[b];
        arr[b] = temp;
    }

    // This function sorts arr[0..n-1] in wave form, i.e.,
    // arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4]..
    void sortInWave(int arr[], int n)
    {
        // Sort the input array
        Arrays.sort(arr);

        // Swap adjacent elements
        for (int i=0; i<n-1; i += 2)
            swap(arr, i, i+1);
    }

    // Driver method
    public static void main(String args[])
    {
        Test ob = new Test();
        int arr[] = {10, 90, 49, 2, 1, 5, 23};
        int n = arr.length;
        ob.sortInWave(arr, n);
        for (int i : arr)
            System.out.print(i + " ");
    }
}
/*This code is contributed by Rajat Mishra*/


Output:
2 1 10 5 49 23 90

The time complexity of the above solution is O(nLogn) if a O(nLogn) sorting algorithm like Merge Sort, Heap Sort, .. etc is used.

This can be done in O(n) time by doing a single traversal of given array. The idea is based on the fact that if we make sure that all even positioned (at index 0, 2, 4, ..) elements are greater than their adjacent odd elements, we don’t need to worry about odd positioned element. Following are simple steps.
1) Traverse all even positioned elements of input array, and do following.
….a) If current element is smaller than previous odd element, swap previous and current.
….b) If current element is smaller than next odd element, swap next and current.

Below are implementations of above simple algorithm.

C++

// A O(n) program to sort an input array in wave form
#include<iostream>
using namespace std;

// A utility method to swap two numbers.
void swap(int *x, int *y)
{
    int temp = *x;
    *x = *y;
    *y = temp;
}

// This function sorts arr[0..n-1] in wave form, i.e., arr[0] >= 
// arr[1] <= arr[2] >= arr[3] <= arr[4] >= arr[5] ....
void sortInWave(int arr[], int n)
{
    // Traverse all even elements
    for (int i = 0; i < n; i+=2)
    {
        // If current even element is smaller than previous
        if (i>0 && arr[i-1] > arr[i] )
            swap(&arr[i], &arr[i-1]);

        // If current even element is smaller than next
        if (i<n-1 && arr[i] < arr[i+1] )
            swap(&arr[i], &arr[i + 1]);
    }
}

// Driver program to test above function
int main()
{
    int arr[] = {10, 90, 49, 2, 1, 5, 23};
    int n = sizeof(arr)/sizeof(arr[0]);
    sortInWave(arr, n);
    for (int i=0; i<n; i++)
       cout << arr[i] << " ";
    return 0;
}

Python

# Python function to sort the array arr[0..n-1] in wave form,
# i.e., arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4] >= arr[5]
def sortInWave(arr, n):
    
    # Traverse all even elements
    for i in range(0, n, 2):
        
        # If current even element is smaller than previous
        if (i> 0 and arr[i] < arr[i-1]):
            arr[i],arr[i-1] = arr[i-1],arr[i]
        
        # If current even element is smaller than next
        if (i < n-1 and arr[i] < arr[i+1]):
            arr[i],arr[i+1] = arr[i+1],arr[i]

# Driver program
arr = [10, 90, 49, 2, 1, 5, 23]
sortInWave(arr, len(arr))
for i in range(0,len(arr)):
    print arr[i],
    
# This code is contributed by __Devesh Agrawal__

Java

// A O(n) Java program to sort an input array in wave form
class SortWave
{
    // A utility method to swap two numbers.
    void swap(int arr[], int a, int b)
    {
        int temp = arr[a];
        arr[a] = arr[b];
        arr[b] = temp;
    }

    // This function sorts arr[0..n-1] in wave form, i.e.,
    // arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4]....
    void sortInWave(int arr[], int n)
    {
        // Traverse all even elements
        for (int i = 0; i < n; i+=2)
        {
            // If current even element is smaller
            // than previous
            if (i>0 && arr[i-1] > arr[i] )
                swap(arr, i-1, i);

            // If current even element is smaller
            // than next
            if (i<n-1 && arr[i] < arr[i+1] )
                swap(arr, i, i + 1);
        }
    }

    // Driver program to test above function
    public static void main(String args[])
    {
        SortWave ob = new SortWave();
        int arr[] = {10, 90, 49, 2, 1, 5, 23};
        int n = arr.length;
        ob.sortInWave(arr, n);
        for (int i : arr)
            System.out.print(i+" ");
    }
}
/*This code is contributed by Rajat Mishra*/

Output:

90 10 49 1 5 2 23

Asked in: Amazon, Goldman Sachs, Paytm

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

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