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Sort an array in wave form

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  • Difficulty Level : Medium
  • Last Updated : 26 Jul, 2022

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] >= …..

what is a wave array?

well, you have seen waves right? how do they look? if you will form a graph of them it would be some in some up-down fashion.

that is what you have to do here, you are supposed to arrange numbers in such a way that if we will form a graph it will be in an up-down fashion rather than a straight line.

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
                 
                 here you can see {10, 5, 6, 2, 20, 3, 100, 80} first element is larger than the second and the same
                 thing is repeated again and again. large element - small element-large element -small element
                 and so on .
                 it can be small element-larger element - small element-large element -small element too.
                 all you need to maintain is the up-down fashion which represents a wave.
                  there can be multiple answers.
                 

 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
 
Recommended Practice

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 is implementation of the above idea. 

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;
}

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[])
    {
        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*/

Python3




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

C#




// C# implementation of naive method
// for sorting an array in wave form.
using System;
 
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
        Array.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()
    {
        SortWave ob = new SortWave();
        int[] arr = { 10, 90, 49, 2, 1, 5, 23 };
        int n = arr.Length;
         
        ob.sortInWave(arr, n);
        for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
    }
}
 
// This code is contributed by vt_m.

Javascript




<script>
 
// A JavaScript program to sort an array
// in wave form using a sorting function
 
// A utility method to swap two numbers.
function swap(arr, x, y)
{
    let temp = arr[x];
    arr[x] = arr[y];
    arr[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]..
function sortInWave(arr, n)
{
     
    // Sort the input array
    arr.sort((a, b) => a - b);
 
    // Swap adjacent elements
    for(let i = 0; i < n - 1; i += 2)
        swap(arr, i, i + 1);
}
 
// Driver code
let arr = [ 10, 90, 49, 2, 1, 5, 23 ];
let n = arr.length;
 
sortInWave(arr, n);
 
for(let i = 0; i < n; i++)
    document.write(arr[i] + " ");
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output: 

2 1 10 5 49 23 90

Time Complexity: O(nlogn)

Auxiliary Space: O(1)
The time complexity of the above solution is O(nLogn) if an O(nLogn) sorting algorithm like Merge Sort, Heap Sort, etc are used.
This can be done in O(n) time by doing a single traversal of the 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 oddly positioned elements. 

The following are simple steps. 
1) Traverse all even positioned elements of the input array, and do the following. 
….a) If the current element is smaller than the previous odd element, swap previous and current. 
….b) If the current element is smaller than the next odd element, swap next and current.

Below is implementation of the above idea.  

C++14




// 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;
}

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-1; i+=2){
            //swap odd and even positions
            if(i > 0 && arr[i - 1] > arr[i])
              swap(arr, i, i+1);
              if(i < n-1 && arr[i + 1] > arr[i])
              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*/

Python3




# 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],end=" ")
     
# This code is contributed by __Devesh Agrawal__

C#




// A O(n) C# program to sort an
// input array in wave form
using System;
 
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()
    {
        SortWave ob = new SortWave();
        int[] arr = { 10, 90, 49, 2, 1, 5, 23 };
        int n = arr.Length;
         
        ob.sortInWave(arr, n);
        for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
    }
}
 
// This code is contributed by vt_m.

Javascript




<script>
 
// A O(n) program to sort
// an input array in wave form
 
// A utility method to swap two numbers.
function swap(arr, a, b)
    {
        let 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]....
function sortInWave( arr, n)
    {
        // Traverse all even elements
        for (let 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 code
     
    let arr = [10, 90, 49, 2, 1, 5, 23];
    let n = arr.length;
    sortInWave(arr, n);
    for (let i=0; i<n; i++)
      document.write(arr[i] + " ");
     
</script>

Output: 

90 10 49 1 5 2 23

Time Complexity: O(n)

Auxiliary Space: O(1)

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


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