Wave Array Optimized Approach
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.
Follow the steps mentioned below to implement the idea:
- Traverse all even positioned elements of the input array, and do the following.
- If the current element is smaller than the previous odd element, swap the previous and current.
- If the current element is smaller than the next odd element, swap next and current.
Below is the implementation of the above approach:
// A O(n) program to sort an input array in wave form
#include<iostream>
using namespace std;
// 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;
}
// A O(n) Java program to sort an input array in wave form
public 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*/
# 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 - 1, 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__
// 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.
<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)
Sort an array in wave form
Given an unsorted array of integers, sort the array into a wave 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}
Explanation:
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}