Hello Everyone,
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.
// 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;
}
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.
- 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.
// 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;
}
Output:
90 10 49 1 5 2 23