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; } |
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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*/ |
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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__ |
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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. |
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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; } |
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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*/ |
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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__ |
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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. |
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Output:
90 10 49 1 5 2 23
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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