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}
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.
Wave Array using sorting
A idea is to use sorting. First sort the input array, then swap all adjacent elements.
Follow the steps mentioned below to implement the idea:
- Sort the array.
- Traverse the array from index 0 to N-1, and increase the value of the index by 2.
- While traversing the array swap arr[i] with arr[i+1].
- Print the final array.
Below is the implementation of the above 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 functionint 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;} |
C
// A C program to sort an array in wave form using// a bubble sort function#include<stdio.h> // 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,n); // Swap adjacent elements for (int i=0; i<n-1; i += 2) swap(&arr[i], &arr[i+1]);}void Sort(int arr[], int n){ int i, j; for (i = 0; i < n - 1; i++) for (j = 0; j < n - i - 1; j++) if (arr[j] > arr[j + 1]) swap(&arr[j], &arr[j + 1]);} void swap(int *x, int *y){ int temp = *x; *x = *y; *y = temp;}// Driver program to test above functionint 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++) printf("%d \t",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 programarr = [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 codelet 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(N*log(N))
Auxiliary Space: O(1)
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:
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 functionint 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 formpublic 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 - 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 programarr = [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 formusing 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)
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