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.
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
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.
Space Complexity: O(1) as no extra space has been used.
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.
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
Please refer complete article on Sort an array in wave form for more details!
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