Prerequisite : Tail Call Elimination
In QuickSort, partition function is in-place, but we need extra space for recursive function calls. A simple implementation of QuickSort makes two calls to itself and in worst case requires O(n) space on function call stack.
The worst case happens when the selected pivot always divides the array such that one part has 0 elements and other part has n-1 elements. For example, in below code, if we choose last element as pivot, we get worst case for sorted arrays (See this for visualization)
C
/* A Simple implementation of QuickSort that makes two two recursive calls. */void quickSort(int arr[], int low, int high){ if (low < high) { /* pi is partitioning index, arr[p] is now at right place */ int pi = partition(arr, low, high); // Separately sort elements before // partition and after partition quickSort(arr, low, pi - 1); quickSort(arr, pi + 1, high); }}// See below link for complete running code |
Java
// A Simple implementation of QuickSort that // makes two recursive calls. static void quickSort(int arr[], int low, int high){ if (low < high) { // pi is partitioning index, arr[p] is // now at right place int pi = partition(arr, low, high); // Separately sort elements before // partition and after partition quickSort(arr, low, pi - 1); quickSort(arr, pi + 1, high); }}// This code is contributed by rutvik_56 |
Can we reduce the auxiliary space for function call stack?
We can limit the auxiliary space to O(Log n). The idea is based on tail call elimination. As seen in the previous post, we can convert the code so that it makes one recursive call. For example, in the below code, we have converted the above code to use a while loop and have reduced the number of recursive calls.
C
/* QuickSort after tail call elimination using while loop */void quickSort(int arr[], int low, int high){ while (low < high) { /* pi is partitioning index, arr[p] is now at right place */ int pi = partition(arr, low, high); // Separately sort elements before // partition and after partition quickSort(arr, low, pi - 1); low = pi+1; }}// See below link for complete running code |
Although we have reduced number of recursive calls, the above code can still use O(n) auxiliary space in worst case. In worst case, it is possible that array is divided in a way that the first part always has n-1 elements. For example, this may happen when last element is choses as pivot and array is sorted in decreasing order.
We can optimize the above code to make a recursive call only for the smaller part after partition. Below is implementation of this idea.
Further Optimization :
C
/* This QuickSort requires O(Log n) auxiliary space in worst case. */void quickSort(int arr[], int low, int high){ while (low < high) { /* pi is partitioning index, arr[p] is now at right place */ int pi = partition(arr, low, high); // If left part is smaller, then recur for left // part and handle right part iteratively if (pi - low < high - pi) { quickSort(arr, low, pi - 1); low = pi + 1; } // Else recur for right part else { quickSort(arr, pi + 1, high); high = pi - 1; } }}// See below link for complete running code |
In the above code, if left part becomes smaller, then we make recursive call for left part. Else for the right part. In worst case (for space), when both parts are of equal sizes in all recursive calls, we use O(Log n) extra space.
Reference:
http://www.cs.nthu.edu.tw/~wkhon/algo08-tutorials/tutorial2b.pdf
This article is contributed by Dheeraj Jain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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