Question

Write a nonrecursive Quicksort algorithm. Analyze your algorithm, and show the results using order notation. Note that it will be necessary to explicitly maintain a stack in your algorithm.

Short answer

The nonrecursive QuickSort algorithm involves initializing a stack to manage start and end indices of partitions, defining a partition process which involves choosing a pivot to sort smaller and larger elements accordingly, followed by an iterative process that loops until all partitions have been sorted which signifies an empty stack. The time complexity of this algorithm is \(O(n \log{n})\) in the best case scenario and \(O(n^2)\) in the worst case scenario.

Step-by-step solution

QuickSort Algorithm

Start by setting up a stack data structure, which will be used to store the start and end indices of partitions. The algorithm begins by pushing the first range, [0, array.length-1], onto the stack. Then the algorithm enters a loop that continues until the stack is empty. In each iteration, it pops the top range off of the stack, partitions that range, then pushes the two resulting ranges back onto the stack unless they are empty.

Partition Process

Select a pivot randomly within the range. Then, make a pass through the array to move smaller elements to the left and larger elements to the right. The partition function should address the range [start, end]. The function takes the last element as pivot, places the pivot element at its correct position in sorted array, and places all smaller (smaller than pivot) to left of pivot and all greater elements to right of pivot. In the end, return the index of the pivot.

Iterative Process

For the next iteration, two ranges will be pushed onto the stack: [start, pivot-1] and [pivot+1, end] assuming that pivot is the index of the partition. This process will continue until the stack is empty, and since each time we push onto the stack, we’re reducing the size of the range, we can be sure that it will eventually empty out.

Time Complexity Analysis

In the best case, each time we perform a partition, we divide the list into two nearly equal pieces. So each time we perform a partition we reduce the problem size by about half. Hence, the best case time complexity is \(O(n \log{n})\). In the worst case the smallest or largest element becomes pivot so the partition process takes \(O(n^2)\) time. However, the worst case scenario is highly unlikely when we use random pivots.