Implement Quicksort with first element as pivot

QuickSort is a Divide and Conquer algorithm. It picks an element as a pivot and partitions the given array around the pivot. There are many different versions of quickSort that pick the pivot in different ways. 

• Always pick the first element as a pivot.
• Always pick the last element as a pivot.
• Pick a random element as a pivot.
• Pick the median as the pivot.

Note: Here we will be implementing quick sort by picking the first element as the pivot.

The key function in quick sort is a partition. The target of partitions is to put the pivot in its correct position if the array is sorted and the smaller (or equal) to its left and higher elements to its right and do all this in linear time.

Partition Algorithm:

There can be many ways to do partition, the following pseudo-code adopts the method given in the CLRS book.

• We start from the leftmost element and keep track of the index of smaller (or equal) elements as i. 
• While traversing, if we find a smaller (or equal) element, we swap the current element with arr[i]. 
• Otherwise, we ignore the current element.

Pseudo Code for recursive QuickSort function:

// low  –> Starting index,  high  –> Ending index

quickSort(arr[], low, high) {
    if (low < high) {

        // pi is partitioning index, arr[pi] is now at right place
        pi = partition(arr, low, high);
        quickSort(arr, low, pi – 1);  // Before pi
        quickSort(arr, pi + 1, high); // After pi
    }
}

Pseudo code for partition() function

/* This function takes first element as pivot, places the pivot element at its correct position in sorted array, and places all smaller (smaller than or equal to pivot) to left of pivot and all greater elements to right of pivot */

partition (arr[], low, high) {
    // first element as pivot
    pivot = arr[low]
    k = high
    for (i = high; i > low; i--) {
        if (arr[i] > pivot){
            swap arr[i] and arr[k];
            k--;
        }
    }
    swap arr[k] and arr[low]
    return k;
}

Illustration of partition() : 

Consider: arr[] = { 7,   6,   10,   5,   9,   2,   1,   15,   7 }

First Partition: low = 0, high = 8, pivot = arr[low] = 7
Initialize index of right most element k = high = 8.

• Traverse from i = high to low:
  • if arr[i] is greater than pivot:
    • Swap arr[i] and arr[k].
    • Decrement k;
• At the end swap arr[low] and arr[k].

Now the correct position of pivot is index 5

👁 First partition

Second Partition: low = 0, high = 4, pivot = arr[low] = 2
Similarly initialize k = high = 4; 

The correct position of 2 becomes index 1. And the left part is only one element and the right part has {6, 5, 7}.

👁 Partition of the left half

On the other hand partition happens on the segment [6, 8] i.e., the array {10, 9, 15}.
Here low = 6, high = 8, pivot = 10 and k = 8.

The correct position of 10 becomes index 7 and the right and left part both has only one element.

👁 Image

Third partition:  Here partition the segment {6, 5, 7}. The low = 2, high = 4, pivot = 6 and k = 4.
If the same process is applied, we get correct position of 6 as index 3 and the left and the right part is having only one element.

👁 Third partition

The total array becomes sorted in this way. Check the below image for the recursion tree

👁 Recursion tree for partitions

Follow the below steps to implement the approach.

• Use a recursive function (say quickSort) to initialize the function.
• Call the partition function to partition the array and inside the partition function do the following
  • Take the first element as pivot and initialize and iterator k = high.
  • Iterate in a for loop from i = high to low+1:
    • If arr[i] is greater than pivot then swap arr[i] and arr[k] and decrement k.
  • After the iteration is swap the pivot with arr[k].
  • Return k-1 as the point of partition.
• Now recursively call quickSort for the left half and right half of the partition index.

Implementation of the above approach.

Sorted array: 
1 2 5 6 7 7 9 10 15 

Complexity Analysis:

• Time Complexity:
  • Average Case: O(N * logN), where N is the length of the array.
  • Best Case: O(N * logN)
  • Worst Case: O(N²)
• Auxiliary Space: O(1)