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# quicksort

pranav1-program

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

Always pick first element as pivot.
Always pick last element as pivot (implemented below)
Pick a random element as pivot.
Pick median as pivot.
The key process in quickSort is partition(). Target of partitions is, given an array and an element x of array as pivot, put x at its correct position in sorted array and put all smaller elements (smaller than x) before x, and put all greater elements (greater than x) after x. All this should be done in linear time.

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
}


}
quicksort

Partition Algorithm
There can be many ways to do partition, following pseudo code adopts the method given in CLRS book. The logic is simple, we start from the leftmost element and keep track of index of smaller (or equal to) elements as i. While traversing, if we find a smaller element, we swap current element with arr[i]. Otherwise we ignore current element.

/* 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()

/* This function takes 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 */
partition (arr[], low, high)
{
// pivot (Element to be placed at right position)
pivot = arr[high];

i = (low - 1)  // Index of smaller element and indicates the
// right position of pivot found so far

for (j = low; j <= high- 1; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;    // increment index of smaller element
swap arr[i] and arr[j]
}
}
swap arr[i + 1] and arr[high])
return (i + 1)


}
Illustration of partition() :

arr[] = {10, 80, 30, 90, 40, 50, 70}
Indexes: 0 1 2 3 4 5 6

low = 0, high = 6, pivot = arr[h] = 70
Initialize index of smaller element, i = -1

Traverse elements from j = low to high-1
j = 0 : Since arr[j] <= pivot, do i++ and swap(arr[i], arr[j])
i = 0
arr[] = {10, 80, 30, 90, 40, 50, 70} // No change as i and j
// are same

j = 1 : Since arr[j] > pivot, do nothing
// No change in i and arr[]

j = 2 : Since arr[j] <= pivot, do i++ and swap(arr[i], arr[j])
i = 1
arr[] = {10, 30, 80, 90, 40, 50, 70} // We swap 80 and 30

j = 3 : Since arr[j] > pivot, do nothing
// No change in i and arr[]

j = 4 : Since arr[j] <= pivot, do i++ and swap(arr[i], arr[j])
i = 2
arr[] = {10, 30, 40, 90, 80, 50, 70} // 80 and 40 Swapped
j = 5 : Since arr[j] <= pivot, do i++ and swap arr[i] with arr[j]
i = 3
arr[] = {10, 30, 40, 50, 80, 90, 70} // 90 and 50 Swapped

We come out of loop because j is now equal to high-1.
Finally we place pivot at correct position by swapping
arr[i+1] and arrhigh
arr[] = {10, 30, 40, 50, 70, 90, 80} // 80 and 70 Swapped

Now 70 is at its correct place. All elements smaller than
70 are before it and all elements greater than 70 are after
it.
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Implementation:
Following are the implementations of QuickSort:

/* C++ implementation of QuickSort */

# include

using namespace std;

// A utility function to swap two elements
void swap(int* a, int* b)
{
int t = *a;
*a = *b;
*b = t;
}

/* This function takes 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 */
int partition (int arr[], int low, int high)
{
int pivot = arr[high]; // pivot
int i = (low - 1); // Index of smaller element and indicates the right position of pivot found so far

for (int j = low; j <= high - 1; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++; // increment index of smaller element
swap(&arr[i], &arr[j]);
}
}
swap(&arr[i + 1], &arr[high]);
return (i + 1);


}

/* The main function that implements QuickSort
arr[] --> Array to be sorted,
low --> Starting index,
high --> Ending index /
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);
}


}

/* Function to print an array */
void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
cout << arr[i] << " ";
cout << endl;
}

// Driver Code
int main()
{
int arr[] = {10, 7, 8, 9, 1, 5};
int n = sizeof(arr) / sizeof(arr[0]);
quickSort(arr, 0, n - 1);
cout << "Sorted array: \n";
printArray(arr, n);
return 0;
}

// This code is contributed by rathbhupendra
Output
Sorted array:
1 5 7 8 9 10

Best Case: The best case occurs when the partition process always picks the middle element as pivot. Following is recurrence for best case.

T(n) = 2T(n/2) + \theta(n)
The solution of above recurrence is \theta (nLogn). It can be solved using case 2 of Master Theorem.