Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid element or right element is encountered . 3) Compare list1[i] and list[j]. i) If list1[i]<=list2[j] c[k++]=list1[i++] else c[k++]=list2[j++] count = count + mid-i; 4) add rest elements of list1 and list2 in c. 5) copy sorted list c back to original list. 6) return count. void quickSort(int arr[], int left, int right) { int i = left, j = right; int tmp; int pivot = arr[(left + right) / 2]; /* partition */ while (i <= j) { while (arr[i] < pivot) i++; while (arr[j] > pivot) j--; if (i <= j) { tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; i++; j--; } }; /* recursion */ if (left < j) quickSort(arr, left, j); if (i < right) quickSort(arr, i, right); } Solution Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid elem.

1. Quicksort Algorithm:
Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two
smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort
the sub-arrays.
The steps are:
The base case of the recursion is arrays of size zero or one, which never need to be sorted.
The pivot selection and partitioning steps can be done in several different ways; the choice of
specific implementation schemes greatly affects the algorithm's performance.
This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In
this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the
problem in two parts. And then two parts are solved recursively. The key concept is two count
the number of inversion in merge procedure. In merge procedure we pass two sub-list. The
element is sorted and inversion is found as follows
a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n].
b)Conquer : Conquer the sub-problem by solving them recursively.
1) Set count=0,0,i=left,j=mid. C is the sorted list.
2) Traverse list1 and list2 until mid element or right element is encountered .
3) Compare list1[i] and list[j].
i) If list1[i]<=list2[j]
c[k++]=list1[i++]
else
c[k++]=list2[j++]
count = count + mid-i;
4) add rest elements of list1 and list2 in c.
5) copy sorted list c back to original list.
6) return count.
void quickSort(int arr[], int left, int right) {
int i = left, j = right;
int tmp;
int pivot = arr[(left + right) / 2];
/* partition */
while (i <= j) {
while (arr[i] < pivot)
i++;
while (arr[j] > pivot)

2. j--;
if (i <= j) {
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
i++;
j--;
}
};
/* recursion */
if (left < j)
quickSort(arr, left, j);
if (i < right)
quickSort(arr, i, right);
}
Solution
Quicksort Algorithm:
Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two
smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort
the sub-arrays.
The steps are:
The base case of the recursion is arrays of size zero or one, which never need to be sorted.
The pivot selection and partitioning steps can be done in several different ways; the choice of
specific implementation schemes greatly affects the algorithm's performance.
This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In
this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the
problem in two parts. And then two parts are solved recursively. The key concept is two count
the number of inversion in merge procedure. In merge procedure we pass two sub-list. The
element is sorted and inversion is found as follows
a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n].
b)Conquer : Conquer the sub-problem by solving them recursively.
1) Set count=0,0,i=left,j=mid. C is the sorted list.
2) Traverse list1 and list2 until mid element or right element is encountered .
3) Compare list1[i] and list[j].

3. i) If list1[i]<=list2[j]
c[k++]=list1[i++]
else
c[k++]=list2[j++]
count = count + mid-i;
4) add rest elements of list1 and list2 in c.
5) copy sorted list c back to original list.
6) return count.
void quickSort(int arr[], int left, int right) {
int i = left, j = right;
int tmp;
int pivot = arr[(left + right) / 2];
/* partition */
while (i <= j) {
while (arr[i] < pivot)
i++;
while (arr[j] > pivot)
j--;
if (i <= j) {
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
i++;
j--;
}
};
/* recursion */
if (left < j)
quickSort(arr, left, j);
if (i < right)
quickSort(arr, i, right);
}