A divide and conquer based approach is being used to count far away an array is from completely being sorted. A modification in merge sort helps in counting the number of inversions in O(N log N) time. Consider the following array of sumber of inversions, number inversions is a metric of how mu elements 8, 10, 15, 9 the element 9 should have come before 10 therefore (10, 9) is an inversion, similarty (15, 9) is also an inversion, these are the only two inversions present in this array. int mergeSort(int arri], int templ], int left, int right) if (right > left) mid = (right + left) / 2; left_inv_count = mergeSort(arr, temp, left, mid); right_inv_count=mergeSort(arr, temp, mid +1, right); merge_count-merge(arr, temp, left, mid +1, right); inv_count= J/Blank A return inv_count; int merge(int arrl], int temp[], int left, int mid, int right) int i, j, k; int inv_count = 0; j=left:
A divide and conquer based approach is being used to count far away an array is from completely being sorted. A modification in merge sort helps in counting the number of inversions in O(N log N) time. Consider the following array of sumber of inversions, number inversions is a metric of how mu elements 8, 10, 15, 9 the element 9 should have come before 10 therefore (10, 9) is an inversion, similarty (15, 9) is also an inversion, these are the only two inversions present in this array. int mergeSort(int arri], int templ], int left, int right) if (right > left) mid = (right + left) / 2; left_inv_count = mergeSort(arr, temp, left, mid); right_inv_count=mergeSort(arr, temp, mid +1, right); merge_count-merge(arr, temp, left, mid +1, right); inv_count= J/Blank A return inv_count; int merge(int arrl], int temp[], int left, int mid, int right) int i, j, k; int inv_count = 0; j=left:
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Do not run the code. Don't copy from chegg it is wrong there ,
solve if you had knowldege else i dislike
Give me the proper explanation what to be filled at Blank A And Blank B
![k= left;
while (i<mid - 1) && (j<= right))
if (arrli) <= arrij])
temp[k+] = arrli++);
else
{
temp[k++] = arrij++];
inv_count =
J/Blank B
while (i<=mid - 1)
temp[k++] = arr[i+];
while (j <= right)
temp[k++] = arrij++];
for (i = left; ieright; i++)
arr(i) = temp[i);
return inv_count;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3316840f-2811-49a3-93c3-908592b89a60%2F1d4929d9-44ed-4f95-863f-a2d861d733a7%2Fbo1f5ih_processed.jpeg&w=3840&q=75)
Transcribed Image Text:k= left;
while (i<mid - 1) && (j<= right))
if (arrli) <= arrij])
temp[k+] = arrli++);
else
{
temp[k++] = arrij++];
inv_count =
J/Blank B
while (i<=mid - 1)
temp[k++] = arr[i+];
while (j <= right)
temp[k++] = arrij++];
for (i = left; ieright; i++)
arr(i) = temp[i);
return inv_count;
![A divide and conquer based approach is being used to count the number of inversions, the number of inversions is a metric of how much
far away an array is from completely being sorted. A modification in merge sort helps in counting the number of inversions in O(N log N) time. Consider the following array of
elements 8, 10, 15, 9 the element 9 should have come before 10 therefore (10, 9) is an inversion, similarty (15, 9) is also an inversion, these are the only two inversions present in this
array.
int mergeSort(int arr], int temp[], int left, int right)
if (right > left)
mid = (right + left) / 2;
left_inv_count = mergeSort(arr, temp, left, mid);
right_inv_count = mergeSort(arr, temp, mid + 1, right);
merge_count=merge(arr, temp, left, mid + 1, right);
inv_count=
J/Blank A
return inv_count;
int merge(int arr(], int temp[], int left, int mid, int right)
int i, j, k;
int inv_count = 0;
i= left;
j= mid;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3316840f-2811-49a3-93c3-908592b89a60%2F1d4929d9-44ed-4f95-863f-a2d861d733a7%2F6vxwgfa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A divide and conquer based approach is being used to count the number of inversions, the number of inversions is a metric of how much
far away an array is from completely being sorted. A modification in merge sort helps in counting the number of inversions in O(N log N) time. Consider the following array of
elements 8, 10, 15, 9 the element 9 should have come before 10 therefore (10, 9) is an inversion, similarty (15, 9) is also an inversion, these are the only two inversions present in this
array.
int mergeSort(int arr], int temp[], int left, int right)
if (right > left)
mid = (right + left) / 2;
left_inv_count = mergeSort(arr, temp, left, mid);
right_inv_count = mergeSort(arr, temp, mid + 1, right);
merge_count=merge(arr, temp, left, mid + 1, right);
inv_count=
J/Blank A
return inv_count;
int merge(int arr(], int temp[], int left, int mid, int right)
int i, j, k;
int inv_count = 0;
i= left;
j= mid;
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