Question

please answer all parts and code thanks

. - PART 1-Introduction to Sorting, 21 points Use this array of integer for the problems 1A, 1B, and 1c: 9 57 8324761 Each of these problems is worth 3 points A. Show the contents of the array each time a selection sort changes it while sorting the array into B. Show the contents of the array each time an insertion sort changes it while sorting C. Show the contents of the array each time a Shell sort ascending order the array into ascending order changes it while sorting the array into ascending order D. 2 pointsSuppose you want to find the largest entry in an unsorted array of n entries. Algorithm A se entire array sequentially and records the largest entry seen so far. AlgorithmTheay as inieiaiay B sorts the array into descending order and then reports the first entry as the largest. Compare the time efficiency of the two approaches. E. 10 points Consider ann by n array of integer values. Write asn algorithm to sort the rows of the array by their first value Implement your algorithm. The code for this problem is provided in the Assignment-03-Code.tip archive. The azzay after sozting 123 45 Your output must be identical to the output to the right. 451 PART 2-Queues, Deques, and Priority Queues, 9 points After the following statements execute, what are the contents of the queue? QueueinterfacecString myQueue new LinkedQueue> myQueue.enqueuel"Jane"l: myQueue.enqueuel"Jess"l myQueue.enqueue("Jon") myQueue.enqueuelmyQueue.dequeue)) myQueue.enqueuelmyQueue.getFront): myQueue.enqueue("Jim") String name myQueue.dequeue) myQueue.enqueue(myQueue.getFrontÜ): Updated: 3/18/201973

Answer 1

Part 1 . <A>

Step 1 : 9,5,7,8,3,2,4,7,6,1

Step 2 : 5,7,8,3,2,4,7,6,1, 9

Step 3 : 5,7,3,2,4,7,6,1, 8,9

Step 4 : 5,3,2,4,7,6,1, 7,8,9

Step 5 : 5,3,2,4,6,1, 7,7,8,9

Step 6 : 5,3,2,4,1, 6,7,7,8,9

Step 7 : 3,2,4,1, 5,6,7,7,8,9

Step 8 : 3,2,1, 4,5,6,7,7,8,9

Step 9 : 2,1, 3,4,5,6,7,7,8,9

Step 10 : 1, 2,3,4,5,6,7,7,8,9

Step 11 : 1,2,3,4,5,6,7,7,8,9 Completed

Part 1 . <B>

Step 1 : 9,5,7,8,3,2,4,7,6,1

Step 2 : { 1 },{ 9,5,7,8,3,2,4,7,6 }

Step 3 : { 1,2 },{ 9,5,7,8,3,4,7,6 }

Step 4 : { 1,2,3 },{ 9,5,7,8,4,7,6 }

Step 5 : { 1,2,3,4 },{ 9,5,7,8,7,6 }

Step 6 : { 1,2,3,4,5 },{ 9,7,8,7,6 }

Step 7 : { 1,2,3,4,5,6 },{ 9,7,8,7 }

Step 8 :  { 1,2,3,4,5,6,7 },{ 9,8,7 }

Step 9 : { 1,2,3,4,5,6,7,7 },{ 9,8 }

Step 10 : { 1,2,3,4,5,6,7,7,8 },{ 9 }

Step 11 : { 1,2,3,4,5,6,7,7,8,9 } Completed

Part 1 . <C>

Step 1 : 9,5,7,8,3,2,4,7,6,1

Step 2 : 2,5,7,8,3, 9,4,7,6,1 compare 9,2 and sort

Step 3 : 2,4,7,8,3, 9,5,7,6,1 compare 5,4 and sort

Step 4 : 2,4,7,8,3, 9,5,7,6,1 compare 7,7 and sort

Step 5 : 2,4,7,6,3, 9,5,7,8,1 compare 8,6 and sort

Step 6 : 2,4,7,6,1, 9,5,7,8,3 compare 3,1 and sort

Step 7 : 1,4,7,6, 2,9,5,7, 8,3 compare 2,1,8 and sort

Step 8 :  1,4,7,6, 2,3,5,7, 8,9 compare 4,9,3 and sort

Step 9 : 1,4,5,6, 2,3,7,7, 8,9 compare 7,5 and sort

Step 10 : 1,4,5,6, 2,3,7,7, 8,9 compare 6,7 and sort

Likewise we will get { 1,2,3,4,5,6,7,7,8,9 } Completed

Part 1 . <D>

Here the algorithm A is just searching in the list but the algorithm B is focusing over the sorting and then searching so as algorithm A includes only linear search so its very fast as comparison to the algorithm B.

Part 1 . <E>

This is the c language code for given problem

#include <stdio.h>

int main()
{
int i,j,c,d,swap,x;
int n=5;
int position=0;
printf("Here the program starts: \n");
int disp[5][5] = {
{1,2,3,4,5},
{3,4,5,1,2},
{5,2,3,4,1},
{2,3,1,4,5},
{4,2,3,1,5}
};
for(i=0; i<5; i++)
{
for(j=0;j<5;j++)
{
printf("%d ", disp[i][j]);
}
printf("\n");
}
for (c = 0; c < (n - 1); c++)
{
position = c;

for (d = c + 1; d < n; d++)
{
if (disp[position][0] > disp[d][0])
position = d;
}
if (position != c)
{
for(x=0;x<n;x++)
{
swap = disp[c][x];
disp[c][x] = disp[position][x];
disp[position][x] = swap;
};
}
}
printf("Here the array is sorted : \n");
for(i=0; i<5; i++)
{
for(j=0;j<5;j++)
{
printf("%d ", disp[i][j]);
}
printf("\n");
}
return 0;
}

Part 2 . <A>

Here output be like

step 1 : first queue starts

step 2 : queue : Jane

step 3 : queue : Jane, Jess

step 4 : queue : Jane, Jess, Jon

step 5 : Jane removed from queue

step 6 : Jess will be printed on screen

step 7 : Jim added to queue and its like queue : Jess, Jon, Jim

step 8 : jess removed from queue and gone to my string

step 9 : Jess will be printed on screen Jon