We sometimes use arrays to represent sets. For example S = {\'b\', \'f, \'z\', \

Question

We sometimes use arrays to represent sets. For example S = {'b', 'f, 'z', 'a'} is represented by the following array: You can use zero as a character, 'O', when an entry does not have any data. For example if we store S in an array of size 6: You can assume that arrays have a maximum size: 30 Write a C program that takes two arrays containing alphabetical characters and computes the following operations: Union of the two arrays and output the result in a new array Intersection of two arrays and output the result in a new array, Is a set empty? Pleases note that you would an array to store the result that has twice the dimension of each array. If your input arrays have a dimension n, the resulting array should have 2n dimension. (a) Write a pseudocode for this program (b) Implement your pseudocode in C. (c) Test You program using the following code: a. Input two sets where all the elements are the same. b. Input two sets that have different elements c. One input empty set.

Explanation / Answer

#include<stdio.h>

/* Function prints union of arr1[] and arr2[]
m is the number of elements in arr1[]
n is the number of elements in arr2[] */
int printUnion(int arr1[], int arr2[], int m, int n)
{
int i = 0, j = 0;
while (i < m && j < n)
{
   if (arr1[i] < arr2[j])
   printf(" %d ", arr1[i++]);
   else if (arr2[j] < arr1[i])
   printf(" %d ", arr2[j++]);
   else
   {
   printf(" %d ", arr2[j++]);
   i++;
   }
}

/* Print remaining elements of the larger array */
while(i < m)
printf(" %d ", arr1[i++]);
while(j < n)
printf(" %d ", arr2[j++]);
}

/* Driver program to test above function */
int main()
{
int arr1[] = {1, 2, 4, 5, 6};
int arr2[] = {2, 3, 5, 7};
int m = sizeof(arr1)/sizeof(arr1[0]);
int n = sizeof(arr2)/sizeof(arr2[0]);
printUnion(arr1, arr2, m, n);
getchar();
return 0;
}

#include<stdio.h>

/* Function prints Intersection of arr1[] and arr2[]
m is the number of elements in arr1[]
n is the number of elements in arr2[] */
int printIntersection(int arr1[], int arr2[], int m, int n)
{
int i = 0, j = 0;
while (i < m && j < n)
{
   if (arr1[i] < arr2[j])
   i++;
   else if (arr2[j] < arr1[i])
   j++;
   else /* if arr1[i] == arr2[j] */
   {
   printf(" %d ", arr2[j++]);
   i++;
   }
}
}

/* Driver program to test above function */
int main()
{
int arr1[] = {1, 2, 4, 5, 6};
int arr2[] = {2, 3, 5, 7};
int m = sizeof(arr1)/sizeof(arr1[0]);
int n = sizeof(arr2)/sizeof(arr2[0]);
printIntersection(arr1, arr2, m, n);
getchar();
return 0;
}

If input array is of n dimension then rsulting array is of 2n dimension:

Source Code:

#include <stdio.h>

/* Assuming -1 is filled for the places where element
is not available */
#define NA -1

/* Function to move m elements at the end of array mPlusN[] */
void moveToEnd(int mPlusN[], int size)
{
int i = 0, j = size - 1;
for (i = size-1; i >= 0; i--)
   if (mPlusN[i] != NA)
   {
   mPlusN[j] = mPlusN[i];
   j--;
   }
}

/* Merges array N[] of size n into array mPlusN[]
of size m+n*/
int merge(int mPlusN[], int N[], int m, int n)
{
int i = n; /* Current index of i/p part of mPlusN[]*/
int j = 0; /* Current index of N[]*/
int k = 0; /* Current index of of output mPlusN[]*/
while (k < (m+n))
{
   /* Take an element from mPlusN[] if
   a) value of the picked element is smaller and we have
       not reached end of it
   b) We have reached end of N[] */
   if ((i < (m+n) && mPlusN[i] <= N[j]) || (j == n))
   {
   mPlusN[k] = mPlusN[i];
   k++;
   i++;
   }
   else // Otherwise take element from N[]
   {
   mPlusN[k] = N[j];
   k++;
   j++;
   }
}
}

/* Utility that prints out an array on a line */
void printArray(int arr[], int size)
{
int i;
for (i=0; i < size; i++)
   printf("%d ", arr[i]);

printf(" ");
}

/* Driver function to test above functions */
int main()
{
/* Initialize arrays */
int mPlusN[] = {2, 8, NA, NA, NA, 13, NA, 15, 20};
int N[] = {5, 7, 9, 25};
int n = sizeof(N)/sizeof(N[0]);
int m = sizeof(mPlusN)/sizeof(mPlusN[0]) - n;

/*Move the m elements at the end of mPlusN*/
moveToEnd(mPlusN, m+n);

/*Merge N[] into mPlusN[] */
merge(mPlusN, N, m, n);

/* Print the resultant mPlusN */
printArray(mPlusN, m+n);

return 0;
}

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