Problem Solving: Difference between Sets (Using Functions)

Learn to calculate the difference between sets using the functions.

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main.cpp
Sets.txt
void loadSet(const char Msg[], int S[ ], int &size, ifstream &rdr);
void sort(int S[ ], int size);
void setPrint(const char Msg[ ], int S[ ], int size);
bool isPresent(int S[ ], int size, int t);
/*
	Assume you have the implementation of the above 4 functions.
*/
void setMinus()  // Add parameter list
{
 // Add code here.
}
int main()
{
    const int capacity = 500;
	int A[capacity] = { 0 }, B[capacity] = { 0 }, sizeA, sizeB;
	ifstream rdr("Sets.txt");
	cout << "Size of Set A = ";
	rdr >> sizeA; cout << sizeA<<"\n";
	cout << "Size of Set B = ";
	rdr >> sizeB; cout << sizeA<<"\n";
	loadSet("A",A, sizeA, rdr);
    loadSet("B",B, sizeB, rdr);
	
	//Displaying the two sets
	setPrint("A", A, sizeA);
	setPrint("B", B, sizeB);
	int BMinusA[500], AMinusB[500];
	int minusSize;
	
	// Write code for computing A-B
	// Printing A-B result
	// Write code for computing B-A
	// Printing B-A
	
	return 0;
}

Difference between sets

Let’s make two functions:

void setMinus(int A1[ ], int sizeA1,int A2[ ], int sizeA2,int R[ ], int &sizeR)

In the setMinus function, we have the following six parameters:

  • The first parameter is set A1[].
  • The second parameter is set sizeA1.
  • The third parameter is set A2.
  • The fourth parameter is set sizeA2.
  • The fifth parameter is R[], which stores the result of difference (the elements of A1 that are not present in set A2).
  • The last parameter is the size of the R[] set, sizeR, which will be passed by reference.

This function will call the isPresent() function and check if an element A1[ai] of set A1 is not present in set A2. If not present, it’ll simply add it into R[] and increment the count maintained in mi.

Let’s compare the two implementations:

The previous implementation:

// A-B algorithm
int AMinusB[500];
bool found;
int mi = 0;
for (int ai = 0; ai <= sizeA - 1; ai++)
{
    found = false;
    for (int bi = 0; bi <= sizeB - 1; bi++)
    {
        if (A[ai] == B[bi])
        {
            found = true; break;
        }
    }
    if (!found)
    {
        AMinusB[mi] = A[ai];
        mi++;
    }
}

// **Set B-A algorithm**
int BMinusA[500];
mi = 0;
bool present;
for (int bi = 0; bi <= sizeB - 1; bi++)
{
    present = false;
    for (int ai = 0; ai <= sizeA - 1; ai++)
    {
        if (B[bi] == A[ai])
        {
            present=true;
            break;
        }
    }
    if (!present)
    {
        BMinusA[mi] = B[bi];
        mi++;
    }
}

Implementation with functions:



void setMinus(int A1[ ], int sizeA1, 
        int A2[ ], int sizeA2, 
        int R[ ], int &sizeR)
{
    bool found;
    int mi = 0;
    for (int ai = 0; ai <= sizeA1-1; ai++)
    {
     if(!isPresent(A2, sizeA2, A1[ai]))   
       {
          R[mi] = A1[ai];           
          mi++;
        }
    }
    sizeR = mi;
}

Call functions from main:


.
.
.
    setMinus(A, sizeA, B, sizeB, 
        AMinusB, minusSize);
	setPrint("A-B", AMinusB, minusSize);
	
    setMinus(B, sizeB, A, sizeA, 
        BMinusA, minusSize);
	setPrint("B-A", BMinusA, minusSize);
.
.
.

Instruction: Add the function setMinus() and test them inside the main() and see the output.