Question

Write a function that will accept one integer matrix E and one column vector F by reference parameters, and one integer I as a value parameter. The function will return an integer v, which is the i-th coefficient of row vector denoted by E*F (multiplication of E and F). For example, if V = E*F, the function will return v, which is equal to V[i].

Explain the time complexity of this function inside of the function code as a comment.

in c++ program

Answer 1

Program :

//here this program is not with dynamically allocated memory because we can not pass by reference an array with unknown bound so we need to pass the size also

//we can make this dynamic by use of pointers

//time complexity of the function will be O(n) where n is no of cols in matrix E or no of rows in col vector F

as we just need to return the coefficient i of V where V = E*F

so don't need to find all the coefficients just need find the coefficient for ith row of E

1 2 3 1 14 = (1*1 + 2*2 + 3*3)

4 5 6 X 2 = 32 = (4*1+ 5*2+6*3)

7 8 9 3 50 = (7*1+8*2+9*3)

main.cpp

#include <iostream>
#include <vector>
using namespace std;

//Required function
//we can not pass reference array of unkown bound
int findCoefficient(int (&E)[3][3],vector<int> &F,int i)
{
    int row1 = sizeof(E)/sizeof(E[0]);  //no of rows in matrix E
    int col1 = sizeof(E[0])/sizeof(int); // no of cols in matrix E
    int row2 = F.size();  // no of rows in F
    
    if(i < 0 || i>= row1)
    {
        cout<<"\nCoefficient index is out of range!";
        return -1;
    }
    if(col1 != row2)
    {
        cout<<"\nMultipliction is not possible!";
        return -1;
    }
    int v=0;  // to store value of V[i] where V = E*F index i = 0 to row1-1
    /***
    * here loop is going from 1 to (no of cols in E or no of rows in F) 
    *therefore the time complexity = O(n)   where n is no of cols in matrix E or no of rows in vector F
    *here we don't need to find all multiplications we just need to find the multiplication for the 
    ith Coefficient of E*F so we will consider all cols of ith row in matrix E and all rows of vector F 
    ***/
    
    for(int j=0; j<col1; j++)
    {
        v = v + E[i][j]*F[j];
    }

    return v;
}

//main program for testing the function
int main()
{
    int E[][3]= {{1,2,3},{4,5,6},{7,8,9}};
    vector<int> F;
    F.push_back(1);
    F.push_back(2);
    F.push_back(3);
    
    //entring the index to find v[i]
    int index;
    cout<<"\nEnter the Coefficient index i = (0 to row-1) to find V[i] where V = E*F : ";
    cin>>index;
    
    //calling the function
    int v = findCoefficient(E,F,index);
    cout<<"\nValue of V["<<index<<"]= "<<v;
    
    return 0;
}

Output :

Program screenshot :