Question & Answer: Suppose f(x):R rightarrow R be a function given by Find maximum and minimum of functional values using an…..

Step1 : start

Step2 : function declaration of max
Step3 : function declaration of min
Step4 : array declaration
Step5 : for(x=0;x<8;x++)
Step6 : print the elements of the F
Step7 : calling function to max
Step8 : calling function to min
Step9 : print the maximum element of the F
Step10 : print the minimum element of the F
Step11 : called function of max
Step12 : maxInt=F[0];
Step13 : for(x=0; x<8; x++)
Step14 : temp=F[x];
Step15 : if(temp>maxInt)
Step16 : maxInt=temp;
Step17 : return maximum value
Step18 : called function of min
Step19 : minInt=F[0];
Step20 : for(x=0; x<8; x++)
Step21 : temp=F[x];
Step22 : if(temp<minInt)
Step23 : minInt=temp;
Step24 : return minimum value

Step25 : stop

Program :

#include <iostream>
using namespace std;
int findMax(int F[], int index); // function declaration
int findMin(int F[], int index); // function declaration
int main()
{
int F[8]={33,25,71,83,22,47,51,36}; // F declaration
int x,max,min;
for(x=0;x<8;x++)
{
cout<<“F[“<<x<<“] = “<<F[x]<<endl; // print the elements of the F
}
max=findMax(F,8); // calling function
min=findMin(F,8); // calling function
cout<<“The maximum element is “<<max<<endl; // print the maximum element of the F
cout<<“The minimum element is “<<min<<endl; // print the minimum element of the F
return 0;
}
int findMax(int F[], int index) // called function
{
int maxInt,temp,x;
maxInt=F[0];
for(x=0; x<8; x++) // loop to find maximum element of an F
{
temp=F[x]; // assign F element in temp
if(temp>maxInt) // if the F element is less than the maximum value
{
maxInt=temp; // change maximum value
}
}
return maxInt; // return maximum value
}
int findMin(int F[], int index) // called function
{
int minInt,temp,x;
minInt=F[0];
for(x=0; x<8; x++) // loop to find minimum element of an F
{
temp=F[x]; // assign F element in temp
if(temp<minInt) // if the F element is less than the minimum value
{
minInt=temp; // change minimum value
}
}
return minInt; // return minimum value
}

Output :

Complexity of algorithm :

• Here for loop for n times.
• The time complexity of for loop is O(n).
• Since the for loop consumes more time which is O(n) and remaining code consumes less than O(n).

So Time complexity of algorithm is O(n).