Answer:
#include <iostream>
using namespace std;
int * reverse(int a[],int n)//function to reverse the array.
{
int i;
for(i=0;i<n/2;i++)
{
int temp=a[i];
a[i]=a[n-i-1];
a[n-i-1]=temp;
}
return a;//return pointer to the array.
}
int main() {
int array[50],* arr,N;//declaring three variables.
cin>>N;//taking input of size..
if(N>50||N<0)//if size greater than 50 or less than 0 then terminating the program..
return 0;
for(int i=0;i<N;i++)
{
cin>>array[i];//prompting array elements..
}
arr=reverse(array,N);//function call.
for(int i=0;i<N;i++)
cout<<arr[i]<<endl;//printing reversed array..
cout<<endl;
return 0;
}
Output:-
5
4 5 6 7 8
8
7
6
5
4
Explanation:
I have created a function reverse which reverses the array and returns pointer to an array.I have also considered edge cases where the function terminates if the value of the N(size) is greater than 50 or less than 0.
Answer:
B.O(1)
Explanation:
When we are implementing ADT stack using linked chain we can pop an entry from the stack having O(1) time complexity because in linked chain we have the head or top pointer in linked chain only.Popping and pushing in stack happens on only one end that is top.So we have move to move top in linked chain to the next and delete prev node.
Answer:
A) ALU
Explanation:
The arithmetic logic unit (ALU) performs the arithmetic and logical functions that are the work of the computer.
So, the correct option is - A) ALU
Answer:
The correct option is A
Explanation:
In project management, earliest finish time for activity A refers to the earliest start time for succeeding activities such as B and C to start.
Assume that activities A and B comes before C, the earliest finish time for C can be arrived at by computing the earliest start-finish (critical path) of the activity with the largest EF.
That is, if two activities (A and B) come before activity C, one can estimate how long it's going to take to complete activity C if ones knows how long activity B will take (being the activity with the largest earliest finish time).
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