Where now stands that small knot of villages known as the
Endians, a mighty forest once stood. Indeed, legand has it
that you could have stoodon the edge of the wood and seen
it stretch out for miles, were it not for the trees getting
in the way.
In one section of the forest, the trees stood in a row and
were of hight from 1 to n, each hight occurring once and
once only.
A tree was only visible if there were no higher trees
before it in the row.
For example, if the heights were 324165, the only visible
trees would have been those of height 3,4 & 6.
Write a Program that takes an array of integers
representing the heights of the trees in the row as input
and prints the list of the visible trees.
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#include<iostream.h>
#include<conio.h>
void main()
{
int n,i,tree[50],show_t;
clrscr();
cout<<"Enter the value of N: ";
cin>>n;
for(i=0;i<n;i++)
{
cin>>tree[i];
}
cout<<"Tree which will be visible : ";
show_t=tree[0];
cout<<tree[0];
for(i=1;i<n;i++)
{
if(show_t < tree[i])
{
cout<<", "<<tree[i];
show_t=tree[i];
}
}
getch();
}
| Is This Answer Correct ? | 22 Yes | 8 No |
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A research student is given a singly-linked list. Each node of the list has a color, which is either “Black” or “White”. He must find if there are more black nodes than white nodes, or vice versa. His advisor gives him 5,000Rs to buy a computer to do the work. He goes to the computer store and finds a slightly defective computer which costs a mere 3,000Rs. This computer has the small problem of not being able to do arithmetic. This means that he cannot use a counter to count the nodes in the list to determine the majority color. The computer is otherwise fully functional. He has the evil idea that he could buy the defective computer and somehow use it to do his work, so that he can use the rest of the money on enjoyment. Show how he can accomplish this amazing task. Write code for an algorithm called ‘findMajorityColor’ which takes a singly-linked list, L, with n nodes and returns the majority color among nodes of L. This algorithm should have the same asymptotic running time as counting the nodes (O(n)). Note: No arithmetic is allowed.
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