Count possible ways to construct buildings

Dynamic ProgrammingAlgorithmsData Structure

Here n number of sections are given, in each section, there are two sides on the road to constructing buildings. If there is one empty space between two houses are needed, then how many possible ways to construct buildings in the plot.

There are four possibilities to construct buildings

•  One side of the road
•  Another side of the road
•  No building can be constructed
•  Both sides of the road

Input and Output

Input:
It takes the number of sections to construct buildings. Say the input is 3.
Output:
Enter Number of sections: 3
Buildings can be constructed in 25 different ways.

Algorithm

constructionWays(n)

Input: There are n number of section.

Output − Number of possible ways.

Begin
if n = 1, then
return 4
countEnd := 1
countEndSpace := 1

for i := 2 to n, do
prevCountEnd := countEnd
prevCountEndSpace := countEndSpace
countEndSpace := countEnd + prevCountEndSpace
countEnd := prevCountEndSpace
done

End

Example

#include<iostream>
using namespace std;

int constructionWays(int n) {
if (n == 1)        //if there is one section
return 4;       //4 possible ways to construct building in that section

//set counting values for place at the end and end with space
int countEnd=1, countEndSpace=1, prevCountEnd, prevCountEndSpace;

for (int i=2; i<=n; i++) {       //fot the second section to nth section
prevCountEnd = countEnd;
prevCountEndSpace = countEndSpace;

countEndSpace = countEnd + prevCountEndSpace;
countEnd = prevCountEndSpace;
}

//possible ways to end with space and building at the end
int answer = countEndSpace + countEnd;

}

int main() {
int n;
cout << "Enter Number of sections: ";
cin >> n;
cout << "Buildings can be constructed in " << constructionWays(n) <<" different ways." ;
}

Output

Enter Number of sections: 3
Buildings can be constructed in 25 different ways.
Published on 11-Jul-2018 09:45:38