Skip to content
Related Articles

Related Articles

Number of cards needed build a House of Cards of a given level N
  • Last Updated : 23 Mar, 2020

Given a number N, the task is to find the number of cards needed to make a House of Cards of N levels.

Examples:

Input: N = 3
Output: 15
From the above image, it is clear that for the House of Cards for 3 levels 15 cards are needed

Input: N = 2
Output: 7



Approach:

  1. If we observe carefully, then a series will be formed as shown below in which i-th item denotes the number of triangular cards needed to make a pyramid of i levels:

    2, 7, 15, 26, 40, 57, 77, 100, 126, 155………and so on.

  2. The above series is a method of difference series where differences are in AP as 5, 8, 11, 14……. and so on.
  3. Therefore nth term of the seris will be:
    nth term = 2 + {5 + 8 + 11 +14 +.....(n-1) terms}
             = 2 + (n-1)*(2*5+(n-1-1)*3)/2
             = 2 + (n-1)*(10+(n-2)*3)/2
             = 2 + (n-1)*(10+3n-6)/2
             = 2 + (n-1)*(3n+4)/2
             = n*(3*n+1)/2;
    
  4. Therefore the number of cards needed for building a House of Cards of N levels will be:
    n*(3*n+1)/2

Below is the implementation of the above approach:

CPP

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find number of cards needed
int noOfCards(int n)
{
    return n * (3 * n + 1) / 2;
}
  
// Driver Code
int main()
{
    int n = 3;
    cout << noOfCards(n) << ", ";
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach
import java.lang.*; 
  
class GFG 
    // Function to find number of cards needed
    public static int noOfCards(int n)
    {
        return n * (3 * n + 1) / 2;
    }
      
    // Driver Code
    public static void main(String args[]) 
    {
        int n = 3;
        System.out.print(noOfCards(n));
    }
}
  
// This code is contributed by shubhamsingh10

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the above approach
  
# Function to find number of cards needed
def noOfCards(n):
    return n * (3 * n + 1) // 2
  
# Driver Code
n = 3
print(noOfCards(n))
  
# This code is contributed by mohit kumar 29

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach
using System;
  
class GFG 
    // Function to find number of cards needed
    public static int noOfCards(int n)
    {
        return n * (3 * n + 1) / 2;
    }
       
    // Driver Code
    public static void Main(String []args) 
    {
        int n = 3;
        Console.Write(noOfCards(n));
    }
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

15

Time Complexity: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :