Best Way to Build House of Cards With Least Amount
Number of cards needed build a House of Cards of a given level Due north
Given a number N, the task is to find the number of cards needed to brand a House of Cards of N levels.
Examples:
Input: Due north = iii
Output: 15
From the above prototype, it is clear that for the Business firm of Cards for 3 levels xv cards are needed
Input: N = two
Output: 7
Approach:
- If we find carefully, so a series will be formed as shown below in which i-th detail denotes the number of triangular cards needed to make a pyramid of i levels:
two, seven, 15, 26, 40, 57, 77, 100, 126, 155………and then on.
- The above serial is a method of deviation serial where differences are in AP as 5, eight, 11, fourteen……. and and so on.
- Therefore nth term of the series volition be:
nth term = 2 + {5 + 8 + 11 +14 +.....(n-1) terms} = two + (north-1)*(2*5+(n-1-1)*3)/2 = two + (due north-ane)*(10+(n-2)*3)/ii = ii + (due north-i)*(10+3n-6)/ii = ii + (northward-1)*(3n+4)/2 = due north*(3*northward+1)/2; - Therefore the number of cards needed for building a House of Cards of Northward levels will be:
Below is the implementation of the above approach:
CPP
#include <bits/stdc++.h>
using namespace std;
int noOfCards( int due north)
{
render north * (3 * n + 1) / 2;
}
int main()
{
int due north = 3;
cout << noOfCards(n) << ", " ;
return 0;
}
Coffee
import java.lang.*;
class GFG
{
public static int noOfCards( int due north)
{
return n * ( three * n + 1 ) / ii ;
}
public static void main(Cord args[])
{
int n = 3 ;
Organization.out.impress(noOfCards(n));
}
}
Python3
def noOfCards(north):
return n * ( 3 * northward + 1 ) / / 2
n = iii
print (noOfCards(north))
C#
using System;
class GFG
{
public static int noOfCards( int n)
{
render n * (iii * n + ane) / 2;
}
public static void Main(String []args)
{
int n = 3;
Console.Write(noOfCards(n));
}
}
Javascript
<script>
function noOfCards(northward)
{
return parseInt(n * (3 * n + ane) / two);
}
var n = three;
document.write(noOfCards(north));
</script>
Fourth dimension Complexity: O(1)
Auxiliary Space: O(1)
Source: https://www.geeksforgeeks.org/number-of-cards-needed-build-a-house-of-cards-of-a-given-level-n/
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