Check if Pascal’s Triangle is possible with a complete layer by using numbers upto N
Last Updated :
05 Apr, 2021
Given a number N, the task is to determine if it is possible to make Pascal’s triangle with a complete layer by using total number N integer if possible print Yes otherwise print No.
Note: Pascal’s triangle is a triangular array of the binomial coefficients. Following are the first 6 rows of Pascal’s Triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
In Pascal’s Triangle from the topmost layer there is 1 integer, at every next layer from top to bottom size of the layer increased by 1.
Examples:
Input: N = 10
Output: Yes
Explanation:
You can use 1, 2, 3 and 4 integers to make first, second, third, and fourth layer of pascal’s triangle respectively and also N = 10 satisfy by using (1 + 2 + 3 + 4) integers on each layer = 10.
Input: N = 5
Output: No
Explanation:
You can use 1 and 2 integers to make first and second layer respectively and after that you have only 2 integers left and you can’t make 3rd layer complete as that layer required 3 integers.
Approach: Here we are using integer 1, 2, 3, … on every layer starting from first layer, so we can only make Pascal’s triangle complete if it’s possible to represent N by the sum of 1 + 2 +…
- The sum of first X integers is given by
- We can only make pascal’s triangle by using N integers if and only if where X must be a positive integer. So we have to check is there any positive integer value of x exist or not.
- To determine value of X from second step we can deduced the formula as:
- If the value of X integer for the given value of N then we can make Pascal Triangle. Otherwise, we can’t make Pascal Triangle.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void checkPascaltriangle( int N)
{
double x = ( sqrt (8 * N + 1) - 1) / 2;
if ( ceil (x) - x == 0)
cout << "Yes" ;
else
cout << "No" ;
}
int main()
{
int N = 10;
checkPascaltriangle(N);
return 0;
}
|
Java
class GFG{
static void checkPascaltriangle( int N)
{
double x = (Math.sqrt( 8 * N + 1 ) - 1 ) / 2 ;
if (Math.ceil(x) - x == 0 )
System.out.print( "Yes" );
else
System.out.print( "No" );
}
public static void main(String[] args)
{
int N = 10 ;
checkPascaltriangle(N);
}
}
|
Python3
import math
def checkPascaltriangle(N):
x = (math.sqrt( 8 * N + 1 ) - 1 ) / 2
if (math.ceil(x) - x = = 0 ):
print ( "Yes" )
else :
print ( "No" )
N = 10
checkPascaltriangle(N)
|
C#
using System;
class GFG{
static void checkPascaltriangle( int N)
{
double x = (Math.Sqrt(8 * N + 1) - 1) / 2;
if (Math.Ceiling(x) - x == 0)
Console.Write( "Yes" );
else
Console.Write( "No" );
}
public static void Main(String[] args)
{
int N = 10;
checkPascaltriangle(N);
}
}
|
Javascript
<script>
function checkPascaltriangle(N) {
var x = (Math.sqrt(8 * N + 1) - 1) / 2;
if (Math.ceil(x) - x == 0)
document.write( "Yes" );
else
document.write( "No" );
}
var N = 10;
checkPascaltriangle(N);
</script>
|
Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)
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