# Check if Pascal’s Triangle is possible with a complete layer by using numbers upto N

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 +…

1. The sum of first X integers is given by 2. 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.
3. To determine value of X from second step we can deduced the formula as: 4. 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++

 // C++ program for the above approach  #include  using namespace std;     // Function to check if Pascaltriangle  // can be made by N integers  void checkPascaltriangle(int N)  {      // Find X      double x = (sqrt(8 * N + 1) - 1) / 2;         // If x is integer      if (ceil(x) - x == 0)          cout << "Yes";         else         cout << "No";  }     // Driver Code  int main()  {      // Given number N      int N = 10;         // Function Call      checkPascaltriangle(N);      return 0;  }

## Java

 // Java program for the above approach  class GFG{     // Function to check if Pascaltriangle  // can be made by N integers  static void checkPascaltriangle(int N)  {             // Find X      double x = (Math.sqrt(8 * N + 1) - 1) / 2;         // If x is integer      if (Math.ceil(x) - x == 0)          System.out.print("Yes");      else         System.out.print("No");  }     // Driver Code  public static void main(String[] args)  {             // Given number N      int N = 10;         // Function call      checkPascaltriangle(N);  }  }     // This code is contributed by amal kumar choubey

Output:

Yes


Time Complexity: O(sqrt(N))
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Amal Kumar Choubey