# 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

## Python3

 # Python3 program for the above approach  import math      # Function to check if Pascaltriangle  # can be made by N integers  def checkPascaltriangle(N):             # Find X      x = (math.sqrt(8 * N + 1) - 1) / 2        # If x is integer      if (math.ceil(x) - x == 0):          print("Yes")      else:          print("No")     # Driver Code     # Given number N  N = 10    # Function call  checkPascaltriangle(N)     # This code is contributed by sanjoy_62

## C#

 // C# program for the above approach  using System;     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.Ceiling(x) - x == 0)          Console.Write("Yes");      else         Console.Write("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.

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