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# 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 +…

1. The sum of first X integers is given by 1. 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.
2. To determine value of X from second step we can deduced the formula as: 1. 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 integersvoid 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 Codeint main(){    // Given number N    int N = 10;     // Function Call    checkPascaltriangle(N);    return 0;}

## Java

 // Java program for the above approachclass GFG{ // Function to check if Pascaltriangle// can be made by N integersstatic 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 Codepublic 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 approachimport math # Function to check if Pascaltriangle# can be made by N integersdef 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 NN = 10 # Function callcheckPascaltriangle(N) # This code is contributed by sanjoy_62

## C#

 // C# program for the above approachusing System; class GFG{ // Function to check if Pascaltriangle// can be made by N integersstatic 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 Codepublic static void Main(String[] args){         // Given number N    int N = 10;     // Function call    checkPascaltriangle(N);}} // This code is contributed by amal kumar choubey

## Javascript

 
Output:
Yes

Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)

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