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 = 10Output:YesExplanation:

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 = 5Output:NoExplanation:

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++

`// C++ program for the above approach ` `#include <bits/stdc++.h> ` `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; ` `} ` |

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## 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 ` |

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**Output:**

Yes

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

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