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: Pascalâ€™s triangle is a triangular array of the binomial coefficients. Following are the first 6 rows of Pascalâ€™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 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

Javascript

 

Output:

Yes

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

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