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 

    1 + 2 + 3 + 4 + .... + X = \frac{X*(X + 1)}{2}

  2. We can only make pascal’s triangle by using N integers if and only if N = \frac{X*(X + 1)}{2} 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: 

    X = \frac{\sqrt{8*N + 1} - 1}{2}

  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++

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// 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

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// 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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Python3

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# 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

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C#

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// 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

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Output: 

Yes

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

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