Check if Pascal’s Triangle is possible with a complete layer by using numbers upto N

0 0 3 minutes read

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 +…

The sum of first X integers is given by

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

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.
To determine value of X from second step we can deduced the formula as:

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

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;
}

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

<script>
 
      // JavaScript program for the above approach
 
      // Function to check if Pascaltriangle
      // can be made by N integers
      function checkPascaltriangle(N) {
        // Find X
        var x = (Math.sqrt(8 * N + 1) - 1) / 2;
 
        // If x is integer
        if (Math.ceil(x) - x == 0) 
        document.write("Yes");
        else
        document.write("No");
      }
 
      // Driver Code
 
      // Given number N
      var N = 10;
 
      // Function Call
      checkPascaltriangle(N);
       
    </script>

Output:

Yes

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!