Sum of all the numbers present at given level in Modified Pascal’s triangle

Given a level N, the task is to find the sum of all the integers present at this given level in an Alternating Pascal’s triangle. 
A Modified Pascal triangle with 5 levels is shown below.

     1
   -1 1
   1 -2 1
 -1 3 -3 1
1 -4 6 -4 1

Examples:

Input: N = 1
Output: 1

Input: N = 2
Output: 0

Approach: As we can observe for even level sum is 0 and for odd level except for 1 sum is also 0. So There can be at most 2 cases:  

• If L = 1, then the answer is 1.
• Otherwise, the answer will always be 0.

Below is the implementation of the above approach: 

C++

// C++ program to calculate sum of
// all the numbers present at given
// level in an Modified Pascal’s triangle

#include <bits/stdc++.h>
using namespace std;

// Function to calculate sum
void ans(int n)
{
    if (n == 1)
        cout << "1";
    else
        cout << "0";
}

// Driver Code
int main()
{
    int n = 2;
    ans(n);

    return 0;
}

Java

// Java program to calculate sum of
// all the numbers present at given
// level in an Modified Pascal's triangle
import java.io.*;

public class GFG {

    // Function to calculate sum
    static void ans(int n)
    {
        if (n == 1)
            System.out.println("1");
        else
            System.out.println("0");
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 2;
        ans(n);
    }
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to calculate sum of
# all the numbers present at given
# level in an Modified Pascal’s triangle

# Function to calculate sum
def ans(n) :

    if (n == 1) :
        print("1",end="");
    else :
        print("0",end="");

# Driver Code
if __name__ == "__main__" :

    n = 2;
    ans(n);

# This code is contributed by AnkitRai01

C#

// C# program to calculate sum of
// all the numbers present at given
// level in an Modified Pascal's triangle
using System;

class GFG
{

// Function to calculate sum
static void ans(int n)
{
    if (n == 1)
        Console.WriteLine("1");
    else
        Console.WriteLine("0");
}

// Driver Code
public static void Main(String[] args)
{
    int n = 2;
    ans(n);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to calculate sum of
// all the numbers present at given
// level in an Modified Pascal’s triangle

// Function to calculate sum
function ans(n)
{
    if (n == 1)
        document.write("1");
    else
        document.write("0");
}

// Driver Code
var n = 2;

ans(n);

// This code is contributed by rutvik_56

</script>

Output:

0

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