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Sum of all the numbers present at given level in Modified Pascal’s triangle
  • Last Updated : 07 Apr, 2021

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. 
An Modified Pascal triangle with 5 levels is as 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 that 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 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
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)
 

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