Given a level L. The task is to find the sum of all the integers present at the given level in Pascal’s triangle .
A Pascal triangle with 6 levels is shown below:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Examples:
Input: L = 3
Output: 4
1 + 2 + 1 = 4Input: L = 2
Output:2
Approach: If we observe carefully the series of the sum of levels will go on like 1, 2, 4, 8, 16…., which is a GP series with a = 1 and r = 2.
Therefore, sum of Lth level is L’th term in the above series.
Lth term = 2L-1
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to find sum of numbers at // Lth level in Pascals Triangle int sum( int h)
{ return pow (2, h - 1);
} // Driver Code int main()
{ int L = 3;
cout << sum(L);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to find sum of numbers at
// Lth level in Pascals Triangle
static int sum( int h)
{
return ( int )Math.pow( 2 , h - 1 );
}
// Driver Code
public static void main (String[] args)
{
int L = 3 ;
System.out.println(sum(L));
}
} // This code is contributed by AnkitRai01 |
# Python3 implementation of the above approach # Function to find sum of numbers at # Lth level in Pascals Triangle def summ(h):
return pow ( 2 , h - 1 )
# Driver Code L = 3
print (summ(L))
# This code is contributed by mohit kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to find sum of numbers at
// Lth level in Pascals Triangle
static int sum( int h)
{
return ( int )Math.Pow(2, h - 1);
}
// Driver Code
public static void Main ()
{
int L = 3;
Console.WriteLine(sum(L));
}
} // This code is contributed by anuj_67.. |
<script> // Javascript implementation of the above approach // Function to find sum of numbers at // Lth level in Pascals Triangle function sum(h)
{ return Math.pow(2, h - 1);
} // Driver Code var L = 3;
document.write(sum(L)); </script> |
4
Time Complexity: O(log2L) because it is using pow function
Auxiliary Space: O(1)