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Odd numbers in N-th row of Pascal’s Triangle
  • Difficulty Level : Hard
  • Last Updated : 22 May, 2018

Given N, the row number of Pascal’s triangle(row starting from 0). Find the count of odd numbers in N-th row of Pascal’s Triangle.

Prerequisite : Pascal’s Triangle | Count number of 1’s in binary representation of N

Examples :

Input : 11
Output : 8

Input : 20
Output : 4 

Approach : It appears the answer is always a power of 2. In fact, the following theorem exists :
THEOREM : The number of odd entries in row N of Pascal’s Triangle is 2 raised to the number of 1’s in the binary expansion of N.
Example: Since 83 = 64 + 16 + 2 + 1 has binary expansion (1010011), then row 83 has pow(2, 4) = 16 odd numbers.



Below is the implementation of above approach :

C++




    
// CPP code to find the count of odd numbers
// in n-th row of Pascal's Triangle
#include <bits/stdc++.h>     
using namespace std ;
  
/* Function to get no of set
   bits in binary representation 
   of positive integer n */
int countSetBits(int n)
{
    unsigned int count = 0;
    while (n)
    {
        count += n & 1;
        n >>= 1;
    }
      
    return count;
}
  
int countOfOddsPascal(int n)
{
    // Count number of 1's in binary
    // representation of n.
    int c = countSetBits(n);
      
    // Number of odd numbers in n-th
    // row is 2 raised to power the count.
    return pow(2, c);
}
  
// Driver code
int main()
{
    int n = 20;    
    cout << countOfOddsPascal(n) ;    
    return 0;
}

Java




// Java code to find the count of odd
// numbers in n-th row of Pascal's 
// Triangle
import java.io.*;
  
class GFG {
      
    /* Function to get no of set
    bits in binary representation 
    of positive integer n */
    static int countSetBits(int n)
    {
        long count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
          
        return (int)count;
    }
      
    static int countOfOddsPascal(int n)
    {
          
        // Count number of 1's in binary
        // representation of n.
        int c = countSetBits(n);
          
        // Number of odd numbers in n-th
        // row is 2 raised to power the
        // count.
        return (int)Math.pow(2, c);
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int n = 20
        System.out.println(
                     countOfOddsPascal(n));
    }
}
  
// This code is contributed by anuj_67.

Python3




# Python code to find the count of
# odd numbers in n-th row of 
# Pascal's Triangle
  
# Function to get no of set
# bits in binary representation
# of positive integer n
def countSetBits(n):
    count =0
    while n:
        count += n & 1
        n >>= 1
          
    return count
  
def countOfOddPascal(n):
  
    # Count number of 1's in binary
    # representation of n.
    c = countSetBits(n)
  
    # Number of odd numbers in n-th
    # row is 2 raised to power the count.
    return pow(2, c)
  
# Driver Program
n = 20
print(countOfOddPascal(n))
  
# This code is contributed by Shrikant13

C#




// C# code to find the count of odd numbers
// in n-th row of Pascal's Triangle
using System;
  
class GFG {
      
    /* Function to get no of set
    bits in binary representation 
    of positive integer n */
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
          
        return count;
    }
      
    static int countOfOddsPascal(int n)
    {
        // Count number of 1's in binary
        // representation of n.
        int c = countSetBits(n);
          
        // Number of odd numbers in n-th
        // row is 2 raised to power the
        // count.
        return (int)Math.Pow(2, c);
    }
      
    // Driver code
    public static void Main () 
    {
        int n = 20; 
        Console.WriteLine(
                 countOfOddsPascal(n)) ; 
    }
}
  
// This code is contributed by anuj_67.

PHP




<?php
// PHP code to find the 
// count of odd numbers
// in n-th row of Pascal's 
// Triangle
  
/* Function to get no of set
   bits in binary representation 
   of positive integer n */
function countSetBits($n)
{
    $count = 0;
    while ($n)
    {
        $count += $n & 1;
        $n >>= 1;
    }
      
    return $count;
}
  
function countOfOddsPascal($n)
{
      
    // Count number of 1's in binary
    // representation of n.
    $c = countSetBits($n);
      
    // Number of odd numbers in n-th
    // row is 2 raised to power the count.
    return pow(2, $c);
}
  
    // Driver code
    $n = 20; 
    echo countOfOddsPascal($n) ; 
  
// This code is contributed by mits. 
?>
Output:
4

Time Complexity : O(L), where L is the length of binary representation of given N.
Reference : https://www.math.hmc.edu/funfacts/ffiles/30001.4-5.shtml

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