Odd numbers in N-th row of Pascal’s Triangle

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.

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       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

 >= 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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