Space and time efficient Binomial Coefficient (original) (raw)

Last Updated : 11 Sep, 2023

Here the function takes two parameters n and k and returns the value of Binomial Coefficient C(n, k).

Example:

Input: n = 4 and k = 2 Output: 6 Explanation: 4 C 2 is 4!/(2!*2!) = 6

Input: n = 5 and k = 2 Output: 10 Explanation: 5 C 2 is 5!/(3!*2!) = 10

We have discussed O(n*k) time and O(k) extra space algorithm in this post. The value of C(n, k) can be calculated in O(k) time and O(1) extra space.

Approach:

1. Change r to n-r if r is greater than n-r. and create a variable to store the answer.
2. Run a loop from 0 to r-1
3. In every iteration update ans as (ans*(n-i))/(i+1) where i is the loop counter.
4. So the answer will be equal to ((n/1)*((n-1)/2)*...*((n-r+1)/r) which is equal to nCr.

C(n, k) = n! / (n-k)! * k! = [n * (n-1) .... 1] / [ ( (n-k) * (n-k-1) * .... * 1) * ( k * (k-1) * .... * 1 ) ] After simplifying, we get C(n, k) = [n * (n-1) * .... * (n-k+1)] / [k * (k-1) * .... * 1]

Also, C(n, k) = C(n, n-k)
// r can be changed to n-r if r > n-r

Following implementation uses the above formula to calculate C(n, k).

C++ `

// Program to calculate C(n, k)

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

// Returns value of Binomial Coefficient C(n, k) int binomialCoeff(int n, int k) { int res = 1;

// Since C(n, k) = C(n, n-k)
if (k > n - k)
    k = n - k;

// Calculate value of
// [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
}

return res;

}

// Driver Code int main() { int n = 8, k = 2; cout << "Value of C(" << n << ", " << k << ") is " << binomialCoeff(n, k); return 0; }

// This is code is contributed by rathbhupendra

C

// Program to calculate C(n, k) #include <stdio.h>

// Returns value of Binomial Coefficient C(n, k) int binomialCoeff(int n, int k) { int res = 1;

// Since C(n, k) = C(n, n-k)
if (k > n - k)
    k = n - k;

// Calculate value of
// [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
}

return res;

}

/* Driver program to test above function*/ int main() { int n = 8, k = 2; printf("Value of C(%d, %d) is %d ", n, k, binomialCoeff(n, k)); return 0; }

Java

// Program to calculate C(n, k) in java class BinomialCoefficient { // Returns value of Binomial Coefficient C(n, k) static int binomialCoeff(int n, int k) { int res = 1;

    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;

    // Calculate value of
    // [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }

    return res;
}

/* Driver program to test above function*/
public static void main(String[] args)
{
    int n = 8;
    int k = 2;
    System.out.println("Value of C(" + n + ", " + k
                       + ") "
                       + "is"
                       + " " + binomialCoeff(n, k));
}

} // This Code is Contributed by Saket Kumar

Python3

Python program to calculate C(n, k)

Returns value of Binomial Coefficient

C(n, k)

def binomialCoefficient(n, k): # since C(n, k) = C(n, n - k) if(k > n - k): k = n - k # initialize result res = 1 # Calculate value of # [n * (n-1) --- (n-k + 1)] / [k * (k-1) ---- 1] for i in range(k): res = res * (n - i) res = res // (i + 1) return res

Driver program to test above function

n = 8 k = 2 res = binomialCoefficient(n, k) print("Value of C(% d, % d) is % d" %(n, k, res))

This code is contributed by Aditi Sharma

C#

// C# Program to calculate C(n, k) using System;

class BinomialCoefficient {

// Returns value of Binomial
// Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
    int res = 1;

    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;

    // Calculate value of [n * ( n - 1) *---* (
    // n - k + 1)] / [k * (k - 1) *----* 1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }

    return res;
}

// Driver Code
public static void Main()
{
    int n = 8;
    int k = 2;
    Console.Write("Value of C(" + n + ", " + k + ") "
                  + "is"
                  + " " + binomialCoeff(n, k));
}

}

// This Code is Contributed by // Smitha Dinesh Semwal.

PHP

JavaScript

`

Output

Value of C(8, 2) is 28

Complexity Analysis:

Time Complexity: O(r) A loop has to be run from 0 to r. So, the time complexity is O(r).

Auxiliary Space: O(1) As no extra space is required.

This article is compiled by Aashish Barnwal and reviewed by the GeeksforGeeks team.