Coeficiente binomial | DP-9 – Part 1

Posted on by Rudeus Greyrat

Las siguientes son las definiciones comunes de coeficientes binomiales . 

Un coeficiente binomial C(n, k) se puede definir como el coeficiente de x^k en la expansión de (1 + x)^n.

Un coeficiente binomial C(n, k) también da el número de formas, sin tener en cuenta el orden, en que pueden elegirse k objetos entre n objetos. De manera más formal, el número de subconjuntos de k elementos (o k combinaciones) de un elemento n establecer.

El problema  
Escriba una función que tome dos parámetros n y k y devuelva el valor del coeficiente binomial C(n, k). Por ejemplo, su función debería devolver 6 para n = 4 y k = 2, y debería devolver 10 para n = 5 y k = 2 .

1) Subestructura óptima 
El valor de C(n, k) se puede calcular recursivamente utilizando la siguiente fórmula estándar para coeficientes binomiales.  

   C(n, k) = C(n-1, k-1) + C(n-1, k)
   C(n, 0) = C(n, n) = 1

La siguiente es una implementación recursiva simple que simplemente sigue la estructura recursiva mencionada anteriormente.  

C++

// A naive recursive C++ implementation
#include <bits/stdc++.h>
using namespace std;
 
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    // Base Cases
    if (k > n)
        return 0;
    if (k == 0 || k == n)
        return 1;
 
    // Recur
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}
 
/* Driver code*/
int main()
{
    int n = 5, k = 2;
    cout << "Value of C(" << n << ", " << k << ") is "
         << binomialCoeff(n, k);
    return 0;
}
 
// This is code is contributed by rathbhupendra

C

// A Naive Recursive Implementation
#include <stdio.h>
 
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    // Base Cases
    if (k > n)
        return 0;
    if (k == 0 || k == n)
        return 1;
 
    // Recur
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}
 
/* Driver program to test above function*/
int main()
{
    int n = 5, k = 2;
    printf("Value of C(%d, %d) is %d ", n, k,
           binomialCoeff(n, k));
    return 0;
}

Java

// JAVA Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
import java.util.*;
 
class GFG {
 
    // Returns value of Binomial
    // Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {
 
        // Base Cases
        if (k > n)
            return 0;
        if (k == 0 || k == n)
            return 1;
 
        // Recur
        return binomialCoeff(n - 1, k - 1)
            + binomialCoeff(n - 1, k);
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 5, k = 2;
        System.out.printf("Value of C(%d, %d) is %d ", n, k,
                          binomialCoeff(n, k));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3

# A naive recursive Python implementation
 
 
def binomialCoeff(n, k):
 
    if k > n:
        return 0
    if k == 0 or k == n:
        return 1
 
    # Recursive Call
    return binomialCoeff(n-1, k-1) + binomialCoeff(n-1, k)
 
 
# Driver Program to test ht above function
n = 5
k = 2
print ("Value of C(%d,%d) is (%d)" % (n, k,
                                     binomialCoeff(n, k)))
 
# This code is contributed by Nikhil Kumar Singh (nickzuck_007)

C#

// C# Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
using System;
 
class GFG {
 
    // Returns value of Binomial
    // Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {
 
        // Base Cases
        if (k > n)
            return 0;
        if (k == 0 || k == n)
            return 1;
 
        // Recur
        return binomialCoeff(n - 1, k - 1)
            + binomialCoeff(n - 1, k);
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 5, k = 2;
        Console.Write("Value of C(" + n + "," + k + ") is "
                      + binomialCoeff(n, k));
    }
}
 
// This code is contributed by Sam007.

PHP

<?php
// PHP Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
 
// Returns value of
// Binomial Coefficient C(n, k)
function binomialCoeff($n, $k)
{
    // Base Cases
    if ($k > $n)
        return 0;
    if ($k==0 || $k==$n)
        return 1;
     
    // Recur
    return binomialCoeff($n - 1, $k - 1) +
               binomialCoeff($n - 1, $k);
}
 
    // Driver Code
    $n = 5;
    $k = 2;
    echo "Value of C","(",$n ,$k,") is "
               , binomialCoeff($n, $k);
 
// This code is contributed by aj_36
?>

JavaScript

<script>
// javascript Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
 
   // Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff(n , k)
{
 
    // Base Cases
    if (k > n)
        return 0;
    if (k == 0 || k == n)
        return 1;
 
    // Recur
    return binomialCoeff(n - 1, k - 1)
        + binomialCoeff(n - 1, k);
}
 
/* Driver program to test above function */
var n = 5, k = 2;
document.write("Value of C("+n+", "+k+") is "+binomialCoeff(n, k));
 
// This code is contributed by Amit Katiyar
</script>
Producción

Value of C(5, 2) is 10

Complejidad del tiempo: O(n*max(k,nk)) 

Espacio auxiliar: O(n*max(k,nk))

2) Subproblemas superpuestos 
Cabe señalar que la función anterior calcula los mismos subproblemas una y otra vez. Vea el siguiente árbol de recursión para n = 5 an k = 2. La función C(3, 1) se llama dos veces. Para valores grandes de n, habrá muchos subproblemas comunes. 
 

Binomial Coefficients Recursion tree

Árbol de recursión de coeficientes binomiales para C(5,2)

Dado que se vuelven a llamar los mismos subproblemas, este problema tiene la propiedad Superposición de subproblemas. Entonces, el problema del coeficiente binomial tiene las dos propiedades (ver this y this ) de un problema de programación dinámica. Al igual que otros problemas típicos de Programación Dinámica (DP) , los nuevos cálculos de los mismos subproblemas se pueden evitar mediante la construcción de una array 2D temporal C[][] de forma ascendente. A continuación se muestra la implementación basada en la programación dinámica. 

C++

// A Dynamic Programming based solution that uses
// table C[][] to calculate the Binomial Coefficient
#include <bits/stdc++.h>
using namespace std;
 
// Prototype of a utility function that
// returns minimum of two integers
int min(int a, int b);
 
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;
 
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= min(i, k); j++) {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
 
    return C[n][k];
}
 
// A utility function to return
// minimum of two integers
int min(int a, int b) { return (a < b) ? a : b; }
 
// Driver Code
int main()
{
    int n = 5, k = 2;
    cout << "Value of C[" << n << "][" << k << "] is "
         << binomialCoeff(n, k);
}
 
// This code is contributed by Shivi_Aggarwal

C

// A Dynamic Programming based solution
// that uses table C[][] to
// calculate the Binomial Coefficient
#include <stdio.h>
 
// Prototype of a utility function that
// returns minimum of two integers
int min(int a, int b);
 
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;
 
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= min(i, k); j++) {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using
            // previously stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
 
    return C[n][k];
}
 
// A utility function to return
// minimum of two integers
int min(int a, int b) { return (a < b) ? a : b; }
 
/* Drier program to test above function*/
int main()
{
    int n = 5, k = 2;
    printf("Value of C(%d, %d) is %d ", n, k,
           binomialCoeff(n, k));
    return 0;
}

Java

// A Dynamic Programming based
// solution that uses table C[][] to
// calculate the Binomial Coefficient
 
class BinomialCoefficient {
    // Returns value of Binomial
    // Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {
        int C[][] = new int[n + 1][k + 1];
        int i, j;
 
        // Calculate  value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= min(i, k); j++) {
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;
 
                // Calculate value using
                // previously stored values
                else
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }
 
        return C[n][k];
    }
 
    // A utility function to return
    // minimum of two integers
    static int min(int a, int b) { return (a < b) ? a : b; }
 
    /* Driver program to test above function*/
    public static void main(String args[])
    {
        int n = 5, k = 2;
        System.out.println("Value of C(" + n + "," + k
                           + ") is " + binomialCoeff(n, k));
    }
}
/*This code is contributed by Rajat Mishra*/

Python3

# A Dynamic Programming based Python
# Program that uses table C[][]
# to calculate the Binomial Coefficient
 
# Returns value of Binomial Coefficient C(n, k)
 
 
def binomialCoef(n, k):
    C = [[0 for x in range(k+1)] for x in range(n+1)]
 
    # Calculate value of Binomial
    # Coefficient in bottom up manner
    for i in range(n+1):
        for j in range(min(i, k)+1):
            # Base Cases
            if j == 0 or j == i:
                C[i][j] = 1
 
            # Calculate value using
            # previously stored values
            else:
                C[i][j] = C[i-1][j-1] + C[i-1][j]
 
    return C[n][k]
 
 
# Driver program to test above function
n = 5
k = 2
print("Value of C[" + str(n) + "][" + str(k) + "] is "
      + str(binomialCoef(n, k)))
 
# This code is contributed by Bhavya Jain

C#

// A Dynamic Programming based solution that
// uses table C[][] to calculate the Binomial
// Coefficient
using System;
 
class GFG {
 
    // Returns value of Binomial Coefficient
    // C(n, k)
    static int binomialCoeff(int n, int k)
    {
        int[, ] C = new int[n + 1, k + 1];
        int i, j;
 
        // Calculate value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= Math.Min(i, k); j++) {
                // Base Cases
                if (j == 0 || j == i)
                    C[i, j] = 1;
 
                // Calculate value using previously
                // stored values
                else
                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j];
            }
        }
 
        return C[n, k];
    }
 
    // A utility function to return minimum
    // of two integers
    static int min(int a, int b) { return (a < b) ? a : b; }
 
    /* Driver program to test above function*/
    public static void Main()
    {
        int n = 5, k = 2;
        Console.WriteLine("Value of C(" + n + "," + k
                          + ") is " + binomialCoeff(n, k));
    }
}
 
// This code is contributed by anuj_67.

PHP

<?php
// A Dynamic Programming based
// solution that uses table C[][] to
// calculate the Binomial Coefficient
 
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff( $n, $k)
{
    $C = array(array());
    $i; $j;
 
    // Calculate value of Binomial
    // Coefficient in bottom up manner
    for ($i = 0; $i <= $n; $i++)
    {
        for ($j = 0; $j <= min($i, $k); $j++)
        {
             
            // Base Cases
            if ($j == 0 || $j == $i)
                $C[$i][$j] = 1;
 
            // Calculate value using
            // previously stored values
            else
                $C[$i][$j] = $C[$i - 1][$j - 1] +
                                 $C[$i - 1][$j];
        }
    }
 
    return $C[$n][$k];
}
 
    // Driver Code
    $n = 5;
    $k = 2;
    echo "Value of C(" ,$n," ",$k, ") is"," "
                 , binomialCoeff($n, $k) ;
 
// This code is contributed by anuj_67.
?>

JavaScript

<script>
 
// A Dynamic Programming based
// solution that uses table C to
// calculate the Binomial Coefficient
 
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff(n, k)
{
    var C = Array(n + 1).fill(0).map(
      x => Array(k + 1).fill(0));;
    var i, j;
 
    // Calculate  value of Binomial
    // Coefficient in bottom up manner
    for(i = 0; i <= n; i++)
    {
        for(j = 0; j <= min(i, k); j++)
        {
             
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using
            // previously stored values
            else
                C[i][j] = C[i - 1][j - 1] +
                          C[i - 1][j];
        }
    }
    return C[n][k];
}
 
// A utility function to return
// minimum of two integers
function min(a, b)
{
    return (a < b) ? a : b;
}
 
// Driver code
var n = 5, k = 2;
document.write("Value of C(" + n + "," + k +
                     ") is " + binomialCoeff(n, k));
 
// This code is contributed by 29AjayKumar
 
</script>
Producción

Value of C[5][2] is 10

Complejidad temporal: O(n*k) 
Espacio auxiliar: O(n*k)

A continuación se muestra una versión optimizada para el espacio del código anterior. El siguiente código solo usa O(k). Gracias a AK por sugerir este método. 

C++

// C++ program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
#include <bits/stdc++.h>
using namespace std;
 
int binomialCoeff(int n, int k)
{
    int C[k + 1];
    memset(C, 0, sizeof(C));
 
    C[0] = 1; // nC0 is 1
 
    for (int i = 1; i <= n; i++)
    {
       
        // Compute next row of pascal triangle using
        // the previous row
        for (int j = min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}
 
/* Driver code*/
int main()
{
    int n = 5, k = 2;
    cout << "Value of C(" << n << "," << k << ")"<< "is " <<binomialCoeff(n, k);
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C

// C++ program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
#include <stdio.h>
 
int binomialCoeff(int n, int k)
{
    int C[k + 1];
    memset(C, 0, sizeof(C));
 
    C[0] = 1; // nC0 is 1
 
    for (int i = 1; i <= n; i++) {
        // Compute next row of pascal triangle using
        // the previous row
        for (int j = min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}
 
/* Driver code*/
int main()
{
    int n = 5, k = 2;
    printf("Value of C(%d, %d) is %d ", n, k,
           binomialCoeff(n, k));
    return 0;
}

Java

// JAVA Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
import java.util.*;
 
class GFG {
 
    static int binomialCoeff(int n, int k)
    {
        int C[] = new int[k + 1];
 
        // nC0 is 1
        C[0] = 1;
 
        for (int i = 1; i <= n; i++) {
            // Compute next row of pascal
            // triangle using the previous row
            for (int j = Math.min(i, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }
        return C[k];
    }
 
    /* Driver code  */
    public static void main(String[] args)
    {
        int n = 5, k = 2;
        System.out.printf("Value of C(%d, %d) is %d ", n, k,
                          binomialCoeff(n, k));
    }
}

Python3

# Python program for Optimized
# Dynamic Programming solution to
# Binomial Coefficient. This one
# uses the concept of pascal
# Triangle and less memory
 
 
def binomialCoeff(n, k):
 
    # Declaring an empty array
    C = [0 for i in range(k+1)]
    C[0] = 1  # since nC0 is 1
 
    for i in range(1, n+1):
 
        # Compute next row of pascal triangle using
        # the previous row
        j = min(i, k)
        while (j > 0):
            C[j] = C[j] + C[j-1]
            j -= 1
 
    return C[k]
 
 
# Driver Code
n = 5
k = 2
print ("Value of C(%d,%d) is %d" % (n, k, binomialCoeff(n, k)))
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
using System;
 
class GFG {
 
    static int binomialCoeff(int n, int k)
    {
        int[] C = new int[k + 1];
 
        // nC0 is 1
        C[0] = 1;
 
        for (int i = 1; i <= n; i++) {
            // Compute next row of pascal
            // triangle using the previous
            // row
            for (int j = Math.Min(i, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }
        return C[k];
    }
 
    /* Driver Code */
    public static void Main()
    {
        int n = 5, k = 2;
        Console.WriteLine("Value of C(" + n + " " + k
                          + ") is " + binomialCoeff(n, k));
    }
}
 
// This code is contributed by anuj_67.

PHP

<?php
// PHP program for space optimized
// Dynamic Programming Solution of
// Binomial Coefficient
function binomialCoeff($n, $k)
{
    $C = array_fill(0, $k + 1, 0);
 
    $C[0] = 1; // nC0 is 1
 
    for ($i = 1; $i <= $n; $i++)
    {
        // Compute next row of pascal
        // triangle using the previous row
        for ($j = min($i, $k); $j > 0; $j--)
            $C[$j] = $C[$j] + $C[$j - 1];
    }
    return $C[$k];
}
 
// Driver Code
$n = 5; $k = 2;
echo "Value of C[$n, $k] is ".
        binomialCoeff($n, $k);
     
// This code is contributed by mits.
?>

JavaScript

<script>
 
// Javascript program for space optimized
// Dynamic Programming
// Solution of Binomial Coefficient
function binomialCoeff(n, k)
{
    let C = new Array(k + 1);
    C.fill(0);
 
    // nC0 is 1
    C[0] = 1;
 
    for(let i = 1; i <= n; i++)
    {
         
        // Compute next row of pascal
        // triangle using the previous
        // row
        for(let j = Math.min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}
 
// Driver code
let n = 5, k = 2;
document.write("Value of C(" + n + " " +
               k + ") is " + binomialCoeff(n, k));
                
// This code is contributed by divyesh072019
 
</script>
Producción

Value of C(5, 2) is 10

Complejidad temporal: O(n*k) 
Espacio auxiliar: O(k)

Explicación: 
1==========>> n = 0, C(0,0) = 1 
1–1========>> n = 1, C(1,0) = 1, C(1,1) = 1 
1–2–1======>> n = 2, C(2,0) = 1, C(2,1) = 2, C(2, 2) = 1 
1–3–3–1====>> n = 3, C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C( 3,3)=1 
1–4–6–4–1==>> n = 4, C(4,0) = 1, C(4,1) = 4, C(4,2) = 6, C(4,3)=4, C(4,4)=1 
Así que aquí cada ciclo en i, construye la i-ésima fila del triángulo pascal, usando (i-1)-ésima fila
En cualquier momento, cada elemento de la array C tendrá algún valor (CERO o más) y en la próxima iteración, el valor de esos elementos proviene de la iteración anterior. 
En declaración, 
C[j] = C[j] + C[j-1] 
El lado derecho representa el valor proveniente de la iteración anterior (una fila del triángulo de Pascal depende de la fila anterior). El lado izquierdo representa el valor de la iteración actual que se obtendrá con esta declaración. 

Let's say we want to calculate C(4, 3), 
i.e. n=4, k=3:

All elements of array C of size 4 (k+1) are
initialized to ZERO.

i.e. C[0] = C[1] = C[2] = C[3] = C[4] = 0;
Then C[0] is set to 1

For i = 1:
C[1] = C[1] + C[0] = 0 + 1 = 1 ==>> C(1,1) = 1

For i = 2:
C[2] = C[2] + C[1] = 0 + 1 = 1 ==>> C(2,2) = 1
C[1] = C[1] + C[0] = 1 + 1 = 2 ==>> C(2,1) = 2

For i=3:
C[3] = C[3] + C[2] = 0 + 1 = 1 ==>> C(3,3) = 1
C[2] = C[2] + C[1] = 1 + 2 = 3 ==>> C(3,2) = 3
C[1] = C[1] + C[0] = 2 + 1 = 3 ==>> C(3,1) = 3

For i=4:
C[4] = C[4] + C[3] = 0 + 1 = 1 ==>> C(4,4) = 1
C[3] = C[3] + C[2] = 1 + 3 = 4 ==>> C(4,3) = 4
C[2] = C[2] + C[1] = 3 + 3 = 6 ==>> C(4,2) = 6
C[1] = C[1] + C[0] = 3 + 1 = 4 ==>> C(4,1) = 4

C(4,3) = 4 is would be the answer in our example.

Enfoque de memorización: la idea es crear una tabla de búsqueda y seguir el enfoque recursivo de arriba hacia abajo. Antes de calcular cualquier valor, verificamos si ya está en la tabla de búsqueda. Si es así, devolvemos el valor. De lo contrario, calculamos el valor y lo almacenamos en la tabla de búsqueda. El siguiente es el enfoque de arriba hacia abajo de la programación dinámica para encontrar el valor del coeficiente binomial. 

C++

// A Dynamic Programming based
// solution that uses
// table dp[][] to calculate
// the Binomial Coefficient
// A naive recursive approach
// with table C++ implementation
#include <bits/stdc++.h>
using namespace std;
 
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeffUtil(int n, int k, int** dp)
{
    // If value in lookup table then return
    if (dp[n][k] != -1) //    
        return dp[n][k];
 
    // store value in a table before return
    if (k == 0) {
        dp[n][k] = 1;
        return dp[n][k];
    }
     
    // store value in table before return
    if (k == n) {
        dp[n][k] = 1;
        return dp[n][k];
    }
     
    // save value in lookup table before return
    dp[n][k] = binomialCoeffUtil(n - 1, k - 1, dp) +
               binomialCoeffUtil(n - 1, k, dp);
    return dp[n][k];
}
 
int binomialCoeff(int n, int k)
{
    int** dp; // make a temporary lookup table
    dp = new int*[n + 1];
 
    // loop to create table dynamically
    for (int i = 0; i < (n + 1); i++) {
        dp[i] = new int[k + 1];
    }
 
    // nested loop to initialise the table with -1
    for (int i = 0; i < (n + 1); i++) {
        for (int j = 0; j < (k + 1); j++) {
            dp[i][j] = -1;
        }
    }
 
    return binomialCoeffUtil(n, k, dp);
}
 
/* Driver code*/
int main()
{
    int n = 5, k = 2;
    cout << "Value of C(" << n << ", " << k << ") is "
         << binomialCoeff(n, k) << endl;
    return 0;
}
 
// This is code is contributed by MOHAMMAD MUDASSIR

Java

// A Dynamic Programming based
// solution that uses
// table dp[][] to calculate
// the Binomial Coefficient
// A naive recursive approach
// with table Java implementation
import java.util.*;
class GFG{
 
// Returns value of Binomial
// Coefficient C(n, k)
static int binomialCoeffUtil(int n, int k,
                             Vector<Integer> []dp)
{
  // If value in lookup table
  // then return
  if (dp[n].get(k) != -1)    
    return dp[n].get(k);
 
  // store value in a table
  // before return
  if (k == 0)
  {
    dp[n].add(k, 1);
    return dp[n].get(k);
  }
 
  // store value in table
  // before return
  if (k == n)
  {
    dp[n].add(k, 1);
    return dp[n].get(k);
  }
 
  // save value in lookup table
  // before return
  dp[n].add(k, binomialCoeffUtil(n - 1,
                                 k - 1, dp) +
               binomialCoeffUtil(n - 1,
                                 k, dp));
  return dp[n].get(k);
}
 
static int binomialCoeff(int n, int k)
{
  // Make a temporary lookup table
  Vector<Integer> []dp = new Vector[n+1];
 
  // Loop to create table dynamically
  for (int i = 0; i < (n + 1); i++)
  {
    dp[i] = new Vector<Integer>();
    for(int j = 0; j <= k; j++)
      dp[i].add(-1);
  }
  return binomialCoeffUtil(n, k, dp);
}
 
// Driver code
public static void main(String[] args)
{
  int n = 5, k = 2;
  System.out.print("Value of C(" + n +
                   ", " + k + ") is " +
                   binomialCoeff(n, k) + "\n");
}
}
 
// This code is contributed by Rajput-Ji

Python3

# A Dynamic Programming based solution
# that uses table dp[][] to calculate
# the Binomial Coefficient. A naive
# recursive approach with table
# Python3 implementation
 
# Returns value of Binomial
# Coefficient C(n, k)
def binomialCoeffUtil(n, k, dp):
     
    # If value in lookup table then return
    if dp[n][k] != -1:
        return dp[n][k]
 
    # Store value in a table before return
    if k == 0:
        dp[n][k] = 1
        return dp[n][k]
     
    # Store value in table before return
    if k == n:
        dp[n][k] = 1
        return dp[n][k]
     
    # Save value in lookup table before return
    dp[n][k] = (binomialCoeffUtil(n - 1, k - 1, dp) +
                binomialCoeffUtil(n - 1, k, dp))
                 
    return dp[n][k]
 
def binomialCoeff(n, k):
     
    # Make a temporary lookup table
    dp = [ [ -1 for y in range(k + 1) ]
                for x in range(n + 1) ]
 
    return binomialCoeffUtil(n, k, dp)
 
# Driver code
n = 5
k = 2
 
print("Value of C(" + str(n) +
               ", " + str(k) + ") is",
               binomialCoeff(n, k))
 
# This is code is contributed by Prateek Gupta

C#

// C# program for the
// above approach
 
// A Dynamic Programming based
// solution that uses
// table [,]dp to calculate
// the Binomial Coefficient
// A naive recursive approach
// with table C# implementation
using System;
using System.Collections.Generic;
class GFG{
 
// Returns value of Binomial
// Coefficient C(n, k)
static int binomialCoeffUtil(int n, int k,
                             List<int> []dp)
{
  // If value in lookup table
  // then return
  if (dp[n][k] != -1)    
    return dp[n][k];
 
  // store value in a table
  // before return
  if (k == 0)
  {
    dp[n][k] = 1;
    return dp[n][k];
  }
 
  // store value in table
  // before return
  if (k == n)
  {
    dp[n][k] = 1;
    return dp[n][k];
  }
 
  // save value in lookup table
  // before return
  dp[n][k] = binomialCoeffUtil(n - 1,
                               k - 1, dp) +
             binomialCoeffUtil(n - 1,
                               k, dp);
  return dp[n][k];
}
 
static int binomialCoeff(int n, int k)
{
  // Make a temporary lookup table
  List<int> []dp = new List<int>[n + 1];
 
  // Loop to create table dynamically
  for (int i = 0; i < (n + 1); i++)
  {
    dp[i] = new List<int>();
 
    for(int j = 0; j <= k; j++)
      dp[i].Add(-1);
  }
  return binomialCoeffUtil(n, k, dp);
}
 
// Driver code
public static void Main(String[] args)
{
  int n = 5, k = 2;
  Console.Write("Value of C(" + n +
                ", " + k + ") is " +
                binomialCoeff(n, k) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

JavaScript

<script>
 
// A Dynamic Programming based solution that
// uses table dp[][] to calculate the
// Binomial Coefficient. A naive recursive
// approach with table Javascript implementation
 
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeffUtil(n, k, dp)
{
     
    // If value in lookup table
    // then return
    if (dp[n][k] != -1)   
        return dp[n][k];
     
    // Store value in a table
    // before return
    if (k == 0)
    {
        dp[n][k] = 1;
        return dp[n][k];
    }
     
    // Store value in table
    // before return
    if (k == n)
    {
        dp[n][k] = 1;
        return dp[n][k];
    }
     
    // Save value in lookup table
    // before return
    dp[n][k] = binomialCoeffUtil(n - 1,
                                 k - 1, dp) +
               binomialCoeffUtil(n - 1,
                                 k, dp);
    return dp[n][k];
}
 
function binomialCoeff(n, k)
{
     
    // Make a temporary lookup table
    let dp = new Array(n + 1);
     
    // Loop to create table dynamically
    for(let i = 0; i < (n + 1); i++)
    {
        dp[i] = [];
        for(let j = 0; j <= k; j++)
            dp[i].push(-1);
    }
    return binomialCoeffUtil(n, k, dp);
}
 
// Driver code
let n = 5, k = 2;
document.write("Value of C(" + n +
               ", " + k + ") is " +
               binomialCoeff(n, k) + "\n");
 
// This code is contributed by avanitrachhadiya2155
 
</script>

 Complejidad de tiempo: O(n*k)

Espacio Auxiliar: O(n*k)

Producción

Value of C(5, 2) is 10

Cancelación de factores entre numerador y denominador:

nCr = (n-r+1)*(n-r+2)*….*n / (r!)

Cree una array arr de números de n-r+1 a n que será de tamaño r. Como nCr siempre es un número entero, todos los números en el denominador deben cancelarse con el producto del numerador (representado por arr).

para i = 1 a i = r,

        busque arr, si arr[j] y tengo gcd>1, divida ambos por el gcd y cuando se convierta en 1, detenga la búsqueda

Ahora, la respuesta es solo el producto de arr, cuyo valor mod 10^9+7 se puede encontrar usando un solo paso y la fórmula use (a*b)%mod = (a%mod * b%mod)%mod 

C++

// C++ program to find gcd of
// two numbers in O(log(min(a,b)))
 
#include <bits/stdc++.h>
using namespace std;
 
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
int nCr(int n, int r)
{
    // base case
    if (r > n)
        return 0;
 
    // C(n,r) = C(n,n-r)
    if (r > n - r)
        r = n - r;
 
    int mod = 1000000007;
 
    // array of elements from n-r+1 to n
    int arr[r];
 
    for (int i = n - r + 1; i <= n; i++) {
        arr[i + r - n - 1] = i;
    }
 
    long int ans = 1;
    // for numbers from 1 to r find arr[j],
    // such that gcd(i,arr[j])>1
    for (int k = 1; k < r + 1; k++) {
        int j = 0, i = k;
        while (j < r) {
            int x = gcd(i, arr[j]);
            if (x > 1) {
                // if gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }
 
            // if i becomes 1, no need to search arr
            if (i == 1)
                break;
            j += 1;
        }
    }
 
    // single pass to multiply the numerator
    for (int i : arr)
        ans = (ans * i) % mod;
    return (int)ans;
}
 
int main()
{
    int n = 5, r = 2;
    cout << "Value of C(" << n << ", " << r << ") is "
         << nCr(n, r) << "\n";
    return 0;
}
 
// This code is contributed by rajsanghavi9.

Java

import java.util.*;
class GFG
{
    static int gcd(int a, int b) // function to find gcd of two numbers in O(log(min(a,b)))
    {
        if(b==0) return a;
        return gcd(b,a%b);
    }
    static int nCr(int n, int r)
    {
        if(r>n) // base case
            return 0;
        if(r>n-r) // C(n,r) = C(n,n-r) better time complexity for lesser r value
            r = n-r;
        int mod = 1000000007;
        int[] arr = new int[r]; // array of elements from n-r+1 to n
        for(int i=n-r+1; i<=n; i++)
        {
            arr[i+r-n-1] = i;
        }
        long ans = 1;
        for(int k=1;k<r+1;k++) // for numbers from 1 to r find arr[j] such that gcd(i,arr[j])>1
        {
            int j=0, i=k;
            while(j<arr.length)
            {
                int x = gcd(i,arr[j]);
                if(x>1)
                {
                    arr[j] /= x; // if gcd>1, divide both by gcd
                    i /= x;
                }
                if(i==1)
                    break; // if i becomes 1, no need to search arr
                j += 1;
            }
        }
        for(int i : arr) // single pass to multiply the numerator
            ans = (ans*i)%mod;
        return (int)ans;
    }
    // Driver code
    public static void main(String[] args)
    {
        int n = 5, r = 2;
        System.out.print("Value of C(" +  n+ ", " +  r+ ") is "
             +nCr(n, r) +"\n");
    }
}
// This code is contributed by Gautam Wadhwani

Python3

import math
class GFG:
    def nCr(self, n, r):
        def gcd(a,b): # function to find gcd of two numbers in O(log(min(a,b)))
            if b==0: # base case
                return a
            return gcd(b,a%b)
        if r>n:
            return 0
        if r>n-r: # C(n,r) = C(n,n-r) better time complexity for lesser r value
            r = n-r
        mod = 10**9 + 7
        arr = list(range(n-r+1,n+1)) # array of elements from n-r+1 to n
        ans = 1
        for i in range(1,r+1): # for numbers from 1 to r find arr[j] such that gcd(i,arr[j])>1
            j=0
            while j<len(arr):
                x = gcd(i,arr[j])
                if x>1:
                    arr[j] //= x # if gcd>1, divide both by gcd
                    i //= x
                if arr[j]==1: # if element becomes 1, its of no use anymore so delete from arr
                    del arr[j]
                    j -= 1
                if i==1:
                    break # if i becomes 1, no need to search arr
                j += 1
        for i in arr: # single pass to multiply the numerator
            ans = (ans*i)%mod
        return ans
     # Driver code
n = 5
k = 2
ob = GFG()
print("Value of C(" + str(n) +
               ", " + str(k) + ") is",
               ob.nCr(n, k))
  
# This is code is contributed by Gautam Wadhwani

C#

using System;
 
class GFG{
 
// Function to find gcd of two numbers
// in O(log(min(a,b)))  
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
         
    return gcd(b, a % b);
}
 
static int nCr(int n, int r)
{
     
    // Base case
    if (r > n)
        return 0;
         
    // C(n,r) = C(n,n-r) better time
    // complexity for lesser r value
    if (r > n - r)
        r = n - r;
         
    int mod = 1000000007;
     
    // Array of elements from n-r+1 to n
    int[] arr = new int[r];
    for(int i = n - r + 1; i <= n; i++)
    {
        arr[i + r - n - 1] = i;
    }
    long ans = 1;
     
    // For numbers from 1 to r find arr[j]
    // such that gcd(i,arr[j])>1
    for(int k = 1; k < r + 1; k++)
    {
        int j = 0, i = k;
        while (j < arr.Length)
        {
            int x = gcd(i,arr[j]);
            if (x > 1)
            {
                 
                // If gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }
             
            if (i == 1)
             
                // If i becomes 1, no need
                // to search arr
                break;
                 
            j += 1;
        }
    }
     
    // Single pass to multiply the numerator
    foreach(int i in arr)
        ans = (ans * i) % mod;
         
    return (int)ans;
}
 
// Driver code
static public void Main()
{
    int n = 5, r = 2;
    Console.WriteLine("Value of C(" +  n +
                      ", " +  r + ") is " +
                      nCr(n, r) + "\n");
}
}
 
// This code is contributed by rag2127

JavaScript

<script>
 
// Javascript program to find gcd of
// two numbers in O(log(min(a,b)))
function gcd(a, b)
{
    if (b == 0)
        return a;
         
    return gcd(b, a % b);
}
 
function nCr(n, r)
{
     
    // Base case
    if (r > n)
        return 0;
     
    // C(n,r) = C(n,n-r) better time
    // complexity for lesser r value
    if (r > n - r)
        r = n - r;
         
     mod = 1000000007;
      
    // Array of elements from n-r+1 to n
    var arr = new Array(r);
    for(var i = n - r + 1; i <= n; i++)
    {
        arr[i + r - n - 1] = i;
    }
    var ans = 1;
     
    // For numbers from 1 to r find arr[j]
    // such that gcd(i,arr[j])>1
    for(var k = 1; k < r + 1 ; k++)
    {
        var j = 0, i = k;
        while (j < arr.length)
        {
            var x = gcd(i, arr[j]);
            if (x > 1)
            {
                 
                // If gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }
             
            // If i becomes 1, no need to search arr
            if (i == 1)
                break;
                 
            j += 1;
        }
    }
     
    // Single pass to multiply the numerator
    arr.forEach(function (i)
    {
        ans = (ans * i) % mod;
    });
     
    return ans;
}
 
// Driver code
var n = 5, r = 2;
document.write("Value of C(" + n + ", " +
               r + ") is " + nCr(n, r) + "\n");
 
// This code is contributed by shivanisinghss2110
 
</script>
Producción

Value of C(5, 2) is 10

Complejidad de tiempo: O(( min(r, nr)^2 ) * log(n)),    útil cuando n >> r o n >> (nr)

Espacio auxiliar: O(min(r, nr))

Ver esto para GCD en tiempo logarítmico

Factorización prima de todos los números del 1 al n usando la criba de Eratóstenes:

1. Cree una array SPF de tamaño n+1 al factor primo más pequeño de cada número de 1 a n 

Set SPF[i] = i for all i = 1 to i = n

2. Usar Tamiz de Eratóstenes:

for i = 2 to i = n:
    if i is prime,
       for all multiples j of i, j<=n:
           if SPF[j] equals j, set SPF[j] = i

3. Una vez que conocemos el SPF de cada número del 1 al n, podemos encontrar la descomposición en factores primos de cualquier número del 1 al n en tiempo O(log(n)) usando la división recursiva por el SPF hasta que el número se convierte en 1

Now, nCr  =  (n-r+1)*(r+2)* ... *(n) / (r)!

4. Cree un diccionario (o hashmap) para almacenar la frecuencia de cada número primo en la descomposición en factores primos del valor real de nCr.

5. Entonces, solo calcule la frecuencia de cada número primo en nCr y multiplíquelos elevados a la potencia de su frecuencia.

6. Para el numerador, itere a través de i = n-r+1 hasta i = n, y para todos los factores primos de i, almacene su frecuencia en un diccionario.

( prime_pow[prime_factor] += freq_in_i )

7. Para el denominador, itere desde i = 1 hasta i = r y ahora reste la frecuencia en lugar de sumar.

8. Ahora, recorra el diccionario y multiplique la respuesta a (prime ^ prime_pow[prime]) % (10^9 + 7)

ans = (ans * pow(prime, prime_pow[prime], mod) ) % mod

C++

#include <bits/stdc++.h>
using namespace std;
 
// pow(base,exp,mod) is used to find
// (base^exp)%mod fast -> O(log(exp))
long int pow(long int b, long int exp, long int mod)
{
    long int ret = 1;
 
    while (exp > 0) {
        if ((exp & 1) > 0)
            ret = (ret * b) % mod;
        b = (b * b) % mod;
        exp >>= 1;
    }
 
    return ret;
}
 
int nCr(int n, int r)
{
    // base case
    if (r > n)
        return 0;
 
    // C(n,r) = C(n,n-r) Complexity for
    // this code is lesser for lower n-r
    if (n - r > r)
        r = n - r;
 
    // list to smallest prime factor
    // of each number from 1 to n
    int SPF[n + 1];
 
    // set smallest prime factor of each
    // number as itself
    for (int i = 1; i <= n; i++)
        SPF[i] = i;
 
    // set smallest prime factor of all
    // even numbers as 2
    for (int i = 4; i <= n; i += 2)
        SPF[i] = 2;
 
    for (int i = 3; i * i < n + 1; i += 2) {
 
        // Check if i is prime
        if (SPF[i] == i) {
            // All multiples of i are
            // composite (and divisible by
            // i) so add i to their prime
            // factorization getpow(j,i)
            // times
            for (int j = i * i; j < n + 1; j += i)
                if (SPF[j] == j) {
                    SPF[j] = i;
                }
        }
    }
    // Hash Map to store power of
    // each prime in C(n,r)
    map<int, int> prime_pow;
 
    // For numerator count frequency of each prime factor
    for (int i = r + 1; i < n + 1; i++) {
 
        int t = i;
 
        // Recursive division to find
        // prime factorization of i
        while (t > 1) {
            if (!prime_pow[SPF[t]]) {
                prime_pow[SPF[t]] = 1;
            }
            else
                prime_pow[SPF[t]]++;
            // prime_pow.put(SPF[t],
            // prime_pow.getOrDefault(SPF[t], 0)
            // + 1);
            t /= SPF[t];
        }
    }
 
    // For denominator subtract the power of
    // each prime factor
    for (int i = 1; i < n - r + 1; i++) {
        int t = i;
 
        // Recursive division to find
        // prime factorization of i
        while (t > 1) {
            prime_pow[SPF[t]]--;
            // prime_pow.put(SPF[t],
            // prime_pow.get(SPF[t]) - 1);
            t /= SPF[t];
        }
    }
 
    // long because mod is large and a%mod
    // * b%mod can overflow int
    long int ans = 1, mod = 1000000007;
 
    // use (a*b)%mod = (a%mod * b%mod)%mod
    for (auto it : prime_pow)
 
        // pow(base,exp,mod) is used to
        // find (base^exp)%mod fast
        ans = (ans
               * pow(it.first, prime_pow[it.first], mod))
              % mod;
    return (int)ans;
}
 
int main()
{
    int n = 5, r = 2;
    cout << "Value of C(" << n << ", " << r << ") is "
         << nCr(n, r) << "\n";
    return 0;
}
 
// This code is contributed by rajsanghavi9.

Java

import java.util.*;
class GFG {
   
    // pow(base,exp,mod) is used to find
    // (base^exp)%mod fast -> O(log(exp))
    static long pow(long b, long exp, long mod)
    {
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) > 0)
                ret = (ret * b) % mod;
            b = (b * b) % mod;
            exp >>= 1;
        }
        return ret;
    }
    static int nCr(int n, int r)
    {
        if (r > n) // base case
            return 0;
       
        // C(n,r) = C(n,n-r) Complexity for
        // this code is lesser for lower n-r
        if (n - r > r)
            r = n - r;
       
        // list to smallest prime factor
        // of each number from 1 to n
        int[] SPF = new int[n + 1];
       
        // set smallest prime factor of each
        // number as itself
        for (int i = 1; i <= n; i++)
            SPF[i] = i;
       
        // set smallest prime factor of all
        // even numbers as 2
        for (int i = 4; i <= n; i += 2)
            SPF[i] = 2;
       
        for (int i = 3; i * i < n + 1; i += 2)
        {
             
            // Check if i is prime
            if (SPF[i] == i)
            {
               
                // All multiples of i are
                // composite (and divisible by
                // i) so add i to their prime
                // factorization getpow(j,i)
                // times
                for (int j = i * i; j < n + 1; j += i)
                    if (SPF[j] == j) {
                        SPF[j] = i;
                    }
            }
        }
       
       // Hash Map to store power of
       // each prime in C(n,r)
       Map<Integer, Integer> prime_pow
            = new HashMap<>();
       
        // For numerator count frequency of each prime factor
        for (int i = r + 1; i < n + 1; i++)
        {
            int t = i;
             
            // Recursive division to find
            // prime factorization of i
            while (t > 1)
            {
                prime_pow.put(SPF[t],
                    prime_pow.getOrDefault(SPF[t], 0) + 1);
                t /= SPF[t];
            }
        }
       
        // For denominator subtract the power of
        // each prime factor
        for (int i = 1; i < n - r + 1; i++)
        {
            int t = i;
           
            // Recursive division to find
            // prime factorization of i
            while (t > 1)
            {
                prime_pow.put(SPF[t],
                              prime_pow.get(SPF[t]) - 1);
                t /= SPF[t];
            }
        }
       
        // long because mod is large and a%mod
        // * b%mod can overflow int
        long ans = 1, mod = 1000000007;
       
        // use (a*b)%mod = (a%mod * b%mod)%mod
        for (int i : prime_pow.keySet())
           
            // pow(base,exp,mod) is used to
            // find (base^exp)%mod fast
            ans = (ans * pow(i, prime_pow.get(i), mod))
                  % mod;
        return (int)ans;
    }
   
    // Driver code
    public static void main(String[] args)
    {
        int n = 5, r = 2;
        System.out.print("Value of C(" + n + ", " + r
                         + ") is " + nCr(n, r) + "\n");
    }
}
// This code is contributed by Gautam Wadhwani

Python3

# Python code for the above approach
import math
 
class GFG:
    def nCr(self, n, r):
       
        # Base case
        if r > n: 
            return 0
           
        # C(n,r) = C(n,n-r) Complexity for this
        # code is lesser for lower n-r
        if n - r > r: 
            r = n - r
             
        # List to store smallest prime factor
        # of every number from 1 to n
        SPF = [i for i in range(n+1)]
        for i in range(4, n+1, 2):
           
            # set smallest prime factor of
            # all even numbers as 2
            SPF[i] = 2
     
        for i in range(3, n+1, 2): 
         
            if i*i > n:
                break
               
            # Check if i is prime
            if SPF[i] == i: 
                 
                # All multiples of i are composite
                # (and divisible by i) so add i to
                # their prime factorization getpow(j,i) times
                for j in range(i*i, n+1, i):
                    if SPF[j] == j:
                       
                        # set smallest prime factor
                        # of j to i only if it is
                        # not previously set
                        SPF[j] = i
         
         # dictionary to store power of each prime in C(n,r)
        prime_pow = {} 
         
        # For numerator count frequency
        # of each prime factor
        for i in range(r+1, n+1):
            t = i
             
            # Recursive division to
            # find prime factorization of i
            while t > 1:
                if not SPF[t] in prime_pow:
                    prime_pow[SPF[t]] = 1
                else:
                    prime_pow[SPF[t]] += 1
                t //= SPF[t]
         
        # For denominator subtract the
        # power of each prime factor
        for i in range(1, n-r+1): 
            t = i
             
            # Recursive division to
            # find prime factorization of i
            while t > 1: 
                prime_pow[SPF[t]] -= 1
                t //= SPF[t]
        ans = 1
        mod = 10**9 + 7
         
         # Use (a*b)%mod = (a%mod * b%mod)%mod
        for i in prime_pow:
           
            # pow(base,exp,mod) is used to
            # find (base^exp)%mod fast
            ans = (ans*pow(i, prime_pow[i], mod)) % mod
        return ans
 
 
# Driver code
n = 5
k = 2
ob = GFG()
print("Value of C(" + str(n) +
      ", " + str(k) + ") is",
      ob.nCr(n, k))
 
# This is code is contributed by Gautam Wadhwani

C#

using System;
using System.Collections.Generic;
 
public class GFG {
 
    // pow(base,exp,mod) is used to find
    // (base^exp)%mod fast -> O(log(exp))
    static long pow(long b, long exp, long mod)
    {
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) > 0)
                ret = (ret * b) % mod;
            b = (b * b) % mod;
            exp >>= 1;
        }
        return ret;
    }
    static int nCr(int n, int r)
    {
        if (r > n) // base case
            return 0;
 
        // C(n,r) = C(n,n-r) Complexity for
        // this code is lesser for lower n-r
        if (n - r > r)
            r = n - r;
 
        // list to smallest prime factor
        // of each number from 1 to n
        int[] SPF = new int[n + 1];
 
        // set smallest prime factor of each
        // number as itself
        for (int i = 1; i <= n; i++)
            SPF[i] = i;
 
        // set smallest prime factor of all
        // even numbers as 2
        for (int i = 4; i <= n; i += 2)
            SPF[i] = 2;
 
        for (int i = 3; i * i < n + 1; i += 2) {
 
            // Check if i is prime
            if (SPF[i] == i) {
 
                // All multiples of i are
                // composite (and divisible by
                // i) so add i to their prime
                // factorization getpow(j,i)
                // times
                for (int j = i * i; j < n + 1; j += i)
                    if (SPF[j] == j) {
                        SPF[j] = i;
                    }
            }
        }
 
        // Hash Map to store power of
        // each prime in C(n,r)
        Dictionary<int, int> prime_pow
            = new Dictionary<int, int>();
 
        // For numerator count frequency of each prime
        // factor
        for (int i = r + 1; i < n + 1; i++) {
            int t = i;
 
            // Recursive division to find
            // prime factorization of i
            while (t > 1) {
                if (prime_pow.ContainsKey(SPF[t])) {
                    prime_pow[SPF[t]]
                        = prime_pow[SPF[t]] + 1;
                }
                else {
                    prime_pow.Add(SPF[t], 1);
                }
                t /= SPF[t];
            }
        }
 
        // For denominator subtract the power of
        // each prime factor
        for (int i = 1; i < n - r + 1; i++) {
            int t = i;
 
            // Recursive division to find
            // prime factorization of i
            while (t > 1) {
 
                if (prime_pow.ContainsKey(SPF[t])) {
                    prime_pow[SPF[t]]
                        = prime_pow[SPF[t]] - 1;
                }
 
                t /= SPF[t];
            }
        }
 
        // long because mod is large and a%mod
        // * b%mod can overflow int
        long ans = 1, mod = 1000000007;
 
        // use (a*b)%mod = (a%mod * b%mod)%mod
        foreach(int i in prime_pow.Keys)
 
            // pow(base,exp,mod) is used to
            // find (base^exp)%mod fast
            ans
            = (ans * pow(i, prime_pow[i], mod)) % mod;
        return (int)ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 5, r = 2;
        Console.Write("Value of C(" + n + ", " + r + ") is "
                      + nCr(n, r) + "\n");
    }
}
 
// This code contributed by gauravrajput1

JavaScript

<script>
 
// Javascript program to find gcd of
// two numbers in O(log(min(a,b)))
 
function gcd(a, b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
function nCr(n, r)
{
    // base case
    if (r > n)
        return 0;
 
    // C(n,r) = C(n,n-r)
    if (r > n - r)
        r = n - r;
 
    var mod = 1000000007;
 
    // array of elements from n-r+1 to n
    var arr = new Array(r);
 
    for (var i = n - r + 1; i <= n; i++) {
        arr[i + r - n - 1] = i;
    }
 
    var ans = 1;
    // for numbers from 1 to r find arr[j],
    // such that gcd(i,arr[j])>1
    for (var k = 1; k < r + 1; k++) {
        var j = 0;
        var i = k;
        do {
            var x = gcd(i, arr[j]);
            if (x > 1) {
                // if gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }
 
            // if i becomes 1, no need to search arr
            if (i == 1)
                break;
            j += 1;
        }
        while (j < r);
    }
 
    // single pass to multiply the numerator
     
    arr.forEach(function (i, index) {
          ans = (ans * i) % mod;
    });       
    return ans;
}
 
 
var n = 5;
var r = 2;
document.write("Value of C(" + n + ", " + r + ") is " + nCr(n, r) + "<br>");
 
// This code is contributed by shivani.
 
</script>
Producción

Value of C(5, 2) is 10

Complejidad de tiempo: O(n*log(n)),   muy útil cuando r->n/2

Espacio Auxiliar: O(n)

Vea esto para la factorización prima en O (log (n))

Otro enfoque: (técnica de inversión modular)

1. La fórmula general de nCr es ( n*(n-1)*(n-2)* … *(n-r+1) ) / (r!). Podemos usar directamente esta fórmula para encontrar nCr. Pero eso se desbordará fuera de límite. Necesitamos encontrar nCr mod m para que no se desborde. Podemos hacerlo fácilmente con fórmula aritmética modular. 

for the  n*(n-1)*(n-2)* ... *(n-r+1) part we can use the formula,
(a*b) mod m = ((a % m) * (b % m)) % m

2. y para el 1/r! parte, necesitamos encontrar el inverso modular de cada número del 1 al r. Luego use la misma fórmula anterior con un inverso modular de 1 a r. Podemos encontrar el inverso modular en tiempo O(r) usando la fórmula, 

inv[1] = 1
inv[i] = − ⌊m/i⌋ * inv[m mod i] mod m
To use this formula, m has to be a prime.

En el problema de práctica, necesitamos mostrar la respuesta con módulo 1000000007 que es un número primo. 

Entonces, esta técnica funcionará.
 

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to find binomial
// coefficient
int binomialCoeff(int n, int r)
{
 
    if (r > n)
        return 0;
    long long int m = 1000000007;
    long long int inv[r + 1] = { 0 };
    inv[0] = 1;
    if(r+1>=2)
    inv[1] = 1;
 
    // Getting the modular inversion
    // for all the numbers
    // from 2 to r with respect to m
    // here m = 1000000007
    for (int i = 2; i <= r; i++) {
        inv[i] = m - (m / i) * inv[m % i] % m;
    }
 
    int ans = 1;
 
    // for 1/(r!) part
    for (int i = 2; i <= r; i++) {
        ans = ((ans % m) * (inv[i] % m)) % m;
    }
 
    // for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for (int i = n; i >= (n - r + 1); i--) {
        ans = ((ans % m) * (i % m)) % m;
    }
    return ans;
}
 
/* Driver code*/
int main()
{
    int n = 5, r = 2;
    cout << "Value of C(" << n << ", " << r << ") is "
         << binomialCoeff(n, r) << endl;
    return 0;
}

Java

// JAVA program for the above approach
import java.util.*;
class GFG
{
 
// Function to find binomial
// coefficient
static int binomialCoeff(int n, int r)
{
 
    if (r > n)
    return 0;
    long  m = 1000000007;
    long  inv[] = new long[r + 1];
    inv[0] = 1;
    if(r+1>=2)
    inv[1] = 1;
 
    // Getting the modular inversion
    // for all the numbers
    // from 2 to r with respect to m
    // here m = 1000000007
    for (int i = 2; i <= r; i++) {
        inv[i] = m - (m / i) * inv[(int) (m % i)] % m;
    }
 
    int ans = 1;
 
    // for 1/(r!) part
    for (int i = 2; i <= r; i++) {
        ans = (int) (((ans % m) * (inv[i] % m)) % m);
    }
 
    // for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for (int i = n; i >= (n - r + 1); i--) {
        ans = (int) (((ans % m) * (i % m)) % m);
    }
    return ans;
}
 
/* Driver code*/
public static void main(String[] args)
{
    int n = 5, r = 2;
    System.out.print("Value of C(" +  n+ ", " +  r+ ") is "
         +binomialCoeff(n, r) +"\n");
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 program for the above approach
 
# Function to find binomial
# coefficient
def binomialCoeff(n, r):
     
    if (r > n):
        return 0
         
    m = 1000000007
    inv = [0 for i in range(r + 1)]
    inv[0] = 1;
    if(r+1>=2)
    inv[1] = 1;
 
    # Getting the modular inversion
    # for all the numbers
    # from 2 to r with respect to m
    # here m = 1000000007
    for i in range(2, r + 1):
        inv[i] = m - (m // i) * inv[m % i] % m
 
    ans = 1
 
    # for 1/(r!) part
    for i in range(2, r + 1):
        ans = ((ans % m) * (inv[i] % m)) % m
 
    # for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for i in range(n, n - r, -1):
        ans = ((ans % m) * (i % m)) % m
         
    return ans
 
# Driver code
n = 5
r = 2
 
print("Value of C(" ,n , ", " , r ,
      ") is ",binomialCoeff(n, r))
 
# This code is contributed by rohan07

C#

// C# program for the above approach
using System;
 
public class GFG
{
 
// Function to find binomial
// coefficient
static int binomialCoeff(int n, int r)
{
 
    if (r > n)
    return 0;
    long  m = 1000000007;
    long  []inv = new long[r + 1];
    inv[0] = 1;
    if(r+1>=2)
    inv[1] = 1;
 
    // Getting the modular inversion
    // for all the numbers
    // from 2 to r with respect to m
    // here m = 1000000007
    for (int i = 2; i <= r; i++) {
        inv[i] = m - (m / i) * inv[(int) (m % i)] % m;
    }
 
    int ans = 1;
 
    // for 1/(r!) part
    for (int i = 2; i <= r; i++) {
        ans = (int) (((ans % m) * (inv[i] % m)) % m);
    }
 
    // for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for (int i = n; i >= (n - r + 1); i--) {
        ans = (int) (((ans % m) * (i % m)) % m);
    }
    return ans;
}
 
/* Driver code*/
public static void Main(String[] args)
{
    int n = 5, r = 2;
    Console.Write("Value of C(" +  n+ ", " +  r+ ") is "
         +binomialCoeff(n, r) +"\n");
}
}
 
 
 
// This code is contributed by 29AjayKumar

JavaScript

<script>
        // JavaScript Program for the above approach
 
        // Function to find binomial
        // coefficient
        function binomialCoeff(n, r) {
 
            if (r > n)
                return 0;
            let m = 1000000007;
            let inv = new Array(r + 1).fill(0);
            inv[0] = 1;
            if (r + 1 >= 2)
                inv[1] = 1;
 
            // Getting the modular inversion
            // for all the numbers
            // from 2 to r with respect to m
            // here m = 1000000007
            for (let i = 2; i <= r; i++) {
                inv[i] = m - Math.floor(m / i) * inv[m % i] % m;
            }
 
            let ans = 1;
 
            // for 1/(r!) part
            for (let i = 2; i <= r; i++) {
                ans = ((ans % m) * (inv[i] % m)) % m;
            }
 
            // for (n)*(n-1)*(n-2)*...*(n-r+1) part
            for (let i = n; i >= (n - r + 1); i--) {
                ans = ((ans % m) * (i % m)) % m;
            }
            return ans;
        }
 
        /* Driver code*/
        let n = 5, r = 2;
        document.write("Value of C(" + n + ", " + r + ") is "
            + binomialCoeff(n, r) + "<br>");
 
    // This code is contributed by Potta Lokesh
    </script>
Producción

Value of C(5, 2) is 10

Complejidad temporal: O(n+k)

Espacio Auxiliar: O(k) 

Vea esto para Coeficiente binomial eficiente en espacio y tiempo 

Referencias: 
http://www.csl.mtu.edu/cs4321/www/Lectures/Lecture%2015%20-%20Dynamic%20Programming%20Binomial%20Coficients.htm 

https://cp-algorithms.com/algebra/module-inverse.html

Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.  

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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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