A binomial coefficient C(n, k) can be defined as the coefficient of X^k in the expansion of (1 + X)^n.
A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) of an n-element set.
The Problem Write a function that takes two parameters n and k and returns the value of Binomial Coefficient C(n, k). For example, your function should return 6 for n = 4 and k = 2, and it should return 10 for n = 5 and k = 2.
1) Optimal Substructure
The value of C(n, k) can be recursively calculated using following standard formula for Binomial Coefficients.
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)
usingSystem;
classGFG {
// Returns value of Binomial
// Coefficient C(n, k)
staticintbinomialCoeff(intn, intk)
{
// Base Cases
if(k == 0 || k == n)
return1;
// Recur
returnbinomialCoeff(n - 1, k - 1) +
binomialCoeff(n - 1, k);
}
/* Driver program to test above function */
publicstaticvoidMain()
{
intn = 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)
functionbinomialCoeff($n, $k)
{
// Base Cases
if($k==0 || $k==$n)
return1;
// Recur
returnbinomialCoeff($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
?>
Output:
Value of C(52) is 10
2) Overlapping Subproblems
It should be noted that the above function computes the same subproblems again and again. See the following recursion tree for n = 5 an k = 2. The function C(3, 1) is called two times. For large values of n, there will be many common subproblems.
Since same suproblems are called again, this problem has Overlapping Subproblems property. So the Binomial Coefficient problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, re-computations of same subproblems can be avoided by constructing a temporary array C[][] in bottom up manner. Following is Dynamic Programming based implementation.
C++
// A Dynamic Programming based solution that uses
// table C[][] to calculate the Binomial Coefficient
#include<bits/stdc++.h>
usingnamespacestd;
// Prototype of a utility function that
// returns minimum of two integers
intmin(inta, intb);
// Returns value of Binomial Coefficient C(n, k)
intbinomialCoeff(intn, intk)
{
intC[n + 1][k + 1];
inti, j;
// Caculate 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 previosly
// stored values
else
C[i][j] = C[i - 1][j - 1] +
C[i - 1][j];
}
}
returnC[n][k];
}
// A utility function to return
// minimum of two integers
intmin(inta, intb)
{
return(a < b) ? a : b;
}
// Driver Code
intmain()
{
intn = 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
intmin(inta, intb);
// Returns value of Binomial Coefficient C(n, k)
intbinomialCoeff(intn, intk)
{
intC[n+1][k+1];
inti, j;
// Caculate 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 previosly stored values
else
C[i][j] = C[i-1][j-1] + C[i-1][j];
}
}
returnC[n][k];
}
// A utility function to return minimum of two integers
intmin(inta, intb)
{
return(a<b)? a: b;
}
/* Drier program to test above function*/
intmain()
{
intn = 5, k = 2;
printf("Value of C(%d, %d) is %d ", n, k, binomialCoeff(n, k) );
return0;
}
Java
// A Dynamic Programming based solution that uses table C[][] to
// calculate the Binomial Coefficient
classBinomialCoefficient
{
// Returns value of Binomial Coefficient C(n, k)
staticintbinomialCoeff(intn, intk)
{
intC[][] = newint[n+1][k+1];
inti, 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 previosly stored values
else
C[i][j] = C[i-1][j-1] + C[i-1][j];
}
}
returnC[n][k];
}
// A utility function to return minimum of two integers
staticintmin(inta, intb)
{
return(a<b)? a: b;
}
/* Driver program to test above function*/
publicstaticvoidmain(String args[])
{
intn = 5, k = 2;
System.out.println("Value of C("+n+","+k+") is "+binomialCoeff(n, k));
}
}
/*This code is contributed by Rajat Mishra*/
Python
# A Dynamic Programming based Python Program that uses table C[][]
# to calculate the Binomial Coefficient
# Returns value of Binomial Coefficient C(n, k)
defbinomialCoef(n, k):
C =[[0forx inrange(k+1)] forx inrange(n+1)]
# Calculate value of Binomial Coefficient in bottom up manner
fori inrange(n+1):
forj inrange(min(i, k)+1):
# Base Cases
ifj ==0orj ==i:
C[i][j] =1
# Calculate value using previosly stored values
else:
C[i][j] =C[i-1][j-1] +C[i-1][j]
returnC[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
usingSystem;
classGFG {
// Returns value of Binomial Coefficient
// C(n, k)
staticintbinomialCoeff(intn, intk)
{
int[,]C = newint[n+1,k+1];
inti, 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 previosly
// stored values
else
C[i,j] = C[i-1,j-1] + C[i-1,j];
}
}
returnC[n,k];
}
// A utility function to return minimum
// of two integers
staticintmin(inta, intb)
{
return(a < b) ? a : b;
}
/* Driver program to test above function*/
publicstaticvoidMain()
{
intn = 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)
functionbinomialCoeff( $n, $k)
{
$C= array(array());
$i; $j;
// Caculate value of Binomial
// Coefficient in bottom up manner
for($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
// previosly 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.
?>
Output:
Value of C[5][2] is 10
Time Complexity: O(n*k) Auxiliary Space: O(n*k)
Following is a space optimized version of the above code. The following code only uses O(k). Thanks to AK for suggesting this method.
C/C++
// C++ program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
#include<bits/stdc++.h>
usingnamespacestd;
intbinomialCoeff(intn, intk)
{
intC[k+1];
memset(C, 0, sizeof(C));
C[0] = 1; // nC0 is 1
for(inti = 1; i <= n; i++)
{
// Compute next row of pascal triangle using
// the previous row
for(intj = min(i, k); j > 0; j--)
C[j] = C[j] + C[j-1];
}
returnC[k];
}
/* Drier program to test above function*/
intmain()
{
intn = 5, k = 2;
printf("Value of C(%d, %d) is %d ",
n, k, binomialCoeff(n, k) );
return0;
}
Java
// JAVA Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
importjava.util.*;
classGFG {
staticintbinomialCoeff(intn, intk)
{
intC[] = newint[k + 1];
// nC0 is 1
C[0] = 1;
for(inti = 1; i <= n; i++)
{
// Compute next row of pascal
// triangle using the previous row
for(intj = Math.min(i, k); j > 0; j--)
C[j] = C[j] + C[j-1];
}
returnC[k];
}
/* Driver program */
publicstaticvoidmain(String[] args)
{
intn = 5, k = 2;
System.out.printf("Value of C(%d, %d) is %d "
, n, k, binomialCoeff(n, k));
}
}
Python
# Python program for Optimized Dynamic Programming solution to
# Binomail Coefficient. This one uses the concept of pascal
# Triangle and less memory
defbinomialCoeff(n , k):
# Declaring an empty array
C =[0fori inxrange(k+1)]
C[0] =1#since nC0 is 1
fori inrange(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
returnC[k]
# Driver Program to test the above function
n =5
k =2
print"Value of C(%d,%d) is %d"%(n,k,binomialCoeff(n,k))
# This code is contribtued by Nikhil Kumar Singh(nickzuck_007)
C#
// C# Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
usingSystem;
classGFG {
staticintbinomialCoeff(intn, intk)
{
int[] C = newint[k + 1];
// nC0 is 1
C[0] = 1;
for(inti = 1; i <= n; i++)
{
// Compute next row of pascal
// triangle using the previous
// row
for(intj = Math.Min(i, k);
j > 0; j--)
C[j] = C[j] + C[j-1];
}
returnC[k];
}
/* Driver program */
publicstaticvoidMain()
{
intn = 5, k = 2;
Console.WriteLine("Value of C("
+ n + " "+ k + ") is "
+ binomialCoeff(n, k));
}
}
// This code is contribtued by anuj_67.
PHP
<?php
// PHP program for space optimized
// Dynamic Programming Solution of
// Binomial Coefficient
functionbinomialCoeff($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.
?>
Output:
Value of C[5][2] is 10
Time Complexity: O(n*k) Auxiliary Space: O(k)
Explanation:
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
So here every loop on i, builds i’th row of pascal triangle, using (i-1)th row
At any time, every element of array C will have some value (ZERO or more) and in next iteration, value for those elements comes from previous iteration.
In statement,
C[j] = C[j] + C[j-1]
Right hand side represents the value coming from previous iteration (A row of Pascal’s triangle depends on previous row). Left Hand side represents the value of current iteration which will be obtained by this statement.
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(2,2) = 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.
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