Dynamic Programming | Set 9 (Binomial Coefficient) | GeeksforGeeks


1) Optimal Substructure
The value of C(n, k) can recursively calculated using following standard formula for Binomial Cofficients.
   C(n, k) = C(n-1, k-1) + C(n-1, k) 
   C(n, 0) = C(n, n) = 1
int binomialCoeff(int n, int k)
{
    int C[n+1][k+1];
    int 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];
}
Following is a space optimized version of the above code. The following code only uses O(k). 
// A space optimized Dynamic Programming Solution
int binomialCoeff(int n, int k)
{
    int* C = (int*)calloc(k+1, sizeof(int));
    int i, j, res;
    C[0] = 1;
    for(i = 1; i <= n; i++)
    {
        for(j = min(i, k); j > 0; j--)
            C[j] = C[j] + C[j-1];
    }
    res = C[k];  // Store the result before freeing memory
    free(C);  // free dynamically allocated memory to avoid memory leak
    return res;
}
Space and time efficient Binomial Coefficient
The value of C(n, k) can be calculated in O(k) time and O(1) extra space.
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)  // we can change r to n-r if r > n-r 
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;
}
UVa 530: Binomial Showdown
public static long Factorial(int n) {
 
    long res = 1;
 
    // n! = 1 * 2 * 3 * ... * (n - 1) * n
    for (int i = 1; i <= n; i++) {
        res *= i;
    }
 
    return res;
}
 
// computes n choose k
public static long Choose(int n, int k) {
 
    // n choose k = n! / (k! * (n - k)!)
    return Factorial(n) / (Factorial(k) * Factorial(n - k));
}

public static long Choose(int n, int k) {
 
    long res = 1;
 
    // n choose k = the product of (n - k + i) / i for i = 1..k
    for (int i = 1; i <= k; i++) {
        res = res * (n - k + i) / i;
    }
 
    return res;
}

\displaystyle \binom{n}{k} = \binom{n}{n - k}
It can easily be derived:
\displaystyle\binom{n}{n - k} = \frac{n!}{(n - k)! (n - (n - k))!} =
\displaystyle \frac{n!}{(n - k)! (k)!} = \binom{n}{k}
public static long Choose(int n, int k) {
 
    // n choose k is equal to n choose (n - k),
    // so we pick the one that is easier to compute,
    // which is the smaller one
    k = Math.min(k, n - k);
 
    long res = 1;
 
    // n choose k = the product of (n - k + i) / i for i = 1..k
    for (int i = 1; i <= k; i++) {
        res = res * (n - k + i) / i;
    }
 
    return res;
}
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