# Compute nCr % p | Set 1 (Introduction and Dynamic Programming Solution)

Given three numbers n, r and p, compute value of nCr mod p.
Example:

Input:  n = 10, r = 2, p = 13
Output: 6
Explanation: 10C2 is 45 and 45 % 13 is 6.

## We strongly recommend that you click here and practice it, before moving on to the solution.

#### METHOD 1: (Using Dynamic Programming)

A Simple Solution is to first compute nCr, then compute nCr % p. This solution works fine when the value of nCr is small.
What if the value of nCr is large?
The value of nCr%p is generally needed for large values of n when nCr cannot fit in a variable, and causes overflow. So computing nCr and then using modular operator is not a good idea as there will be overflow even for slightly larger values of n and r. For example the methods discussed here and here cause overflow for n = 50 and r = 40.
The idea is to compute nCr using below formula

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

Working of Above formula and Pascal Triangle:
Let us see how above formula works for C(4, 3)
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
Extension of above formula for modular arithmetic:
We can use distributive property of modulor operator to find nCr % p using above formula.

   C(n, r)%p = [ C(n-1, r-1)%p + C(n-1, r)%p ] % p
C(n, 0) = C(n, n) = 1

The above formula can implemented using Dynamic Programming using a 2D array.
The 2D array based dynamic programming solution can be further optimized by constructing one row at a time. See Space optimized version in below post for details.
Binomial Coefficient using Dynamic Programming
Below is implementation based on the space optimized version discussed in above post.

 // A Dynamic Programming based solution to compute nCr % p #include  using namespace std;   // Returns nCr % p int nCrModp(int n, int r, int p) {     // Optimization for the cases when r is large     if (r > n - r)         r = n - r;       // The array C is going to store last row of     // pascal triangle at the end. And last entry     // of last row is nCr     int C[r + 1];     memset(C, 0, sizeof(C));       C = 1; // Top row of Pascal Triangle       // One by constructs remaining rows of Pascal     // Triangle from top to bottom     for (int i = 1; i <= n; i++) {           // Fill entries of current row using previous         // row values         for (int j = min(i, r); j > 0; j--)               // nCj = (n-1)Cj + (n-1)C(j-1);             C[j] = (C[j] + C[j - 1]) % p;     }     return C[r]; }   // Driver program int main() {     int n = 10, r = 2, p = 13;     cout << "Value of nCr % p is " << nCrModp(n, r, p);     return 0; }

 // A Dynamic Programming based // solution to compute nCr % p import java.io.*; import java.util.*; import java.math.*;   class GFG {       // Returns nCr % p     static int nCrModp(int n, int r, int p)     {         if (r > n - r)             r = n - r;           // The array C is going to store last         // row of pascal triangle at the end.         // And last entry of last row is nCr         int C[] = new int[r + 1];           C = 1; // Top row of Pascal Triangle           // One by constructs remaining rows of Pascal         // Triangle from top to bottom         for (int i = 1; i <= n; i++) {               // Fill entries of current row using previous             // row values             for (int j = Math.min(i, r); j > 0; j--)                   // nCj = (n-1)Cj + (n-1)C(j-1);                 C[j] = (C[j] + C[j - 1]) % p;         }         return C[r];     }       // Driver program     public static void main(String args[])     {         int n = 10, r = 2, p = 13;         System.out.println("Value of nCr % p is "                            + nCrModp(n, r, p));     } }   // This code is contributed by Nikita Tiwari.

 # A Dynamic Programming based solution to compute nCr % p   # Returns nCr % p def nCrModp(n, r, p):       # Optimization for the cases when r is large     # compared to n-r      if (r > n- r):         r = n - r         # The array C is going to store last row of     # pascal triangle at the end. And last entry     # of last row is nCr.     C = [0 for i in range(r + 1)]       C = 1 # Top row of Pascal Triangle       # One by constructs remaining rows of Pascal     # Triangle from top to bottom     for i in range(1, n + 1):           # Fill entries of current row          # using previous row values         for j in range(min(i, r), 0, -1):               # nCj = (n - 1)Cj + (n - 1)C(j - 1)             C[j] = (C[j] + C[j-1]) % p       return C[r]   # Driver Program n = 10 r = 2 p = 13 print('Value of nCr % p is', nCrModp(n, r, p))   # This code is contributed by Soumen Ghosh

 // A Dynamic Programming based // solution to compute nCr % p using System;   class GFG {       // Returns nCr % p     static int nCrModp(int n, int r, int p)     {           // Optimization for the cases when r is large         if (r > n - r)             r = n - r;           // The array C is going to store last         // row of pascal triangle at the end.         // And last entry of last row is nCr         int[] C = new int[r + 1];           for (int i = 0; i < r + 1; i++)             C[i] = 0;           C = 1; // Top row of Pascal Triangle           // One by constructs remaining rows         // of Pascal Triangle from top to bottom         for (int i = 1; i <= n; i++) {               // Fill entries of current row using             // previous row values             for (int j = Math.Min(i, r); j > 0; j--)                   // nCj = (n-1)Cj + (n-1)C(j-1);                 C[j] = (C[j] + C[j - 1]) % p;         }           return C[r];     }       // Driver program     public static void Main()     {         int n = 10, r = 2, p = 13;           Console.Write("Value of nCr % p is "                       + nCrModp(n, r, p));     } }   // This code is contributed by nitin mittal.

  $n - $r)     $r = $n - $r; // The array C is going  // to store last row of // pascal triangle at  // the end. And last entry // of last row is nCr $C = array();   for( $i = 0; $i < $r + 1; $i++)     $C[$i] = 0;   // Top row of Pascal // Triangle $C = 1;  // One by constructs remaining  // rows of Pascal Triangle from  // top to bottom for ($i = 1; $i <= $n; $i++) {    // Fill entries of current   // row using previous row values  for ($j = Min($i, $r); $j > 0; $j--)           // nCj = (n-1)Cj + (n-1)C(j-1);         $C[$j] = ($C[$j] +                    $C[$j - 1]) % $p; } return $C[$r]; } // Driver Code $n = 10; $r = 2;$p = 13;   echo "Value of nCr % p is ",          nCrModp($n, $r, \$p);   // This code is contributed // by anuj_67. ?>

Output

Value of nCr % p is 6

Time complexity of above solution is O(n*r) and it requires O(r) space. There are more and better solutions to above problem.
Compute nCr % p | Set 2 (Lucas Theorem)

#### METHOD 2(Using Pascal Triangle and Dynamic Pro)

Another approach lies in utilizing the concept of the Pascal Triangle. Instead of calculating the nCrvalue for every n starting from n=0 till n=n, the approach aims at using the nth row itself for the calculation. The method proceeds by finding out a general relationship between nCr and nCr-1.

FORMULA: C(n,r)=C(n,r-1)* (n-r+1)/r

Example:

For instance, take n=5 and r=3.

Input:  n = 5, r = 3, p = 1000000007
Output: 6
Explanation: 5C3 is 10 and 10 % 100000007 is 10.

As per the formula,
C(5,3)=5!/(3!)*(2!)
C(5,3)=10

Also,
C(5,2)=5!/(2!)*(3!)
C(5,2)=10

Let's try applying the above formula.

C(n,r)=C(n,r-1)* (n-r+1)/r
C(5,3)=C(5,2)*(5-3+1)/3
C(5,3)=C(5,2)*1
C(5,3)=10*1

The above example shows that C(n,r) can be easily calculated by calculating C(n,r-1) and multiplying the result with the term (n-r+1)/r. But this multiplication may cause integer overflow for large values of n. To tackle this situation, use modulo multiplication, and modulo division concepts in order to achieve optimizations in terms of integer overflow.

Let’s find out how to build Pascal Triangle for the same.

1D array declaration can be further optimized by just the declaration of a single variable to perform calculations. However, integer overflow demands other functions too for the final implementation.

The post below mentions the space and time-optimized implementation for the binary coefficient calculation.

 // C++ program to find the nCr%p  // based on optimal Dynamic // Programming implementation and  // Pascal Triangle concepts #include  using namespace std;   // Returns (a * b) % mod long long moduloMultiplication(long long a, long long b,                                long long mod) {       // Initialize result     long long res = 0;        // Update a if it is more than     // or equal to mod     a %= mod;       while (b) {           // If b is odd, add a with result         if (b & 1)             res = (res + a) % mod;           // Here we assume that doing 2*a         // doesn't cause overflow         a = (2 * a) % mod;         b >>= 1; // b = b / 2     }     return res; }   // C++ function for extended Euclidean Algorithm long long int gcdExtended(long long int a, long long int b,                           long long int* x,                           long long int* y);   // Function to find modulo inverse of b. It returns // -1 when inverse doesn't exists long long int modInverse(long long int b, long long int m) {       long long int x, y; // used in extended GCD algorithm     long long int g = gcdExtended(b, m, &x, &y);       // Return -1 if b and m are not co-prime     if (g != 1)         return -1;       // m is added to handle negative x     return (x % m + m) % m; }   // C function for extended Euclidean Algorithm (used to // find modular inverse. long long int gcdExtended(long long int a, long long int b,                           long long int* x,                           long long int* y) {       // Base Case     if (a == 0) {         *x = 0, *y = 1;         return b;     }       // To store results of recursive call     long long int x1, y1;        long long int gcd = gcdExtended(b % a, a, &x1, &y1);       // Update x and y using results of recursive     // call     *x = y1 - (b / a) * x1;     *y = x1;     return gcd; }   // Function to compute a/b under modlo m long long int modDivide(long long int a, long long int b,                         long long int m) {       a = a % m;     long long int inv = modInverse(b, m);     if (inv == -1)         // cout << "Division not defined";         return 0;     else         return (inv * a) % m; }   // Function to calculate nCr % p int nCr(int n, int r, int p) {       // Edge Case which is not possible     if (r > n)         return 0;       // Optimization for the cases when r is large     if (r > n - r)         r = n - r;       // x stores the current result at     long long int x = 1;          // each iteration     // Initialized to 1 as nC0 is always 1.     for (int i = 1; i <= r; i++) {           // Formula derived for calculating result is         // C(n,r-1)*(n-r+1)/r         // Function calculates x*(n-i+1) % p.         x = moduloMultiplication(x, (n + 1 - i), p);                 // Function calculates x/i % p.         x = modDivide(x, i, p);     }     return x; }   // Driver Code int main() {       long long int n = 5, r = 3, p = 1000000007;     cout << "Value of nCr % p is " << nCr(n, r, p);     return 0; }

Output
Value of nCr % p is 10

#### Complexity Analysis:

• The above code needs an extra of O(1) space for the calculations.
• The time involved in the calculation of nCr % p is of the order O(n).