Given three numbers n, r and p, compute value of nCr mod p.
Input: n = 10, r = 2, p = 13 Output: 6 Explanation: 10C2 is 45 and 45 % 13 is 6.
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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.
Below is implementation based on the space optimized version discussed in above post.
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.
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