Given an array A[] of size N. Solve Q queries. Find the product in the range [L, R] under modulo P ( P is Prime).
Examples:
Input : A[] = {1, 2, 3, 4, 5, 6}
L = 2, R = 5, P = 229
Output : 120
Input : A[] = {1, 2, 3, 4, 5, 6},
L = 2, R = 5, P = 113
Output : 7
Brute Force
For each of the queries, traverse each element in the range [L, R] and calculate the product under modulo P. This will answer each query in O(N).
Python3
# Python3 program to find # Product in range Queries in O(N)# Function to calculate Product # in the given range.def calculateProduct (A, L, R, P): # As our array is 0 based # and L and R are given as # 1 based index. L = L - 1 R = R - 1 ans = 1 for i in range(R + 1): ans = ans * A[i] ans = ans % P return ans # Driver codeA = [ 1, 2, 3, 4, 5, 6 ]P = 229L = 2R = 5print (calculateProduct(A, L, R, P))L = 1R = 3print (calculateProduct(A, L, R, P))# This code is contributed # by "Abhishek Sharma 44" |
Output :
120 6
Efficient Using Modular Multiplicative Inverse:
As P is prime, we can use Modular Multiplicative Inverse. Using dynamic programming, we can calculate a pre-product array under modulo P such that the value at index i contains the product in the range [0, i]. Similarly, we can calculate the pre-inverse product under modulo P. Now each query can be answered in O(1).
The inverse product array contains the inverse product in the range [0, i] at index i. So, for the query [L, R], the answer will be Product[R]*InverseProduct[L-1]
Note: We can not calculate the answer as Product[R]/Product[L-1] because the product is calculated under modulo P. If we do not calculate the product under modulo P there is always a possibility of overflow.
Python3
# Python3 implementation of the# above approach# Returns modulo inverse of a with # respect to m using extended Euclid # Algorithm. Assumption: a and m are # coprimes, i.e., gcd(a, m) = 1 def modInverse(a, m): m0, x0, x1 = m, 0, 1 if m == 1: return 0 while a > 1: # q is quotient q = a // m t = m # m is remainder now, process # same as Euclid's algo m, a = a % m, t t = x0 x0 = x1 - q * x0 x1 = t # Make x1 positive if x1 < 0: x1 += m0 return x1 # calculating pre_product array def calculate_Pre_Product(A, N, P): pre_product[0] = A[0] for i in range(1, N): pre_product[i] = pre_product[i - 1] * A[i] pre_product[i] = pre_product[i] % P # Calculating inverse_product # array. def calculate_inverse_product(A, N, P): inverse_product[0] = modInverse(pre_product[0], P) for i in range(1, N): inverse_product[i] = modInverse(pre_product[i], P) # Function to calculate # Product in the given range. def calculateProduct(A, L, R, P): # As our array is 0 based as # and L and R are given as 1 # based index. L = L - 1 R = R - 1 ans = 0 if L == 0: ans = pre_product[R] else: ans = pre_product[R] * inverse_product[L - 1] return ans # Driver Code if __name__ == "__main__": # Array A = [1, 2, 3, 4, 5, 6] N = len(A) # Prime P P = 113 MAX = 100 pre_product = [None] * (MAX) inverse_product = [None] * (MAX) # Calculating PreProduct # and InverseProduct calculate_Pre_Product(A, N, P) calculate_inverse_product(A, N, P) # Range [L, R] in 1 base index L, R = 2, 5 print(calculateProduct(A, L, R, P)) L, R = 1, 3 print(calculateProduct(A, L, R, P)) # This code is contributed by Rituraj Jain |
Output :
7 6
Please refer complete article on Products of ranges in an array for more details!
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