Python3 Program for Products of ranges in an array
Last Updated :
20 Jan, 2022
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
def calculateProduct (A, L, R, P):
L = L - 1
R = R - 1
ans = 1
for i in range (R + 1 ):
ans = ans * A[i]
ans = ans % P
return ans
A = [ 1 , 2 , 3 , 4 , 5 , 6 ]
P = 229
L = 2
R = 5
print (calculateProduct(A, L, R, P))
L = 1
R = 3
print (calculateProduct(A, L, R, P))
|
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
def modInverse(a, m):
m0, x0, x1 = m, 0 , 1
if m = = 1 :
return 0
while a > 1 :
q = a / / m
t = m
m, a = a % m, t
t = x0
x0 = x1 - q * x0
x1 = t
if x1 < 0 :
x1 + = m0
return x1
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
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)
def calculateProduct(A, L, R, P):
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
if __name__ = = "__main__" :
A = [ 1 , 2 , 3 , 4 , 5 , 6 ]
N = len (A)
P = 113
MAX = 100
pre_product = [ None ] * ( MAX )
inverse_product = [ None ] * ( MAX )
calculate_Pre_Product(A, N, P)
calculate_inverse_product(A, N, P)
L, R = 2 , 5
print (calculateProduct(A, L, R, P))
L, R = 1 , 3
print (calculateProduct(A, L, R, P))
|
Output :
7
6
Please refer complete article on Products of ranges in an array for more details!
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