Javascript Program for Products of ranges in an array
Last Updated :
05 Jul, 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).
Javascript
<script>
function calculateProduct(A, L, R, P)
{
L = L - 1;
R = R - 1;
let ans = 1;
for (let i = L; i <= R; i++)
{
ans = ans * A[i];
ans = ans % P;
}
return ans;
}
let A = [ 1, 2, 3, 4, 5, 6 ];
let P = 229;
let L = 2, R = 5;
document.write(calculateProduct(A, L, R, P) + "</br>" );
L = 1;
R = 3;
document.write(calculateProduct(A, L, R, P) + "</br>" );
</script>
|
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.
Javascript
<script>
let MAX = 100;
let pre_product = new Array(MAX);
let inverse_product = new Array(MAX);
function modInverse(a, m)
{
let m0 = m, t, q;
let x0 = 0, x1 = 1;
if (m == 1)
return 0;
while (a > 1)
{
q = parseInt(a / m, 10);
t = m;
m = a % m;
a = t;
t = x0;
x0 = x1 - q * x0;
x1 = t;
}
if (x1 < 0)
x1 += m0;
return x1;
}
function calculate_Pre_Product(A, N, P)
{
pre_product[0] = A[0];
for (let i = 1; i < N; i++)
{
pre_product[i] =
pre_product[i - 1] *
A[i];
pre_product[i] =
pre_product[i] % P;
}
}
function calculate_inverse_product(A, N, P)
{
inverse_product[0] =
modInverse(pre_product[0], P);
for (let i = 1; i < N; i++)
inverse_product[i] =
modInverse(pre_product[i], P);
}
function calculateProduct(A, L, R, P)
{
L = L - 1;
R = R - 1;
let ans;
if (L == 0)
ans = pre_product[R];
else
ans = pre_product[R] *
inverse_product[L - 1];
return ans;
}
let A = [ 1, 2, 3, 4, 5, 6 ];
let P = 113;
calculate_Pre_Product(A, A.length, P);
calculate_inverse_product(A, A.length, P);
let L = 2, R = 5;
document.write(calculateProduct(A, L, R, P) + "</br>" );
L = 1;
R = 3;
document.write(calculateProduct(A, L, R, P));
</script>
|
Output :
7
6
Please refer complete article on Products of ranges in an array for more details!
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