# Products of ranges in an array

Given an array A[] of size N. Solve Q queries. Find the product in 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Brute Force

For each of the query, traverse each element in the range [L, R] and calculate the product under modulo P. This will answer each query in O(N).

## C++

 `// Product in range  ` `// Queries in O(N) ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate  ` `// Product in the given range. ` `int` `calculateProduct(``int` `A[], ``int` `L,  ` `                     ``int` `R, ``int` `P) ` `{ ` `    ``// As our array is 0 based  ` `    ``// as and L and R are given ` `    ``// as 1 based index. ` `    ``L = L - 1; ` `    ``R = R - 1; ` ` `  `    ``int` `ans = 1; ` `    ``for` `(``int` `i = L; i <= R; i++)  ` `    ``{ ` `        ``ans = ans * A[i]; ` `        ``ans = ans % P; ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `A[] = { 1, 2, 3, 4, 5, 6 }; ` `    ``int` `P = 229; ` `    ``int` `L = 2, R = 5; ` `    ``cout << calculateProduct(A, L, R, P) ` `         ``<< endl; ` ` `  `    ``L = 1, R = 3; ` `    ``cout << calculateProduct(A, L, R, P)  ` `         ``<< endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Product in range Queries in O(N) ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to calculate  ` `    ``// Product in the given range. ` `    ``static` `int` `calculateProduct(``int` `[]A, ``int` `L,  ` `                                ``int` `R, ``int` `P) ` `    ``{ ` `         `  `        ``// As our array is 0 based as  ` `        ``// and L and R are given as 1  ` `        ``// based index. ` `        ``L = L - ``1``; ` `        ``R = R - ``1``; ` `     `  `        ``int` `ans = ``1``; ` `        ``for` `(``int` `i = L; i <= R; i++) ` `        ``{ ` `            ``ans = ans * A[i]; ` `            ``ans = ans % P; ` `        ``} ` `     `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``static` `public` `void` `main (String[] args) ` `    ``{ ` `        ``int` `[]A = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6` `}; ` `        ``int` `P = ``229``; ` `        ``int` `L = ``2``, R = ``5``; ` `        ``System.out.println( ` `            ``calculateProduct(A, L, R, P)); ` `     `  `        ``L = ``1``;  ` `        ``R = ``3``; ` `        ``System.out.println( ` `            ``calculateProduct(A, L, R, P)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## 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 code ` `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)) ` ` `  `# This code is contributed  ` `# by "Abhishek Sharma 44" `

## C#

 `// Product in range Queries in O(N) ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to calculate  ` `    ``// Product in the given range. ` `    ``static` `int` `calculateProduct(``int` `[]A, ``int` `L,      ` `                                ``int` `R, ``int` `P) ` `    ``{ ` `         `  `        ``// As our array is 0 based  ` `        ``// as and L and R are given  ` `        ``// as 1 based index. ` `        ``L = L - 1; ` `        ``R = R - 1; ` `     `  `        ``int` `ans = 1; ` `        ``for` `(``int` `i = L; i <= R; i++) ` `        ``{ ` `            ``ans = ans * A[i]; ` `            ``ans = ans % P; ` `        ``} ` `     `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `[]A = { 1, 2, 3, 4, 5, 6 }; ` `        ``int` `P = 229; ` `        ``int` `L = 2, R = 5; ` `        ``Console.WriteLine( ` `            ``calculateProduct(A, L, R, P)); ` `     `  `        ``L = 1;  ` `        ``R = 3; ` `        ``Console.WriteLine( ` `            ``calculateProduct(A, L, R, P)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

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 value at index i contains the product in the range [0, i]. Similarly we can calculate the pre-inverse product under modulo P. Now the each query can be answered in O(1).
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 cannot 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.

## C++

 `// Product in range Queries in O(1) ` `#include ` `using` `namespace` `std; ` `#define MAX 100 ` ` `  `int` `pre_product[MAX]; ` `int` `inverse_product[MAX]; ` ` `  `// 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 ` `int` `modInverse(``int` `a, ``int` `m) ` `{ ` `    ``int` `m0 = m, t, q; ` `    ``int` `x0 = 0, x1 = 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 % m, a = t; ` ` `  `        ``t = x0; ` ` `  `        ``x0 = x1 - q * x0; ` ` `  `        ``x1 = t; ` `    ``} ` ` `  `    ``// Make x1 positive ` `    ``if` `(x1 < 0) ` `        ``x1 += m0; ` ` `  `    ``return` `x1; ` `} ` ` `  `// calculating pre_product ` `// array ` `void` `calculate_Pre_Product(``int` `A[],  ` `                           ``int` `N, ``int` `P) ` `{ ` `    ``pre_product = A; ` ` `  `    ``for` `(``int` `i = 1; i < N; i++)  ` `    ``{ ` `        ``pre_product[i] = pre_product[i - 1] *  ` `                                        ``A[i]; ` `        ``pre_product[i] = pre_product[i] % P; ` `    ``} ` `} ` ` `  `// Cacluating inverse_product  ` `// array. ` `void` `calculate_inverse_product(``int` `A[],  ` `                               ``int` `N, ``int` `P) ` `{ ` `    ``inverse_product = modInverse(pre_product, P); ` ` `  `    ``for` `(``int` `i = 1; i < N; i++)  ` `        ``inverse_product[i] = modInverse(pre_product[i], P);  ` `} ` ` `  `// Function to calculate  ` `// Product in the given range. ` `int` `calculateProduct(``int` `A[], ``int` `L,  ` `                     ``int` `R, ``int` `P) ` `{ ` `    ``// As our array is 0 based as  ` `    ``// and L and R are given as 1  ` `    ``// based index. ` `    ``L = L - 1; ` `    ``R = R - 1; ` `    ``int` `ans; ` ` `  `    ``if` `(L == 0) ` `        ``ans = pre_product[R]; ` `    ``else` `        ``ans = pre_product[R] *  ` `              ``inverse_product[L - 1]; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Array ` `    ``int` `A[] = { 1, 2, 3, 4, 5, 6 }; ` ` `  `    ``int` `N = ``sizeof``(A) / ``sizeof``(A); ` ` `  `    ``// Prime P ` `    ``int` `P = 113; ` ` `  `    ``// Calculating PreProduct ` `    ``// and InverseProduct ` `    ``calculate_Pre_Product(A, N, P); ` `    ``calculate_inverse_product(A, N, P); ` ` `  `    ``// Range [L, R] in 1 base index ` `    ``int` `L = 2, R = 5; ` `    ``cout << calculateProduct(A, L, R, P)  ` `         ``<< endl; ` ` `  `    ``L = 1, R = 3; ` `    ``cout << calculateProduct(A, L, R, P) ` `         ``<< endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find Product ` `// in range Queries in O(1) ` `class` `GFG ` `{  ` ` `  `static` `int` `MAX = ``100``; ` `int` `pre_product[] = ``new` `int``[MAX]; ` `int` `inverse_product[] = ``new` `int``[MAX]; ` ` `  `// 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 ` `int` `modInverse(``int` `a, ``int` `m) ` `{ ` `    ``int` `m0 = m, t, q; ` `    ``int` `x0 = ``0``, x1 = ``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 % m; ` `        ``a = t; ` ` `  `        ``t = x0; ` ` `  `        ``x0 = x1 - q * x0; ` ` `  `        ``x1 = t; ` `    ``} ` ` `  `    ``// Make x1 positive ` `    ``if` `(x1 < ``0``) ` `        ``x1 += m0; ` ` `  `    ``return` `x1; ` `} ` ` `  `// calculating pre_product array ` `void` `calculate_Pre_Product(``int` `A[],  ` `                           ``int` `N, ``int` `P) ` `{ ` `    ``pre_product[``0``] = A[``0``]; ` ` `  `    ``for` `(``int` `i = ``1``; i < N; i++)  ` `    ``{ ` `        ``pre_product[i] = pre_product[i - ``1``] * ` `                                        ``A[i]; ` `        ``pre_product[i] = pre_product[i] % P; ` `    ``} ` `} ` ` `  `// Cacluating inverse_product array. ` `void` `calculate_inverse_product(``int` `A[],  ` `                               ``int` `N, ``int` `P) ` `{ ` `    ``inverse_product[``0``] = modInverse(pre_product[``0``], ` `                                                ``P); ` ` `  `    ``for` `(``int` `i = ``1``; i < N; i++)  ` `        ``inverse_product[i] = modInverse(pre_product[i],  ` `                                                     ``P);  ` `} ` ` `  `// Function to calculate Product  ` `// in the given range. ` `int` `calculateProduct(``int` `A[], ``int` `L, ` `                     ``int` `R, ``int` `P) ` `{ ` `    ``// As our array is 0 based as and  ` `    ``// L and R are given as 1 based index. ` `    ``L = L - ``1``; ` `    ``R = R - ``1``; ` `    ``int` `ans; ` ` `  `    ``if` `(L == ``0``) ` `        ``ans = pre_product[R]; ` `    ``else` `        ``ans = pre_product[R] *  ` `              ``inverse_product[L - ``1``]; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] s) ` `{ ` `    ``GFG d = ``new` `GFG(); ` `     `  `    ``// Array ` `    ``int` `A[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6` `}; ` `     `  `    ``// Prime P ` `    ``int` `P = ``113``; ` ` `  `    ``// Calculating PreProduct and  ` `    ``// InverseProduct ` `    ``d.calculate_Pre_Product(A, A.length, P); ` `    ``d.calculate_inverse_product(A, A.length,  ` `                                         ``P); ` ` `  `    ``// Range [L, R] in 1 base index ` `    ``int` `L = ``2``, R = ``5``; ` `    ``System.out.println(d.calculateProduct(A, L,  ` `                                          ``R, P)); ` `    ``L = ``1``; ` `    ``R = ``3``; ` `    ``System.out.println(d.calculateProduct(A, L,  ` `                                          ``R, P)); ` `         `  `} ` `} ` ` `  `// This code is contributed by Prerna Saini `

## 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  ` ` `  `# Cacluating 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 `

## C#

 `// C# program to find Product ` `// in range Queries in O(1) ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `int` `MAX = 100; ` `    ``int` `[]pre_product = ``new` `int``[MAX]; ` `    ``int` `[]inverse_product = ``new` `int``[MAX]; ` `     `  `    ``// 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 ` `    ``int` `modInverse(``int` `a, ``int` `m) ` `    ``{ ` `        ``int` `m0 = m, t, q; ` `        ``int` `x0 = 0, x1 = 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 % m; ` `            ``a = t; ` `            ``t = x0; ` `            ``x0 = x1 - q * x0; ` `            ``x1 = t; ` `        ``} ` `     `  `        ``// Make x1 positive ` `        ``if` `(x1 < 0) ` `            ``x1 += m0; ` `     `  `        ``return` `x1; ` `    ``} ` `     `  `    ``// calculating pre_product array ` `    ``void` `calculate_Pre_Product(``int` `[]A,  ` `                               ``int` `N,  ` `                               ``int` `P) ` `    ``{ ` `        ``pre_product = A; ` `     `  `        ``for` `(``int` `i = 1; i < N; i++) ` `        ``{ ` `            ``pre_product[i] =  ` `                ``pre_product[i - 1] *  ` `                               ``A[i]; ` `                                 `  `            ``pre_product[i] = ` `                ``pre_product[i] % P; ` `        ``} ` `    ``} ` `     `  `    ``// Cacluating inverse_product ` `    ``// array. ` `    ``void` `calculate_inverse_product(``int` `[]A,  ` `                                   ``int` `N,  ` `                                   ``int` `P) ` `    ``{ ` `        ``inverse_product =  ` `                ``modInverse(pre_product, P); ` `     `  `        ``for` `(``int` `i = 1; i < N; i++)  ` `            ``inverse_product[i] =  ` `                ``modInverse(pre_product[i], P);  ` `    ``} ` `     `  `    ``// Function to calculate Product  ` `    ``// in the given range. ` `    ``int` `calculateProduct(``int` `[]A, ``int` `L, ` `                         ``int` `R, ``int` `P) ` `    ``{ ` `         `  `        ``// As our array is 0 based as ` `        ``// and L and R are given as 1 ` `        ``// based index. ` `        ``L = L - 1; ` `        ``R = R - 1; ` `        ``int` `ans; ` `     `  `        ``if` `(L == 0) ` `            ``ans = pre_product[R]; ` `        ``else` `            ``ans = pre_product[R] *  ` `                  ``inverse_product[L - 1]; ` `     `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``GFG d = ``new` `GFG(); ` `         `  `        ``// Array ` `        ``int` `[]A = { 1, 2, 3, 4, 5, 6 }; ` `         `  `        ``// Prime P ` `        ``int` `P = 113; ` `     `  `        ``// Calculating PreProduct and  ` `        ``// InverseProduct ` `        ``d.calculate_Pre_Product(A,  ` `                        ``A.Length, P); ` `                         `  `        ``d.calculate_inverse_product(A, ` `                        ``A.Length, P); ` `     `  `        ``// Range [L, R] in 1 base index ` `        ``int` `L = 2, R = 5; ` `        ``Console.WriteLine( ` `            ``d.calculateProduct(A, L, R, P)); ` `             `  `        ``L = 1; ` `        ``R = 3; ` `        ``Console.WriteLine( ` `            ``d.calculateProduct(A, L, R, P)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

Output :

```7
6
```

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