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Products of ranges in an array
  • Difficulty Level : Medium
  • Last Updated : 23 Jan, 2019

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

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 <bits/stdc++.h>
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




<?php
// Product in range Queries in O(N)
  
// Function to calculate 
// Product in the given range.
function 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;
  
    $ans = 1;
    for ($i = $L; $i <= $R; $i++) 
    {
        $ans = $ans * $A[$i];
        $ans = $ans % $P;
    }
  
    return $ans;
}
  
// Driver code
$A = array( 1, 2, 3, 4, 5, 6 );
$P = 229;
$L = 2; $R = 5;
echo calculateProduct($A, $L, $R, $P),"\n" ;
  
$L = 1; $R = 3;
echo calculateProduct($A, $L, $R, $P),"\n" ;
  
// This code is contributed by ajit.
?>



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 <bits/stdc++.h>
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[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
int main()
{
    // Array
    int A[] = { 1, 2, 3, 4, 5, 6 };
  
    int N = sizeof(A) / sizeof(A[0]);
  
    // 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 //
        t =
  
        # m is remainder now, process 
        # same as Euclid's algo 
        m, a = a % m, t 
        t = x0 
        x0 = x1 - q * x0 
        x1 =
  
    # 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] %
  
# 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[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()
    {
        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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