Product of an array

Given an array, find a product of all array elements.

Examples :

Input: arr[] = [1, 2, 3, 4]
Output: 120
Explanation: Product of array elements is 1 x 2 x 3 x 4 = 24

Input: arr[]= [100000, 100000, 100000]
Output: 993000007
Explanation: The product of the array elements is 100000 × 100000 × 100000 = 1000000000000000. Taking modulo 1000000007, the result is 1000000000000000 % 1000000007 = 993000007

Try It Yourself

Table of Content

Iterative Approach - O(n) Time and O(1) Space
Handling Overflows with Modulo Arithmetic - O(n) Time and O(1) Space

Iterative Approach - O(n) Time and O(1) Space

The idea is to compute the product of all elements by traversing the array once and maintaining a running result.
Initialize ans as 1
One by one traverse array elements and multiply with the ans
Return ans

C++

#include <iostream>
#include <vector>
using namespace std;

int product(vector<int> &arr)
{
    int n = arr.size();
    int ans = 1;
    for (int i = 0; i < n; i++)
        ans = ans * arr[i];
    return ans;
}

int main()
{

    vector<int> arr = { 1, 2, 3, 4 };
    cout << product(arr);
    return 0;
}

#include <stdio.h>

// Function to find product of array elements
int product(int arr[], int n) {
    int ans = 1;
    for (int i = 0; i < n; i++)
        ans = ans * arr[i];
    return ans;
}

int main() {
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d\n", product(arr, n));
    return 0;
}

Java

class GfG {

    // Function to find product of array elements
    static int product(int[] arr) {
        int n = arr.length;
        int result = 1;
        for (int i = 0; i < n; i++)
            ans = ans * arr[i];
        return ans;
    }

    public static void main(String[] args) {
        int[] arr = { 1, 2, 3, 4 };
        
        System.out.println(product(arr));
    }
}

Python

def product(arr):
    n = len(arr)
    ans = 1
    for i in range(n):
        ans *= arr[i]
    return ans

# Driver code
if __name__ == "__main__":
    arr = [1, 2, 3, 4]
    
    print(product(arr))

using System;

class GfG
{
    // Function to find product of array elements
    static int Product(int[] arr)
    {
        int n = arr.Length;
        int ans = 1;
        for (int i = 0; i < n; i++)
            ans = ans * arr[i];
        return ans;
    }

    // Driver code
    static void Main()
    {
        int[] arr = { 1, 2, 3, 4 };
        
        Console.WriteLine(Product(arr));
    }
}

JavaScript

function product(arr)
{
    let n = arr.length;
    let ans = 1;
    for (let i = 0; i < n; i++) {
        ans = ans * arr[i];
    }
    return ans;
}

// Driver code
let arr = [ 1, 2, 3, 4 ];

console.log(product(arr));

Output

Handling Overflows with Modulo Arithmetic - O(n) Time and O(1) Space

The above codes may cause overflow. Therefore, it is always desired to compute product under modulo. The reason for its working is the simple distributive property of modulo
( a * b) % c = ( ( a % c ) * ( b % c ) ) % c

For example

Initial : arr[] = [100000, 100000, 100000], MOD = 1000000007, and = 1
Step 1 : Index = 1
ans = (1 * 100000) % MOD = 100000

Step 2 : Index = 2
result = (100000 × 100000) % MOD
= 10000000000 % 1000000007
= 999999937

Step 3 : Index = 3
result = (999999937 × 100000) % MOD
= 99999993700000 % 1000000007
= 993000007

Below is a program to find and print the product of all the number in this array of Modulo (10^9 +7).

C++

#include <iostream>
using namespace std;

const int MOD = 1000000007;

int product(vector<int> &arr)
{
    int n = arr.size();
    long long ans = 1;

    for (int i = 0; i < n; i++)
        ans = (ans * arr[i]) % MOD;

    return ans;
}

int main()
{

    vector<int> arr = {100000, 100000, 100000};
    cout << product(arr);
    return 0;
}

#include <stdio.h>

const int MOD = 1000000007;

// Function to find product of array elements under modulo
int product(int arr[], int n) {
    long ans = 1;
    for (int i = 0; i < n; i++)
        ans = (ans * arr[i]) % MOD;
    return ans;
}

int main() {
    int arr[] = {100000, 100000, 100000};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d\n", product(arr, n));
    return 0;
}