Program for product of array

Given an array, find product of all array elements.

Examples :

Input  : ar[] = {1, 2, 3, 4, 5}
Output : 120
Product of array elements is 1 x 2
x 3 x 4 x 5 = 120.

Input  : ar[] = {1, 6, 3}
Output : 18

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

C

// C program to find product of array 
// elements. 
#include <stdio.h> 
  
int product(int ar[], int n) 
{ 
    int result = 1; 
    for (int i = 0; i < n; i++) 
        result = result * ar[i]; 
    return result; 
} 
  
// driver code for the above program 
int main() 
{ 
    int ar[] = { 1, 2, 3, 4, 5 }; 
    int n = sizeof(ar) / sizeof(ar[0]); 
    printf("%d", product(a, n)); 
    return 0; 
} 

Java

// Java program to find product of array 
// elements. 
class GFG{ 
  
    static int product(int ar[], int n) 
    { 
        int result = 1; 
        for (int i = 0; i < n; i++) 
            result = result * ar[i]; 
        return result; 
    } 
       
    // driver code for the above program 
    public static void main(String[] args) 
    { 
        int ar[] = { 1, 2, 3, 4, 5 }; 
        int n = ar.length; 
        System.out.printf("%d", product(ar, n)); 
    } 
} 
  
// This code is contributed by Smitha Dinesh Semwal 

Python3

# Python3 program to find  
# product of array elements. 
def product(ar, n): 
  
    result = 1
    for i in range(0, n): 
        result = result * ar[i] 
    return result 
  
  
# Driver Code 
ar = [ 1, 2, 3, 4, 5 ] 
n = len(ar) 
  
print(product(ar, n)) 
  
# This code is contributed by Smitha Dinesh Semwal. 

C#

// C# program to find product of array 
// elements. 
using System; 
  
class GFG { 
  
    static int product(int []ar, int n) 
    { 
        int result = 1; 
          
        for (int i = 0; i < n; i++) 
            result = result * ar[i]; 
              
        return result; 
    } 
      
    // driver code for the above program 
    public static void Main() 
    { 
        int []ar = { 1, 2, 3, 4, 5 }; 
        int n = ar.Length; 
          
        Console.WriteLine(product(ar, n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find product  
// of array elements. 
  
function product($ar, $n) 
{ 
    $result = 1; 
    for ($i = 0; $i < $n; $i++) 
        $result = $result * $ar[$i]; 
    return $result; 
} 
  
// Driver Code 
$ar = array( 1, 2, 3, 4, 5 ); 
$n = count($ar); 
print((int)product($ar, $n)); 
  
// This code is contributed by Sam007 
?> 

Output :

The above code may cause overflow. Therefore it is always desired to compute product under modulo. The reason of its working is simple distributive property of modulo.

( a * b) % c = ( ( a % c ) * ( b % c ) ) % c

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

C

// C code for above program to find product 
// under modulo. 
#include <stdio.h> 
  
const int MOD = 1000000007; 
  
int product(int ar[], int n) 
{ 
    int result = 1; 
    for (int i = 0; i < n; i++) 
        result = (result * ar[i]) % MOD; 
    return result; 
} 
  
// driver code for the above program 
int main() 
{ 
    int ar[] = { 1, 2, 3, 4, 5 }; 
    int n = sizeof(ar) / sizeof(ar[0]); 
    printf("%d", product(ar, n)); 
    return 0; 
} 

Java

// Java code for above program to find product 
// under modulo. 
class GFG { 
      
    static final int MOD = 1000000007; 
  
    static int product(int ar[], int n) 
    { 
        int result = 1; 
        for (int i = 0; i < n; i++) 
            result = (result * ar[i]) % MOD; 
              
        return result; 
    } 
  
    // driver code for the above program 
    public static void main(String[] args) 
    { 
        int ar[] = { 1, 2, 3, 4, 5 }; 
        int n = ar.length; 
          
        System.out.printf("%d", product(ar, n)); 
    } 
} 
  
// This code is contributed by  Smitha Dinesh Semwal. 

Python3

# Python 3 code for above 
# program to find product 
# under modulo. 
  
MOD = 1000000007
  
def product(ar, n): 
  
    result = 1
    for i in range(0, n): 
        result = (result * ar[i]) % MOD 
    return result 
  
  
# driver code for the 
# above program 
ar = [1, 2, 3, 4, 5]  
n = len(ar)  
  
print(product(ar, n)) 
  
# This code is contributed by 
# Smitha Dinesh Semwal 

C#

  // C# code for above program to find product 
// under modulo. 
using System; 
class GFG { 
      
    static  int MOD = 1000000007; 
  
    static int product(int []ar, int n) 
    { 
        int result = 1; 
        for (int i = 0; i < n; i++) 
            result = (result * ar[i]) % MOD; 
              
        return result; 
    } 
  
    // driver code for the above program 
    public static void Main() 
    { 
        int []ar = { 1, 2, 3, 4, 5 }; 
        int n = ar.Length; 
          
        Console.WriteLine(product(ar, n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP code for above program  
// to find product under modulo. 
  
function product($ar, $n) 
{ 
    $result = 1; 
    for ($i = 0; $i < $n; $i++) 
        $result = ($result *  
                   $ar[$i]) % 1000000007; 
    return $result; 
} 
  
// Driver Code 
$ar = array( 1, 2, 3, 4, 5 ); 
$n = count($ar); 
print(product($ar, $n)); 
  
// This code is contributed by Sam007 
?> 

Output :

This article is contributed by Shivani Baghel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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