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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

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

C

Java

Python3

C#

PHP

C

Java

Python3

C#

PHP

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