Program for product of array
Last Updated :
20 Sep, 2024
Given an array, find a 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
Iterative Approach
- Initialize result as 1
- One by one traverse array elements and multiply with the result
- Return result
C++
// C++ program to find product of array elements.
#include <iostream>
using namespace std;
int product(int ar[], int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = result * ar[i];
return result;
}
int main()
{
int ar[] = { 1, 2, 3, 4, 5 };
int n = sizeof(ar) / sizeof(ar[0]);
cout << product(ar, n);
return 0;
}
// This code is contributed by lokeshmvs21.
#include <stdio.h>
// Function to find product of array elements
int product(int ar[], int n) {
int result = 1;
for (int i = 0; i < n; i++)
result = result * ar[i];
return result;
}
int main() {
int ar[] = { 1, 2, 3, 4, 5 };
int n = sizeof(ar) / sizeof(ar[0]);
printf("%d\n", product(ar, n));
return 0;
}
// This code is contributed by lokeshmvs21.
class GfG {
// Function to find product of array elements
static int product(int[] ar, int n) {
int result = 1;
for (int i = 0; i < n; i++)
result = result * ar[i];
return result;
}
public static void main(String[] args) {
int[] ar = { 1, 2, 3, 4, 5 };
int n = ar.length;
System.out.println(product(ar, n));
}
}
// This code is contributed by lokeshmvs21.
# Function to find product of array elements
def product(ar, n):
result = 1
for i in range(n):
result *= ar[i]
return result
# Driver code
if __name__ == "__main__":
ar = [1, 2, 3, 4, 5]
n = len(ar)
print(product(ar, n))
# This code is contributed by lokeshmvs21.
using System;
class GfG
{
// Function to find product of array elements
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
static void Main()
{
int[] ar = { 1, 2, 3, 4, 5 };
int n = ar.Length;
Console.WriteLine(Product(ar, n));
}
}
// This code is contributed by lokeshmvs21.
// Function to find product of array elements
function product(ar, n) {
let result = 1;
for (let i = 0; i < n; i++) {
result = result * ar[i];
}
return result;
}
// Driver code
let ar = [1, 2, 3, 4, 5];
let n = ar.length;
console.log(product(ar, n));
// This code is contributed by lokeshmvs21.
Time Complexity : O(n)
Auxiliary Space : O(1)
Recursive Approach
We can recursively define the problem as
product(arr, n) = arr[n-1) x product(arr, n-1)
With base cases
product(arr, 1) = arr[0]
C++
#include <iostream>
using namespace std;
// Recursive function to find product
// of array elements
int product(int ar[], int n)
{
// Base case: if the array has only
// one element, return it
if (n == 1)
return ar[0];
// Recursive case: multiply the last
// element with the product of the
// remaining array
return ar[n - 1] * product(ar, n - 1);
}
int main()
{
int ar[] = {1, 2, 3, 4, 5};
int n = sizeof(ar) / sizeof(ar[0]);
cout << product(ar, n);
return 0;
}
#include <stdio.h>
// Recursive function to find product
// of array elements
int product(int ar[], int n) {
// Base case: if the array has only
// one element, return it
if (n == 1)
return ar[0];
// Recursive case: multiply the last
// element with the product of the
// remaining array
return ar[n - 1] * product(ar, n - 1);
}
int main() {
int ar[] = {1, 2, 3, 4, 5};
int n = sizeof(ar) / sizeof(ar[0]);
printf("%d\n", product(ar, n));
return 0;
}
class GfG {
// Recursive function to find product
// of array elements
static int product(int[] ar, int n) {
// Base case: if the array has only
// one element, return it
if (n == 1)
return ar[0];
// Recursive case: multiply the last
// element with the product of the
// remaining array
return ar[n - 1] * product(ar, n - 1);
}
public static void main(String[] args) {
int[] ar = {1, 2, 3, 4, 5};
int n = ar.length;
System.out.println(product(ar, n));
}
}
# Recursive function to find product
# of array elements
def product(ar, n):
# Base case: if the array has only
# one element, return it
if n == 1:
return ar[0]
# Recursive case: multiply the last
# element with the product of the
# remaining array
return ar[n - 1] * product(ar, n - 1)
# Driver code
if __name__ == "__main__":
ar = [1, 2, 3, 4, 5]
n = len(ar)
print(product(ar, n))
using System;
class GfG
{
// Recursive function to find product
// of array elements
static int Product(int[] ar, int n)
{
// Base case: if the array has only
// one element, return it
if (n == 1)
return ar[0];
// Recursive case: multiply the last
// element with the product of the
// remaining array
return ar[n - 1] * Product(ar, n - 1);
}
static void Main()
{
int[] ar = {1, 2, 3, 4, 5};
int n = ar.Length;
Console.WriteLine(Product(ar, n));
}
}
// Recursive function to find product
// of array elements
function product(ar, n) {
// Base case: if the array has only
// one element, return it
if (n === 1) {
return ar[0];
}
// Recursive case: multiply the last
// element with the product of the
// remaining array
return ar[n - 1] * product(ar, n - 1);
}
// Driver code
const ar = [1, 2, 3, 4, 5];
const n = ar.length;
console.log(product(ar, n));
Time Complexity : O(n)
Auxiliary Space : O(1)
Which one is better?
Iterative approach is better as the recursive approach requires extra space for recursion call stack and overhead of recursion calls. However writing a recursive code is always a fun exercise.
Handling Overflows with Modulo Arithmetic
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
Below is a program 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 <iostream>
using namespace std;
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;
}
int main()
{
int ar[] = { 1, 2, 3, 4, 5 };
int n = sizeof(ar) / sizeof(ar[0]);
cout << product(ar, n);
return 0;
}
// This code is contributed by lokeshmvs21.
#include <stdio.h>
const int MOD = 1000000007;
// Function to find product of array elements under modulo
int product(int ar[], int n) {
int result = 1;
for (int i = 0; i < n; i++)
result = (result * ar[i]) % MOD;
return result;
}
int main() {
int ar[] = {1, 2, 3, 4, 5};
int n = sizeof(ar) / sizeof(ar[0]);
printf("%d\n", product(ar, n));
return 0;
}
// This code is contributed by lokeshmvs21.
class GfG {
// Modulo constant
static final int MOD = 1000000007;
// Function to find product of array elements under modulo
static int product(int[] ar, int n) {
int result = 1;
for (int i = 0; i < n; i++)
result = (int) ((result * (long) ar[i]) % MOD);
return result;
}
public static void main(String[] args) {
int[] ar = {1, 2, 3, 4, 5};
int n = ar.length;
System.out.println(product(ar, n));
}
}
// This code is contributed by lokeshmvs21.
# Modulo constant
MOD = 1000000007
# Function to find product of array elements under modulo
def product(ar, n):
result = 1
for i in range(n):
result = (result * ar[i]) % MOD
return result
# Driver code
if __name__ == "__main__":
ar = [1, 2, 3, 4, 5]
n = len(ar)
print(product(ar, n))
# This code is contributed by lokeshmvs21.
using System;
class GfG
{
// Modulo constant
const int MOD = 1000000007;
// Function to find product of array elements under modulo
static int Product(int[] ar, int n)
{
int result = 1;
for (int i = 0; i < n; i++)
result = (int)((result * (long)ar[i]) % MOD);
return result;
}
static void Main()
{
int[] ar = { 1, 2, 3, 4, 5 };
int n = ar.Length;
Console.WriteLine(Product(ar, n));
}
}
// This code is contributed by lokeshmvs21.
// Modulo constant
const MOD = 1000000007;
// Function to find product of array elements under modulo
function product(ar, n) {
let result = 1;
for (let i = 0; i < n; i++) {
result = (result * ar[i]) % MOD;
}
return result;
}
// Driver code
let ar = [1, 2, 3, 4, 5];
let n = ar.length;
console.log(product(ar, n));
// This code is contributed by lokeshmvs21.
Time Complexity : O(n)
Auxiliary Space : O(1)
Exercise Problems:
Sign of the product of a Array
Product of Array except itself.
Minimum steps to make the product of the array equal to 1
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