Number of subarrays for which product and sum are equal

Given a array of n numbers. We need to count the number of subarrays having the product and sum of elements are equal
Examples:

Input  : arr[] = {1, 3, 2}
Output : 4
The subarrays are :
[0, 0] sum = 1, product = 1,
[1, 1] sum = 3, product = 3,
[2, 2] sum = 2, product = 2 and 
[0, 2] sum = 1+3+2=6, product = 1*3*2 = 6

Input : arr[] = {4, 1, 2, 1}
Output : 5

The idea is simple, we check for each subarray that if product and sum of its elements are equal or not. If it is then increase the counter variable by 1

C++


[sourcecode language=”CPP” highlight=”6-31″]
// C++ program to count subarrays with
// same sum and product.
#include<bits/stdc++.h>
using namespace std;

// returns required number of subarrays
int numOfsubarrays(int arr[] , int n)
{
int count = 0; // Initialize result

// checking each subarray
for (int i=0; i<n; i++)
{
int product = arr[i];
int sum = arr[i];
for (int j=i+1; j<n; j++)
{
// checking if product is equal
// to sum or not
if (product==sum)
count++;

product *= arr[j];
sum += arr[j];
}

if (product==sum)
count++;
}
return count;
}

// driver function
int main()
{
int arr[] = {1,3,2};
int n = sizeof(arr)/sizeof(arr[0]);
cout << numOfsubarrays(arr , n);
return 0;
}
[/sourcecode]

Java


[sourcecode language=”Java” highlight=”6-31″]
// Java program to count subarrays with
// same sum and product.

class GFG
{
// returns required number of subarrays
static int numOfsubarrays(int arr[] , int n)
{
int count = 0; // Initialize result

// checking each subarray
for (int i=0; i<n; i++)
{
int product = arr[i];
int sum = arr[i];
for (int j=i+1; j<n; j++)
{
// checking if product is equal
// to sum or not
if (product==sum)
count++;

product *= arr[j];
sum += arr[j];
}

if (product==sum)
count++;
}
return count;
}

// Driver function
public static void main(String args[])
{
int arr[] = {1,3,2};
int n = arr.length;
System.out.println(numOfsubarrays(arr , n));
}
}
[/sourcecode]

Python3


[sourcecode language=”Python3″]
# python program to
# count subarrays with
# same sum and product.

# returns required
# number of subarrays
def numOfsubarrays(arr,n):

count = 0 # Initialize result

# checking each subarray
for i in range(n):

product = arr[i]
sum = arr[i]
for j in range(i+1,n):

# checking if product is equal
# to sum or not
if (product==sum):
count+=1

product *= arr[j]
sum += arr[j]

if (product==sum):
count+=1

return count

# Driver code

arr = [1,3,2]
n =len(arr)
print(numOfsubarrays(arr , n))

# This code is contributed
# by Anant Agarwal.
[/sourcecode]

C#


[sourcecode language=”CSHARP” highlight=”6-36″]
// C# program to count subarrays
// with same sum and product.
using System;
class GFG {

// returns required number
// of subarrays
static int numOfsubarrays(int []arr ,
int n)
{

// Initialize result
int count = 0;

// checking each subarray
for (int i = 0; i < n; i++)
{
int product = arr[i];
int sum = arr[i];
for (int j = i + 1; j < n; j++)
{

// checking if product is
// equal to sum or not
if (product == sum)
count++;

product *= arr[j];
sum += arr[j];
}

if (product == sum)
count++;
}
return count;
}

// Driver Code
public static void Main()
{
int []arr = {1,3,2};
int n = arr.Length;
Console.Write(numOfsubarrays(arr , n));
}
}

// This code is contributed by Nitin Mittal.

[/sourcecode]

PHP


[sourcecode language=”php” highlight=”5-33″]
<?php
// PHP program to count subarrays
// with same sum and product.

// function returns required
// number of subarrays
function numOfsubarrays($arr , $n)
{
// Initialize result
$count = 0;

// checking each subarray
for ($i = 0; $i < $n; $i++)
{
$product = $arr[$i];
$sum = $arr[$i];
for ($j = $i + 1; $j < $n; $j++)
{

// checking if product is
// equal to sum or not
if ($product == $sum)
$count++;

$product *= $arr[$j];
$sum += $arr[$j];
}

if ($product == $sum)
$count++;
}
return $count;
}

// Driver Code
$arr = array(1, 3, 2);
$n = sizeof($arr);
echo(numOfsubarrays($arr, $n));

// This code is contributed by Ajit.
?>

[/sourcecode]


Output:

4

Time Complexity : O(n2)

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Improved By : Hariesh, nitin mittal, jit_t



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