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
Implementation:
C++
// 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;
} |
Java
// 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));
}
} |
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. |
C#
// 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. |
PHP
<?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. ?> |
Javascript
<script> // Javascript program to count subarrays // with same sum and product. // Returns required number // of subarrays function numOfsubarrays(arr, n)
{ // Initialize result
let count = 0;
// Checking each subarray
for (let i = 0; i < n; i++)
{
let product = arr[i];
let sum = arr[i];
for (let 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 let arr = [ 1, 3, 2 ]; let n = arr.length; document.write(numOfsubarrays(arr, n)); // This code is contributed by decode2207 </script> |
Output
4
Time Complexity : O(n2)