Number of subarrays for which product and sum are equal
Last Updated :
11 Jul, 2022
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++
#include<bits/stdc++.h>
using namespace std;
int numOfsubarrays( int arr[] , int n)
{
int count = 0;
for ( int i=0; i<n; i++)
{
int product = arr[i];
int sum = arr[i];
for ( int j=i+1; j<n; j++)
{
if (product==sum)
count++;
product *= arr[j];
sum += arr[j];
}
if (product==sum)
count++;
}
return count;
}
int main()
{
int arr[] = {1,3,2};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << numOfsubarrays(arr , n);
return 0;
}
|
Java
class GFG
{
static int numOfsubarrays( int arr[] , int n)
{
int count = 0 ;
for ( int i= 0 ; i<n; i++)
{
int product = arr[i];
int sum = arr[i];
for ( int j=i+ 1 ; j<n; j++)
{
if (product==sum)
count++;
product *= arr[j];
sum += arr[j];
}
if (product==sum)
count++;
}
return count;
}
public static void main(String args[])
{
int arr[] = { 1 , 3 , 2 };
int n = arr.length;
System.out.println(numOfsubarrays(arr , n));
}
}
|
Python3
def numOfsubarrays(arr,n):
count = 0
for i in range (n):
product = arr[i]
sum = arr[i]
for j in range (i + 1 ,n):
if (product = = sum ):
count + = 1
product * = arr[j]
sum + = arr[j]
if (product = = sum ):
count + = 1
return count
arr = [ 1 , 3 , 2 ]
n = len (arr)
print (numOfsubarrays(arr , n))
|
C#
using System;
class GFG {
static int numOfsubarrays( int []arr ,
int n)
{
int count = 0;
for ( int i = 0; i < n; i++)
{
int product = arr[i];
int sum = arr[i];
for ( int j = i + 1; j < n; j++)
{
if (product == sum)
count++;
product *= arr[j];
sum += arr[j];
}
if (product == sum)
count++;
}
return count;
}
public static void Main()
{
int []arr = {1,3,2};
int n = arr.Length;
Console.Write(numOfsubarrays(arr , n));
}
}
|
PHP
<?php
function numOfsubarrays( $arr , $n )
{
$count = 0;
for ( $i = 0; $i < $n ; $i ++)
{
$product = $arr [ $i ];
$sum = $arr [ $i ];
for ( $j = $i + 1; $j < $n ; $j ++)
{
if ( $product == $sum )
$count ++;
$product *= $arr [ $j ];
$sum += $arr [ $j ];
}
if ( $product == $sum )
$count ++;
}
return $count ;
}
$arr = array (1, 3, 2);
$n = sizeof( $arr );
echo (numOfsubarrays( $arr , $n ));
?>
|
Javascript
<script>
function numOfsubarrays(arr, n)
{
let count = 0;
for (let i = 0; i < n; i++)
{
let product = arr[i];
let sum = arr[i];
for (let j = i + 1; j < n; j++)
{
if (product == sum)
count++;
product *= arr[j];
sum += arr[j];
}
if (product == sum)
count++;
}
return count;
}
let arr = [ 1, 3, 2 ];
let n = arr.length;
document.write(numOfsubarrays(arr, n));
</script>
|
Time Complexity : O(n2)
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