## Calculate product of an array efficiently

product of array elements java
product of array except self javascript
write a program to find the sum and product of all elements of an array in java
product of every integer except the integer at that index python
array product ruby
multiply each element of an array by a number in java
replace every element of the array by product of all other elements
a new product leetcode

As a part of a problem I have to multiply around `50000` numbers in list which are smaller than `10**18`.

I've optimized this by doing a divide and conquer approach

```def multnum(niz,l,d):
if(l==d):
return niz[l]
return multnum(niz,l,(l+d)//2)*multnum(niz,(l+d)//2+1,d)
```

However it still is slow, I'm wondering if there is anything I could do to improve the speed.

An example:

```niz =[3,4,5,6]
print(multnum(niz,0,Len(niz)-1)) # Gives 360
```

The l and d variables are just helper variables.

Maybe use numpy?

```l = np.random.randint(1, 1000, size=50000)
%timeit np.prod(l)

# 39.9 µs ± 53.7 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
```

On the other hand, with python

```ll = list(l)
%timeit multnum(ll, 0, 50000-1)
# 22.2 ms ± 108 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
```

So in this case numpy is around 500 times faster.

Multiply all elements in array, How do you find the product of an element in an array? Sum of product of all pairs of array elements. Given an array A [] of integers find sum of product of all pairs of array elements i. e., we need to find of product after execution of following pseudo code. product = 0 for i = 1:n for j = i+1:n product = product + A [i]*A [j]

`math.fsum(np.log(niz))`, and then exponentiate --- or check if you can carry on working in the log space.

Program for product of array, 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​  JavaScript exercises, practice and solution: Write a JavaScript function to calculate the product of values in an array.

I'm a little fuzzy on what l and d are denoting.Numpy has a built in package for this.

```import numpy as np
np.prod(niz)
```

Calculate product of an array efficiently, Maybe use numpy? l = np.random.randint(1, 1000, size=50000) %timeit np.prod(l​) # 39.9 µs ± 53.7 ns per loop (mean ± std. dev. of 7 runs,  Create an array product and initilize its value to 1 and a variable temp = 1. Traverse the array from start to end. For every index i update product [i] as product [i] = temp and temp = temp * array [i], i.e store the product upto i-1 index from the start of array.

Sum and product of an array, Compute the sum and product of an array of integers. An efficient companion filter for computing the product of the items in an array can be  We are given an array and I have to calculate the product of frequencies of numbers in a particular range of the array i,e. [L,R]. How to do it? My approach:- Say, [1,2,2,2,45,45,4]. L=2 and R=6. Answer=3(frequency of 2)*2(frequency of 45)=6.

Google Interview Question: Given an array of numbers, re , In the second pass, multiply this with the product of every number after i. Can't think of a way to do it without the second array. void ArrayMult(int *A, int size) { To compute the product over all dimensions of an array, you can either specify each dimension in the vector dimension argument, or use the 'all' option. B2 = prod (A, [1 2 3]) B2 = -23040 Ball = prod (A, 'all')

Find maximum product of two integers in an array, The maximum product is formed by the (-10, -3) or (5, 6) pair. Naive solution would be to consider every pair of elements and calculate their product. We update  Find element in array that divides all array elements; Divide every element of one array by other array elements; Sum of elements in an array whose difference with the mean of another array is less than k; Check if sum of exactly K elements of the Array can be odd or not; Sum of the elements from index L to R in an array when arr[i] = i * (-1)^i

• Why not something like `np.prod(niz[l:d+1])`, I think this should be much faster, if working with an array