adding consecutive numbers in a vector in matlab - matlab

Question

i have a vector and a scalar as input in a problem , the problem is to compute the largest product of n consecutive number of the vector and output the product and the index of the element of the vector that is the first term of the product.
E.g vector =[ 1 2 3 4 5 6] , n=3
I'm supposed to get '3' (i e n) consecutive number of vector whose product is the largest .
in this case it will be 4*5*6
so output will be 120 and 4 as the index.
now if vector has fewer than 'n' elements, the function outputon returns 0 and -1 as the output.
please i need ideas on how to achieve this

Answer 1

You can make a simple implementation with loops:
a = [1 2 3 4 5 6]
w = 3
n = length(a)
maximum = -1
for i = 1:n-w
p = prod(a(i:i+w))
if (p > maximum)
maximum = p
end
end
maximum
Or, you can use the nlfilter from the image processing palette.
a = []
w = 3
if (length(a) >= w)
products = nlfilter(a, [1 w], #(x) prod(x))
res = max(products)
else
res = -1
end

Answer 2

You can create the moving window from the answer of raryeng for the question Matrix with sliding window elements and then appyl cumprod on the columns and take the max with its index.
myvec = [1 2 2 1 3 1];
n = 3;
ind = bsxfun(#plus, 1:n, (0:1:length(myvec)-n).')';
M = cumprod(myvec(ind));
[val,its_ind] = max(M(end,:));
You can encapsulate this with an if condition checking whether the length of myvec is larger than

Answer 3

You can compute the element-wise logarithm of your vector, so multiplication becomes addition; and sliding addition is just convolution with a window of ones:
function [m, p] = f(v, n)
if numel(v) < n
m = -1;
p = 0;
else
c = conv(log(v(:)), ones(n,1), 'valid'); % convolution with vector of n ones
[~, m] = max(c); % starting index of maximizing window
p = prod(v(m+(0:n-1))); % corresponding product
end
Examples:
>> v = [1 2 3 4 5 6]; n = 3;
>> [m, p] = f(v,n)
m =
4
p =
120
>> v = [1 2 3 4 5 6]; n = 7;
>> [m, p] = f(v,n)
m =
-1
p =
0
>> v = [1 4 6 2 5 3]; n = 3;
>> [m, p] = f(v,n)
m =
3
p =
60