The most efficient implementation of Lagrange Interpolators is O(N).
Calculate the left Product(0..i-1) of each term in left-to-right order, and the Product(i+1..N-1) of each term in right-to-left order. Your mileage may vary on a GPU.
Calculate the left Product(0..i-1) of each term in left-to-right order, and the Product(i+1..N-1) of each term in right-to-left order. Your mileage may vary on a GPU.
Build a table left[j] of product-of-ratios for i < j, and another right[k] for i > k, each in linear-time, and combine the two to find the product-of-ratio for i != m in linear time as well.
What do you mean by left product and right product?