Dot Product of Two Sparse Vectors

This problem asks us to compute the dot product of two sparse vectors. A sparse vector is a vector where most of the elements are zero. This property allows us to optimize the computation and storage.

Naive Solution

The most straightforward approach is to iterate through both vectors and compute the product of corresponding elements, summing the results. This approach works regardless of the sparsity of the vectors.

Code (Python)

def dot_product_naive(vec1, vec2):
    if len(vec1) != len(vec2):
        return 0  # Or raise an exception
    
    result = 0
    for i in range(len(vec1)):
        result += vec1[i] * vec2[i]
    return result

Time Complexity

O(n), where n is the length of the vectors.

Space Complexity

O(1)

Optimal Solution

For sparse vectors, we can optimize by only considering the non-zero elements. We can store the non-zero elements along with their indices in a dictionary or a list of pairs. Then, we can iterate through the smaller representation of one vector and check for corresponding indices in the other vector.

Code (Python)

def dot_product_sparse(vec1, vec2):
    # Assume vec1 and vec2 are dictionaries where keys are indices and values are non-zero values
    result = 0
    for index, value in vec1.items():
        if index in vec2:
            result += value * vec2[index]
    return result

# Example usage:
vec1 = {0: 1, 2: 2, 5: 3}
vec2 = {0: 4, 2: 5, 6: 6}

print(dot_product_sparse(vec1, vec2)) # Output: 14 (1*4 + 2*5)

#If the input is a list, convert to dictionary as follows:
def convert_to_dict(nums):
    dic = {}
    for i in range(len(nums)):
        if nums[i] != 0:
            dic[i] = nums[i]
    return dic

# Example usage with list inputs:
vec_list1 = [1, 0, 2, 0, 0, 3]
vec_list2 = [4, 0, 5, 0, 0, 0, 6]

dic1 = convert_to_dict(vec_list1)
dic2 = convert_to_dict(vec_list2)

print(dot_product_sparse(dic1, dic2))

Time Complexity

O(min(n1, n2)), where n1 and n2 are the number of non-zero elements in the two vectors, respectively. In the worst case, where there are no zeros in both vectors, it becomes O(n).

Space Complexity

O(1), assuming the dictionaries representing the sparse vectors are already given. If we need to create those dictionaries from lists of numbers, the space complexity would be O(n1 + n2) in the worst case (where all elements are non-zero and must be stored in the dictionary).

Edge Cases

• Vectors of different lengths: The dot product is not defined for vectors of different lengths. We should either return 0 or raise an exception.
• Empty vectors: If either vector is empty (all elements are zero), the dot product is 0.
• Very large vectors: For extremely large vectors, consider using libraries optimized for numerical computations, such as NumPy.
• Non-integer values: Consider the cases for float, and potentially complex numbers.