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Dot Product of Two Sparse Vectors

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Topics:

ArraysTwo PointersDynamic Programming

Given two sparse vectors, compute their dot product.

Implement class SparseVector:

SparseVector(nums) Initializes the object with the vector nums
dotProduct(vec) Compute the dot product between the instance vector and vec

A sparse vector is a vector that has mostly zero elements, you should store the index-value pairs of the non-zero elements only.

Implement the SparseVector class using a hash map.

Example 1:

Input: nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0]
Output: 8
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 1*0 + 0*3 + 0*0 + 2*4 + 3*0 = 8

Example 2:

Input: nums1 = [0,1,0,0,0], nums2 = [0,0,0,0,2]
Output: 0
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 0*0 + 1*0 + 0*0 + 0*0 + 0*2 = 0

Example 3:

Input: nums1 = [0,1,0,0,2,0,0], nums2 = [1,0,0,0,3,0,4]
Output: 6

Constraints:

n == nums1.length == nums2.length
1 <= n <= 10⁵
0 <= nums1[i], nums2[i] <= 100

Try coding it in LeetCode

Dot Product of Two Sparse Vectors

Medium

Asked by:

35 views

Topics:

ArraysTwo PointersDynamic Programming

Given two sparse vectors, compute their dot product.

Implement class SparseVector:

SparseVector(nums) Initializes the object with the vector nums
dotProduct(vec) Compute the dot product between the instance vector and vec

A sparse vector is a vector that has mostly zero elements, you should store the index-value pairs of the non-zero elements only.

Implement the SparseVector class using a hash map.

Example 1:

Input: nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0]
Output: 8
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 1*0 + 0*3 + 0*0 + 2*4 + 3*0 = 8

Example 2:

Input: nums1 = [0,1,0,0,0], nums2 = [0,0,0,0,2]
Output: 0
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 0*0 + 1*0 + 0*0 + 0*0 + 0*2 = 0

Example 3:

Input: nums1 = [0,1,0,0,2,0,0], nums2 = [1,0,0,0,3,0,4]
Output: 6

Constraints:

n == nums1.length == nums2.length
1 <= n <= 10⁵
0 <= nums1[i], nums2[i] <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected size of the input vectors, nums1 and nums2? Could they be very large, and if so, what order of magnitude are we talking about?
Can the input arrays contain negative numbers, zero, or non-integer values, such as floats?
What should the function return if either of the input arrays is null or empty?
Are the input arrays guaranteed to be of the same length? If they are not, how should I handle the differing lengths?
What data type should I use to store the dot product result to avoid potential overflow issues given the range of possible integer values in the input arrays?

Brute Force Solution

Approach

The goal is to find the dot product of two lists of numbers, but many numbers are zero. The simplest way is to ignore that they are mostly zero and just calculate the dot product directly.

Here's how the algorithm would work step-by-step:

Go through the first list, one number at a time.
For each number in the first list, find the corresponding number in the second list (the number in the same position).
Multiply these two numbers together.
Keep a running total of all the products you've calculated.
After going through all the numbers, the final total is the dot product.

Code Implementation

def dot_product_brute_force(vector_one, vector_two):
    dot_product = 0
    
    # Iterate over the indices of the vectors
    for index in range(len(vector_one)):

        # Multiply corresponding elements
        element_product = vector_one[index] * vector_two[index]
        
        # Accumulate the product
        dot_product += element_product
        
    return dot_product

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the first list of numbers, which has a length of n. For each number in the first list, it accesses the corresponding number in the second list at the same index. This is a direct access operation that takes constant time, O(1). Since this process is repeated for each of the n elements in the first list, the overall time complexity is O(n * 1), which simplifies to O(n).

Space Complexity

O(1) – The provided algorithm iterates through the input lists using indices and calculates a running product. It only uses a single variable to store the running total of the dot product and loop counters, whose memory usage is independent of the input size (N, where N is the length of the vectors). No auxiliary data structures like lists or hash maps are used to store intermediate results. Therefore, the space complexity is constant.

Optimal Solution

Approach

The problem involves finding the dot product of two vectors, but these vectors have mostly zero values. To avoid unnecessary calculations, we will focus only on the non-zero elements to significantly speed up the process and conserve computing resources.

Here's how the algorithm would work step-by-step:

First, identify and store the positions and values of the non-zero elements in each vector.
Next, find the positions where both vectors have non-zero elements.
For each of these shared positions, multiply the corresponding non-zero values from the two vectors.
Finally, add up all the products you calculated in the previous step. This sum is the dot product of the two sparse vectors.

Code Implementation

def dot_product_of_sparse_vectors(vector_one, vector_two):
    non_zero_elements_vector_one = {}
    non_zero_elements_vector_two = {}

    for index, value in enumerate(vector_one):
        if value != 0:
            non_zero_elements_vector_one[index] = value

    for index, value in enumerate(vector_two):
        if value != 0:
            non_zero_elements_vector_two[index] = value

    dot_product = 0

    # Iterate through keys in vector one to find common indices.
    for index in non_zero_elements_vector_one:
        if index in non_zero_elements_vector_two:

            # Ensures only shared indices contribute to product.
            product = non_zero_elements_vector_one[index] * non_zero_elements_vector_two[index]
            dot_product += product

    return dot_product

Big(O) Analysis

Time Complexity

O(n) – Let n be the size of the input vectors. The first step, identifying and storing the non-zero elements, takes O(n) time in the worst case where all elements are non-zero. The second step, finding shared positions of non-zero elements, also takes O(n) time if we use a hash map or similar data structure to store the non-zero indices of one vector and then iterate through the other vector to check for matches. The remaining steps, multiplication and summation, depend on the number of shared positions but are bounded by n, thus taking at most O(n) time. Therefore, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm first identifies and stores the positions and values of non-zero elements in each vector. In the worst-case scenario, all N elements in each vector could be non-zero, requiring auxiliary storage proportional to the input size to store these non-zero elements and their positions. This gathering of non-zero elements creates implicit data structures, specifically arrays or lists to hold the indices and values. Therefore, the auxiliary space used is directly related to the number of elements, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
Both input arrays are null or empty.	Return 0 since the dot product of two empty vectors is defined as 0.
One array is null or empty while the other is not.	Return 0 since any element multiplied by 0 will be 0 and the sum will be 0.
Arrays have different lengths.	Consider the shorter length to avoid index out-of-bounds errors.
Arrays contain very large positive and negative numbers that could cause integer overflow.	Use a 64-bit integer (long in Java or C++) to store the result to avoid overflow.
Arrays contain only zeros.	The dot product will be zero; the algorithm should handle this correctly.
Arrays contain a mix of zeros and non-zero numbers.	The algorithm should efficiently skip the multiplications involving zero to improve performance with sparse arrays.
Extremely large arrays that could cause memory issues if not handled efficiently (sparse representation).	Use a hash map or dictionary to store the non-zero values and their indices, enabling more efficient calculation in sparse cases.
Arrays contain a very high concentration of non-zero values (almost dense).	Although designed for sparsity, ensure performance is reasonable even when the data approaches a dense representation; the hash map lookup will still have overhead.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected size of the input vectors, nums1 and nums2? Could they be very large, and if so, what order of magnitude are we talking about?
Can the input arrays contain negative numbers, zero, or non-integer values, such as floats?
What should the function return if either of the input arrays is null or empty?
Are the input arrays guaranteed to be of the same length? If they are not, how should I handle the differing lengths?
What data type should I use to store the dot product result to avoid potential overflow issues given the range of possible integer values in the input arrays?

Brute Force Solution

Approach

The goal is to find the dot product of two lists of numbers, but many numbers are zero. The simplest way is to ignore that they are mostly zero and just calculate the dot product directly.

Here's how the algorithm would work step-by-step:

Go through the first list, one number at a time.
For each number in the first list, find the corresponding number in the second list (the number in the same position).
Multiply these two numbers together.
Keep a running total of all the products you've calculated.
After going through all the numbers, the final total is the dot product.

Code Implementation

def dot_product_brute_force(vector_one, vector_two):
    dot_product = 0
    
    # Iterate over the indices of the vectors
    for index in range(len(vector_one)):

        # Multiply corresponding elements
        element_product = vector_one[index] * vector_two[index]
        
        # Accumulate the product
        dot_product += element_product
        
    return dot_product

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, identify and store the positions and values of the non-zero elements in each vector.
Next, find the positions where both vectors have non-zero elements.
For each of these shared positions, multiply the corresponding non-zero values from the two vectors.
Finally, add up all the products you calculated in the previous step. This sum is the dot product of the two sparse vectors.

Code Implementation

def dot_product_of_sparse_vectors(vector_one, vector_two):
    non_zero_elements_vector_one = {}
    non_zero_elements_vector_two = {}

    for index, value in enumerate(vector_one):
        if value != 0:
            non_zero_elements_vector_one[index] = value

    for index, value in enumerate(vector_two):
        if value != 0:
            non_zero_elements_vector_two[index] = value

    dot_product = 0

    # Iterate through keys in vector one to find common indices.
    for index in non_zero_elements_vector_one:
        if index in non_zero_elements_vector_two:

            # Ensures only shared indices contribute to product.
            product = non_zero_elements_vector_one[index] * non_zero_elements_vector_two[index]
            dot_product += product

    return dot_product

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Both input arrays are null or empty.	Return 0 since the dot product of two empty vectors is defined as 0.
One array is null or empty while the other is not.	Return 0 since any element multiplied by 0 will be 0 and the sum will be 0.
Arrays have different lengths.	Consider the shorter length to avoid index out-of-bounds errors.
Arrays contain very large positive and negative numbers that could cause integer overflow.	Use a 64-bit integer (long in Java or C++) to store the result to avoid overflow.
Arrays contain only zeros.	The dot product will be zero; the algorithm should handle this correctly.
Arrays contain a mix of zeros and non-zero numbers.	The algorithm should efficiently skip the multiplications involving zero to improve performance with sparse arrays.
Extremely large arrays that could cause memory issues if not handled efficiently (sparse representation).	Use a hash map or dictionary to store the non-zero values and their indices, enabling more efficient calculation in sparse cases.
Arrays contain a very high concentration of non-zero values (almost dense).	Although designed for sparsity, ensure performance is reasonable even when the data approaches a dense representation; the hash map lookup will still have overhead.