1570. Dot Product of Two Sparse Vectors

Medium


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 of SparseVector and vec

A sparse vector is a vector that has mostly zero values, you should store the sparse vector efficiently and compute the dot product between two SparseVector.

Follow up: What if only one of the vectors is sparse?

 

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 <= 10^5
  • 0 <= nums1[i], nums2[i] <= 100




 class SparseVector:
    def __init__(self, nums: List[int]):
        #self.num1 = nums
        self.num1 = {i:val for i, val in enumerate(nums) if val != 0}

    # Return the dotProduct of two sparse vectors
    def dotProduct(self, vec: 'SparseVector') -> int:


        total = 0
        # for n1, n2 in zip(self.num1, vec.num1):
        #     total += (n1*n2)
        for k, v in self.num1.items():
            if k in vec.num1:
                total+=v*vec.num1[k]
        return total

# Your SparseVector object will be instantiated and called as such:
# v1 = SparseVector(nums1)
# v2 = SparseVector(nums2)
# ans = v1.dotProduct(v2)

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