1570. Dot Product of Two Sparse Vectors

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

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

Solution:

since the vector is sparse, we only store indices with values that are nonzero.

class SparseVector:
	def __init__(self, nums):
		self.seen = {}
		for i, n in enumerate(nums):
			if n != 0:
				self.seen[i] = n              

	# Return the dotProduct of two sparse vectors
	def dotProduct(self, vec):
		res = 0
		for j, n in vec.seen.items():
			if j in self.seen:
				res += n * self.seen[j]
		return res

A very beautiful Pythonic implementation by @claytonjwong in the comments section.
I personally like the longer one better, because it is easier for people to understand regardless of the coding languages they are using. But one-liners are always so cool!

class SparseVector:
	def __init__(self, A):
		self.m = {i: x for i, x in enumerate(A) if x}
	def dotProduct(self, other):
		return sum([x * y for i, x in self.m.items() for j, y in other.m.items() if i == j])

Check out the official solution, it is more detailed than this one.

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