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 vectornums
dotProduct(vec)
Compute the dot product between the instance of SparseVector andvec
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)