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
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.