Given two sparse vectors, compute their dot product.
Implement class SparseVector:
SparseVector(nums) Initializes the object with the vector numsdotProduct(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.length1 <= n <= 1050 <= nums1[i], nums2[i] <= 100
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class SparseVector:
def __init__(self, nums: List[int]):
self.data = {}
for i, v in enumerate(nums):
if v != 0:
self.data[i] = v
# Return the dotProduct of two sparse vectors
def dotProduct(self, vec: 'SparseVector') -> int:
res = 0
for i, v in vec.data.items():
if i in self.data:
res += self.data[i] * v
return res
# Your SparseVector object will be instantiated and called as such:
# v1 = SparseVector(nums1)
# v2 = SparseVector(nums2)
# ans = v1.dotProduct(v2)
'''
Without using a hash table
'''
class SparseVector:
def __init__(self, nums: List[int]):
self.pairs = []
for index, value in enumerate(nums):
if value != 0:
self.pairs.append([index, value])
def dotProduct(self, vec: 'SparseVector') -> int:
result = 0
p, q = 0, 0
while p < len(self.pairs) and q < len(vec.pairs):
if self.pairs[p][0] == vec.pairs[q][0]:
result += self.pairs[p][1] * vec.pairs[q][1]
p += 1
q += 1
elif self.pairs[p][0] < vec.pairs[q][0]:
p += 1
else:
q += 1
return result
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