# 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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