# LeetCode 1570. Dot Product of Two Sparse Vectors

## Description

https://leetcode.com/problems/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`

## Explanation

Do element wise sum between vectors.

## Python Solution

``````class SparseVector:
def __init__(self, nums: List[int]):
self.nums = nums

# Return the dotProduct of two sparse vectors
def dotProduct(self, vec: 'SparseVector') -> int:
result = 0
for n1, n2 in zip(self.nums, vec.nums):
result += n1 * n2

return result

# Your SparseVector object will be instantiated and called as such:
# v1 = SparseVector(nums1)
# v2 = SparseVector(nums2)
# ans = v1.dotProduct(v2)``````
• Time complexity: O(N).
• Space complexity: O(N).