Leetcode #1570: Dot Product of Two Sparse Vectors
In this guide, we solve Leetcode #1570 Dot Product of Two Sparse Vectors in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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 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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Design, Array, Hash Table, Two Pointers
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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
Python Solution
class SparseVector:
def __init__(self, nums: List[int]):
self.d = {i: v for i, v in enumerate(nums) if v}
# Return the dotProduct of two sparse vectors
def dotProduct(self, vec: "SparseVector") -> int:
a, b = self.d, vec.d
if len(b) < len(a):
a, b = b, a
return sum(v * b.get(i, 0) for i, v in a.items())
# Your SparseVector object will be instantiated and called as such:
# v1 = SparseVector(nums1)
# v2 = SparseVector(nums2)
# ans = v1.dotProduct(v2)
Complexity
The time complexity is , and the space complexity is , where is the length of the array. The space complexity is , where is the length of the array.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.