Question

Formatted question description: https://leetcode.ca/all/1570.html

 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

 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

Algorithm

Store in hashmap for only indexes with values.

Code

Java

public class Dot_Product_of_Two_Sparse_Vectors {

    class SparseVector {

        private Map<Integer, Integer> indexToNum = new HashMap<>();

        SparseVector(int[] nums) {
            for (int i = 0; i < nums.length; ++i)
                if (nums[i] != 0) {
                    indexToNum.put(i, nums[i]);
                }
        }

        // Return the dotProduct of two sparse vectors
        public int dotProduct(SparseVector vec) {
            if (indexToNum.size() < vec.indexToNum.size()) {
                return vec.dotProduct(this);
            }

            int ans = 0;

            for (final int index : vec.indexToNum.keySet()) {
                if (indexToNum.containsKey(index)) {
                    ans += vec.indexToNum.get(index) * indexToNum.get(index);
                }
            }

            return ans;
        }
    }
}

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