Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 8

Description

Max Dot Product of Two Subsequences

Given two arrays nums1 and nums2.

Return the maximum dot product between non-empty subsequences of nums1 and nums2 with the same length.

A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, [2,3,5] is a subsequence of [1,2,3,4,5] while [1,5,3] is not).

Example 1:

Input: nums1 = [2,1,-2,5], nums2 = [3,0,-6]
Output: 18
Explanation: Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2.
Their dot product is (2*3 + (-2)*(-6)) = 18.

Example 2:

Input: nums1 = [3,-2], nums2 = [2,-6,7]
Output: 21
Explanation: Take subsequence [3] from nums1 and subsequence [7] from nums2.
Their dot product is (3*7) = 21.

Example 3:

Input: nums1 = [-1,-1], nums2 = [1,1]
Output: -1
Explanation: Take subsequence [-1] from nums1 and subsequence [1] from nums2.
Their dot product is -1.

Constraints:

• 1 <= nums1.length, nums2.length <= 500
• -1000 <= nums1[i], nums2[i] <= 1000

Solution

class Solution {
public:
  int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
    int m = nums1.size();
    int n = nums2.size();

    vector<vector<int>> products(2, vector<int>(n + 1, INT_MIN));
    
    for(int i = 0; i < m; ++i) {
      for(int j = 0; j < n; ++j) {
        int currentProduct = nums1[i] * nums2[j];
        products[i & 1][j + 1] = max({currentProduct, products[((i ^ 1) & 1)][j + 1], products[i & 1][j], currentProduct + max(0, products[((i ^ 1) & 1)][j])});
      }
    }
    return products[(m ^ 1) & 1][n];
  }
};

// Accepted
// 62/62 cases passed (21 ms)
// Your runtime beats 78.51 % of cpp submissions
// Your memory usage beats 91.41 % of cpp submissions (10.1 MB)