README.md

1458. 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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn max_dot_product(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let mut dp = vec![vec![0; nums2.len() + 1]; nums1.len() + 1];
        let mut ret = i32::MIN;

        for i in 0..nums1.len() {
            for j in 0..nums2.len() {
                ret = ret.max(nums1[i] * nums2[j]);
                dp[i + 1][j + 1] = dp[i + 1][j + 1]
                    .max(dp[i][j + 1])
                    .max(dp[i + 1][j])
                    .max(dp[i][j] + nums1[i] * nums2[j]);
            }
        }

        if ret >= 0 {
            ret = ret.max(dp[nums1.len()][nums2.len()]);
        }

        ret
    }
}