1458. Max Dot Product of Two Subsequences

Description

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

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the maximum dot product of two subsequences formed by the first $i$ elements of $\textit{nums1}$ and the first $j$ elements of $\textit{nums2}$. Initially, $f[i][j] = -\infty$.

For $f[i][j]$, we have the following cases:

1. Do not select $\textit{nums1}[i-1]$ or do not select $\textit{nums2}[j-1]$, i.e., $f[i][j] = \max(f[i-1][j], f[i][j-1])$;
2. Select $\textit{nums1}[i-1]$ and $\textit{nums2}[j-1]$, i.e., $f[i][j] = \max(f[i][j], \max(0, f[i-1][j-1]) + \textit{nums1}[i-1] \times \textit{nums2}[j-1])$.

The final answer is $f[m][n]$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of the arrays $\textit{nums1}$ and $\textit{nums2}$, respectively.

Python3JavaC++GoTypeScriptRust

class Solution:
    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        f = [[-inf] * (n + 1) for _ in range(m + 1)]
        for i, x in enumerate(nums1, 1):
            for j, y in enumerate(nums2, 1):
                v = x * y
                f[i][j] = max(f[i - 1][j], f[i][j - 1], max(0, f[i - 1][j - 1]) + v)
        return f[m][n]

class Solution {
    public int maxDotProduct(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        int[][] f = new int[m + 1][n + 1];
        for (var g : f) {
            Arrays.fill(g, Integer.MIN_VALUE);
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int v = nums1[i - 1] * nums2[j - 1];
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
                f[i][j] = Math.max(f[i][j], Math.max(f[i - 1][j - 1], 0) + v);
            }
        }
        return f[m][n];
    }
}

class Solution {
public:
    int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        int f[m + 1][n + 1];
        memset(f, 0xc0, sizeof f);
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int v = nums1[i - 1] * nums2[j - 1];
                f[i][j] = max(f[i - 1][j], f[i][j - 1]);
                f[i][j] = max(f[i][j], max(0, f[i - 1][j - 1]) + v);
            }
        }
        return f[m][n];
    }
};