1458. Max Dot Product of Two Subsequences

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# 題目敘述

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 (23 + (-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.

# 解題思路

# Complexity

• Time complexity: $O(nm)$

• Space complexity: $O(nm)$

# Solution

class Solution {

    public int maxDotProduct(int[] nums1, int[] nums2) {

        int[][] dp = new int[nums1.length + 1][nums2.length + 1];

        for (int i = 0; i <= nums1.length; i++) {

            for (int j = 0; j <= nums2.length; j++) {

                dp[i][j] = Integer.MIN_VALUE / 2;

            }

        }

        for (int i = nums1.length - 1; i >= 0; i--) {

            for (int j = nums2.length - 1; j >= 0; j--) {

                int product = nums1[i] * nums2[j];

                dp[i][j] = Math.max(product, Math.max(dp[i + 1][j], dp[i][j + 1]));

                dp[i][j] = Math.max(dp[i][j], product + dp[i + 1][j + 1]);

            }

        }

        return dp[0][0];

    }

}