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];
    }
}