1458. Max Dot Product of Two Subsequences

Problem Statement

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

Intuition

Approach:

Normal Subsequence problem
Just edge case

1 -1
-1 1
For negative numbers, answer can go wrong, so we introdiuce extra condition in take

Links

https://leetcode.com/problems/max-dot-product-of-two-subsequences/

Video Links

Approach 1:

C++

class Solution {
public:
    vector<vector<int>> dp; 
    int find_ans(vector<int>& nums1, vector<int>& nums2, int i, int j){
        if(i < 0 or j < 0)
            return -1e9;
        if(dp[i][j] != -1)
            return dp[i][j];

        int take, not_take;
        take = max(nums1[i]*nums2[j] ,nums1[i]*nums2[j] + find_ans(nums1, nums2, i-1, j-1));
        not_take = max(find_ans(nums1, nums2, i, j-1), find_ans(nums1, nums2, i-1, j));

        return dp[i][j] =  max(take, not_take);
    }


    int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
        dp = vector<vector<int>> (nums1.size(), vector<int> (nums2.size(), -1));
        return find_ans(nums1, nums2, nums1.size()-1, nums2.size()-1);
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++