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# 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`

```java
class Solution {
    public int maxDotProduct(int[] nums1, int[] nums2) {
        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
        // dp[i][j] -> max dot product sum upto ith element in nums1
        // and upto jth element in nums2
        for (int i = 0; i <= nums1.length; i++) {
            for (int j = 0; j <= nums2.length; j++) {
                // For calculations
                if (i == 0 || j == 0) {
                    dp[i][j] = Integer.MIN_VALUE;
                    continue;
                }
                if (dp[i - 1][j - 1] == Integer.MIN_VALUE)
                    dp[i][j] = nums1[i - 1] * nums2[j - 1];
                else
                    dp[i][j] = Math.max(Math.max(dp[i - 1][j - 1], nums1[i - 1] * nums2[j - 1]),
                            nums1[i - 1] * nums2[j - 1] + dp[i - 1][j - 1]);
                dp[i][j] = Math.max(dp[i][j], Math.max(dp[i][j - 1], dp[i - 1][j]));
            }
        }
        return dp[nums1.length][nums2.length];
    }
}
```