<p>Given two arrays <code>nums1</code>&nbsp;and <code><font face="monospace">nums2</font></code><font face="monospace">.</font></p>

<p>Return the maximum dot product&nbsp;between&nbsp;<strong>non-empty</strong> subsequences of nums1 and nums2 with the same length.</p>

<p>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,&nbsp;<code>[2,3,5]</code>&nbsp;is a subsequence of&nbsp;<code>[1,2,3,4,5]</code>&nbsp;while <code>[1,5,3]</code>&nbsp;is not).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,1,-2,5], nums2 = [3,0,-6]
<strong>Output:</strong> 18
<strong>Explanation:</strong> Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2.
Their dot product is (2*3 + (-2)*(-6)) = 18.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [3,-2], nums2 = [2,-6,7]
<strong>Output:</strong> 21
<strong>Explanation:</strong> Take subsequence [3] from nums1 and subsequence [7] from nums2.
Their dot product is (3*7) = 21.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [-1,-1], nums2 = [1,1]
<strong>Output:</strong> -1
<strong>Explanation: </strong>Take subsequence [-1] from nums1 and subsequence [1] from nums2.
Their dot product is -1.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 500</code></li>
	<li><code>-1000 &lt;= nums1[i], nums2[i] &lt;= 1000</code></li>
</ul>

const int MAXN = 505;
const int INF = 1e9 + 5;
int n, m;
int dp[MAXN][MAXN][2];

class Solution {
public:
    int maxDotProduct(vector<int>& a, vector<int>& b) {
        n = a.size();
        m = b.size();
        memset(dp, 0, sizeof(dp));
        dp[0][0][1] = a[0] * b[0]; 
        for(int i = 1; i < n; i++){ 
            dp[i][0][0] = max(dp[i - 1][0][1], dp[i - 1][0][0]) ;
            dp[i][0][1] = a[i] * b[0];
        }
        for(int j = 1; j < m; j++){ 
            dp[0][j][0] = max(dp[0][j - 1][1], dp[0][j - 1][0]) ;
            dp[0][j][1] = a[0] * b[j];
        }
        for(int i = 1; i < n; i++){
            for(int j = 1; j < m; j++){
                dp[i][j][0] = max({dp[i - 1][j][0], dp[i - 1][j][1], dp[i][j - 1][0], dp[i][j - 1][1], dp[i - 1][j][1], dp[i - 1][j - 1][0], dp[i - 1][j - 1][1]});
                dp[i][j][1] = max({dp[i - 1][j - 1][0], dp[i - 1][j - 1][1]}) + a[i] * b[j];
            }
        }
        int x = INF, y = INF;
        for(int i = 0; i < n; i++){ x = min(x, abs(a[i])); }
        for(int j = 0; j < m; j++){ y = min(y, abs(b[j])); }
        int res = max(dp[n - 1][m - 1][0], dp[n - 1][m - 1][1]);
        if(res == 0){
            return -(x * y);
        }
        return res;
    }
};
14100