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
Solution:
dp[i][j] = max product for subarray [0, i], [0, j]
class Solution {
public int maxDotProduct(int[] nums1, int[] nums2) {
int m = nums1.length, n = nums2.length;
int[][] dp = new int[m][n];
dp[0][0] = nums1[0] * nums2[0];
int max = dp[0][0];
for (int i = 1; i < m; i ++) {
dp[i][0] = Math.max(dp[i - 1][0], nums1[i] * nums2[0]);
max = Math.max(max, dp[i][0]);
}
for (int j = 1; j < n; j ++) {
dp[0][j] = Math.max(dp[0][j - 1], nums1[0] * nums2[j]);
max = Math.max(max, dp[0][j]);
}
for (int i = 1; i < m; i ++) {
for (int j = 1; j < n; j ++) {
int prod = nums1[i] * nums2[j];
dp[i][j] = dp[i - 1][j - 1];
dp[i][j] = Math.max(dp[i][j], prod);
dp[i][j] = Math.max(dp[i][j], prod + dp[i - 1][j - 1]);
dp[i][j] = Math.max(dp[i][j], dp[i - 1][j]);
dp[i][j] = Math.max(dp[i][j], dp[i][j - 1]);
max = Math.max(max, dp[i][j]);
}
}
return max;
}
}