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
classSolution {publicintmaxDotProduct(int[] nums1,int[] nums2) {int[][] dp =newint[nums1.length+1][nums2.length+1];// dp[i][j] -> max dot product sum upto ith element in nums1// and upto jth element in nums2for (int i =0; i <=nums1.length; i++) {for (int j =0; j <=nums2.length; j++) {// For calculationsif (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]; }}