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Hard DP

Microsoft

2020-05-23

1458. Max Dot Product of Two Subsequences

Similar Question: LeetCode Question 1035, LeetCode Question 1143

Question:

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

Solution:

Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. So the DP function is

dp[i][j] = Max(
                dp[i-1][j-1] + num1[i] * num2[j],  // use last numbers from both the first and the second
                dp[i-1][j],  // ignore the last number from first
                dp[i][j-1]   // ignore the last number from second
              );

We are not consider ignore both number(dp[i-1][j-1]) in DP. Becuase it has been progagated via dp[i-1][j] or dp[i][j-1]

dp[i-1][j] = Max(
                dp[i-2][j-1] + num1[i-1] * num2[j],  // use last numbers from both the first and the second
                dp[i-2][j],  // ignore the last number from first
                dp[i-1][j-1]   // ignore the last number from second
              );
dp[i][j-1] = Max(
                dp[i-1][j-2] + num1[i] * num2[j-1],  // use last numbers from both the first and the second
                dp[i-1][j-1],  // ignore the last number from first
                dp[i][j-2]   // ignore the last number from second
              );

From above DP function, we can know that dp[i-1][j] or dp[i][j-1] has included the case when we are not taking both i and j index.

class Solution {
    public int maxDotProduct(int[] A, int[] B) {
        int A_len = A.length, B_len = B.length;
        int[][] dp = new int[A_len][B_len];
        for (int i = 0; i < A_len; i++) {    
            for (int j = 0; j < B_len; j++) {
                // Find current dot product value
                dp[i][j] = A[i] * B[j];
                
                // Case when we are taking both index(i,j) in A and B
                // Taking max with 0 to make sure the previous number is not negative
                if (i > 0 && j > 0) {
                    dp[i][j] += Math.max(0, dp[i-1][j-1]);
                }
                
                // Case when we are not taking the index(i) in A, comparing with curr value
                if (i > 0) {
                    dp[i][j] = Math.max(dp[i][j], dp[i-1][j]);
                }
                
                // Case when we are not taking the index(j) in B, comparing with curr value
                if (j > 0) {
                    dp[i][j] = Math.max(dp[i][j], dp[i][j-1]);
                }
            }
        }
        return dp[A_len-1][B_len-1];
    }
}