Microsoft

2020-05-23

## 1458. Max Dot Product of Two Subsequences

### 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  from nums1 and subsequence  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  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];
}
}
``````