Formatted question description: https://leetcode.ca/all/1458.html

1458. Max Dot Product of Two Subsequences (Hard)

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

Related Topics:
Dynamic Programming

Solution 1. DP

Let dp[i + 1][j + 1] be the answer to the subproblem on s[0..i] and t[0..j].

For dp[0][i] and dp[i][0], they mean that either s or t is empty array. Since the question is asking for non-empty subsequences, they are not valid cases, so we should regard them as -INF.

For dp[i + 1][j + 1], we have three choices:

• We include s[i] and t[j] in the subsequences. We get max(0, dp[i][j]) + s[i] * t[j].
• We ignore s[i] and reuse the result of dp[i][j + 1].
• We ignore t[j] and reuse the result of dp[i + 1][j].

dp[i + 1][j + 1] = max(
                        max(0, dp[i][j]) + s[i] * t[j],   // If we include s[i] and t[j] in the subsequences
                        dp[i][j + 1],                     // If we don't include s[i] in the subsequence
                        dp[i + 1][j]                      // If we don't include t[j] in the subsequence
                      )
dp[0][i] = dp[i][0] = -INF 

// OJ: https://leetcode.com/problems/max-dot-product-of-two-subsequences/

// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int maxDotProduct(vector<int>& s, vector<int>& t) {
        int M = s.size(), N = t.size();
        vector<vector<int>> dp(M + 1, vector<int>(N + 1, INT_MIN));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                dp[i + 1][j + 1] = max({ max(0, dp[i][j]) + s[i] * t[j], dp[i + 1][j], dp[i][j + 1] });
            }
        }
        return dp[M][N];
    }
};

Solution 2. DP with Space Optimization

Since dp[i + 1][j + 1] only depends on dp[i][j], dp[i + 1][j] and dp[i][j + 1], we can reduce the size of the dp array from M * N to 1 * N.

// OJ: https://leetcode.com/problems/max-dot-product-of-two-subsequences/

// Time: O(MN)
// Space: O(min(M, N))
class Solution {
public:
    int maxDotProduct(vector<int>& s, vector<int>& t) {
        int M = s.size(), N = t.size();
        if (M < N) swap(M, N), swap(s, t);
        vector<int> dp(N + 1, INT_MIN);
        for (int i = 0; i < M; ++i) {
            int prev = INT_MIN;
            for (int j = 0; j < N; ++j) {
                int cur = dp[j + 1];
                dp[j + 1] = max({ max(0, prev) + s[i] * t[j], dp[j], dp[j + 1] });
                prev = cur;
            }
        }
        return dp[N];
    }
};

Java

class Solution {
    public int maxDotProduct(int[] nums1, int[] nums2) {
        int length1 = nums1.length, length2 = nums2.length;
        int[][] dp = new int[length1][length2];
        for (int i = 0; i < length1; i++) {
            for (int j = 0; j < length2; j++)
                dp[i][j] = Integer.MIN_VALUE;
        }
        dp[0][0] = nums1[0] * nums2[0];
        for (int i = 1; i < length1; i++) {
            int currProduct = nums1[i] * nums2[0];
            dp[i][0] = Math.max(dp[i - 1][0], currProduct);
        }
        for (int j = 1; j < length2; j++) {
            int currProduct = nums1[0] * nums2[j];
            dp[0][j] = Math.max(dp[0][j - 1], currProduct);
        }
        for (int i = 1; i < length1; i++) {
            for (int j = 1; j < length2; j++) {
                int currProduct = nums1[i] * nums2[j];
                dp[i][j] = Math.max(Math.max(dp[i - 1][j], dp[i][j - 1]), Math.max(Math.max(dp[i - 1][j - 1], currProduct), dp[i - 1][j - 1] + currProduct));
            }
        }
        return dp[length1 - 1][length2 - 1];
    }
}

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