1458

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

Companies: Microsoft

Related Topics:
Array, Dynamic Programming

Hints:

• 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.

Solution 1. DP

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

For dp[0][i] and dp[i][0], they mean that either A or B 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 A[i] and B[j] in the subsequences. We get max(0, dp[i][j]) + A[i] * B[j].
• We ignore A[i] and reuse the result of dp[i][j + 1].
• We ignore B[j] and reuse the result of dp[i + 1][j].

// For 0 <= i < M, 0 <= j < N
dp[i + 1][j + 1] = max(
                        max(0, dp[i][j]) + A[i] * B[j],   // If we include A[i] and B[j] in the subsequences
                        dp[i][j + 1],                     // If we don't include A[i] in the subsequence
                        dp[i + 1][j]                      // If we don't include B[j] in the subsequence
                      )

// For 0 <= j <= N, 0 <= i <= M
dp[0][j] = dp[i][0] = -INF

// OJ: https://leetcode.com/problems/max-dot-product-of-two-subsequences/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int maxDotProduct(vector<int>& A, vector<int>& B) {
        int M = A.size(), N = B.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]) + A[i] * B[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 2 * N.

// OJ: https://leetcode.com/problems/max-dot-product-of-two-subsequences
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(min(M, N))
class Solution {
public:
    int maxDotProduct(vector<int>& A, vector<int>& B) {
        int M = A.size(), N = B.size();
        if (N > M) swap(A, B), swap(M, N);
        vector<int> dp(N + 1, INT_MIN);
        for (int i = 0; i < M; ++i) {
            vector<int> next(N + 1, INT_MIN);
            for (int j = 0; j < N; ++j) {
                next[j + 1] = max({A[i] * B[j] + max(0, dp[j]), dp[j + 1], next[j]});
            }
            swap(dp, next);
        }
        return dp[N];
    }
};

Or further reduce the space to 1 * N.

// OJ: https://leetcode.com/problems/max-dot-product-of-two-subsequences/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(min(M, N))
class Solution {
public:
    int maxDotProduct(vector<int>& A, vector<int>& B) {
        int M = A.size(), N = B.size();
        if (N > M) swap(A, B), swap(M, N);
        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({A[i] * B[j] + max(0, prev), dp[j + 1], dp[j]});
                prev = cur;
            }
        }
        return dp[N];
    }
};