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]
andt[j]
in the subsequences. We getmax(0, dp[i][j]) + s[i] * t[j]
. - We ignore
s[i]
and reuse the result ofdp[i][j + 1]
. - We ignore
t[j]
and reuse the result ofdp[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];
}
}