Description#
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
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
| class Solution:
def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
m, n = len(nums1), len(nums2)
dp = [[-inf] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
v = nums1[i - 1] * nums2[j - 1]
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1], max(dp[i - 1][j - 1], 0) + v)
return dp[-1][-1]
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
| class Solution {
public int maxDotProduct(int[] nums1, int[] nums2) {
int m = nums1.length, n = nums2.length;
int[][] dp = new int[m + 1][n + 1];
for (int[] e : dp) {
Arrays.fill(e, Integer.MIN_VALUE);
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
dp[i][j] = Math.max(
dp[i][j], Math.max(0, dp[i - 1][j - 1]) + nums1[i - 1] * nums2[j - 1]);
}
}
return dp[m][n];
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
| class Solution {
public:
int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size(), n = nums2.size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1, INT_MIN));
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
int v = nums1[i - 1] * nums2[j - 1];
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
dp[i][j] = max(dp[i][j], max(0, dp[i - 1][j - 1]) + v);
}
}
return dp[m][n];
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
| func maxDotProduct(nums1 []int, nums2 []int) int {
m, n := len(nums1), len(nums2)
dp := make([][]int, m+1)
for i := range dp {
dp[i] = make([]int, n+1)
for j := range dp[i] {
dp[i][j] = math.MinInt32
}
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
v := nums1[i-1] * nums2[j-1]
dp[i][j] = max(dp[i-1][j], dp[i][j-1])
dp[i][j] = max(dp[i][j], max(0, dp[i-1][j-1])+v)
}
}
return dp[m][n]
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| impl Solution {
#[allow(dead_code)]
pub fn max_dot_product(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
let n = nums1.len();
let m = nums2.len();
let mut dp = vec![vec![i32::MIN; m + 1]; n + 1];
// Begin the actual dp process
for i in 1..=n {
for j in 1..=m {
dp[i][j] = std::cmp::max(
std::cmp::max(dp[i - 1][j], dp[i][j - 1]),
std::cmp::max(dp[i - 1][j - 1], 0) + nums1[i - 1] * nums2[j - 1]
);
}
}
dp[n][m]
}
}
|
