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 an 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: Dynamic Programming
We define f[i][j] to represent the maximum dot product of two subsequences formed by the first i elements of nums1 and the first j elements of nums2. Initially, f[i][j] = -∞.
For f[i][j], we have the following cases:
- Do not select nums1[i-1] or do not select nums2[j-1], i.e., f[i][j] = max(f[i-1][j], f[i][j-1]);
- Select nums1[i-1] and nums2[j-1], i.e., f[i][j] = max(f[i][j], max(0, f[i-1][j-1]) + nums1[i-1] × nums2[j-1]).
The final answer is f[m][n].
The time complexity is O(m × n), and the space complexity is O(m × n). Here, m and n are the lengths of the arrays nums1 and nums2, respectively.
PythonJavaC++GoTypeScriptRust
class Solution:
def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
m, n = len(nums1), len(nums2)
f = [[-inf] * (n + 1) for _ in range(m + 1)]
for i, x in enumerate(nums1, 1):
for j, y in enumerate(nums2, 1):
v = x * y
f[i][j] = max(f[i - 1][j], f[i][j - 1], max(0, f[i - 1][j - 1]) + v)
return f[m][n](code-box)
class Solution {
public int maxDotProduct(int[] nums1, int[] nums2) {
int m = nums1.length, n = nums2.length;
int[][] f = new int[m + 1][n + 1];
for (var g : f) {
Arrays.fill(g, Integer.MIN_VALUE);
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
int v = nums1[i - 1] * nums2[j - 1];
f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
f[i][j] = Math.max(f[i][j], Math.max(f[i - 1][j - 1], 0) + v);
}
}
return f[m][n];
}
}(code-box)
class Solution {
public:
int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size(), n = nums2.size();
int f[m + 1][n + 1];
memset(f, 0xc0, sizeof f);
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
int v = nums1[i - 1] * nums2[j - 1];
f[i][j] = max(f[i - 1][j], f[i][j - 1]);
f[i][j] = max(f[i][j], max(0, f[i - 1][j - 1]) + v);
}
}
return f[m][n];
}
};(code-box)
func maxDotProduct(nums1 []int, nums2 []int) int {
m, n := len(nums1), len(nums2)
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
for j := range f[i] {
f[i][j] = math.MinInt32
}
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
v := nums1[i-1] * nums2[j-1]
f[i][j] = max(f[i-1][j], f[i][j-1])
f[i][j] = max(f[i][j], max(0, f[i-1][j-1])+v)
}
}
return f[m][n]
}(code-box)
function maxDotProduct(nums1: number[], nums2: number[]): number {
const m = nums1.length;
const n = nums2.length;
const f = Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => -Infinity));
for (let i = 1; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
const v = nums1[i - 1] * nums2[j - 1];
f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
f[i][j] = Math.max(f[i][j], Math.max(0, f[i - 1][j - 1]) + v);
}
}
return f[m][n];
}(code-box)
impl Solution {
pub fn max_dot_product(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
let m = nums1.len();
let n = nums2.len();
let mut f = vec![vec![i32::MIN; n + 1]; m + 1];
for i in 1..=m {
for j in 1..=n {
let v = nums1[i - 1] * nums2[j - 1];
f[i][j] = f[i][j].max(f[i - 1][j]).max(f[i][j - 1]);
f[i][j] = f[i][j].max(f[i - 1][j - 1].max(0) + v);
}
}
f[m][n]
}
}(code-box)
