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1458. Max Dot Product of Two Subsequences
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
Dynamic Programming.
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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]; } } -
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]; } }; -
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] -
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] } -
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] } } -
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]; }