diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README.md b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README.md index 2d5cc8c3efca0..b40b63e101f36 100644 --- a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README.md +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README.md @@ -80,17 +80,16 @@ tags: ### 方法一：动态规划 -定义 $dp[i][j]$ 表示 $nums1$ 前 $i$ 个元素和 $nums2$ 前 $j$ 个元素得到的最大点积。 +我们定义 $f[i][j]$ 表示 $\textit{nums1}$ 的前 $i$ 个元素和 $\textit{nums2}$ 的前 $j$ 个元素构成的两个子序列的最大点积。初始时 $f[i][j] = -\infty$。 -那么有： +对于 $f[i][j]$，我们有以下几种情况： -$$ -dp[i][j]=max(dp[i-1][j], dp[i][j - 1], max(dp[i - 1][j - 1], 0) + nums1[i] \times nums2[j]) -$$ +1. 不选 $\textit{nums1}[i-1]$ 或者不选 $\textit{nums2}[j-1]$，即 $f[i][j] = \max(f[i-1][j], f[i][j-1])$； +2. 选 $\textit{nums1}[i-1]$ 和 $\textit{nums2}[j-1]$，即 $f[i][j] = \max(f[i][j], \max(0, f[i-1][j-1]) + \textit{nums1}[i-1] \times \textit{nums2}[j-1])$。 -答案为 $dp[m][n]$。 +最终答案即为 $f[m][n]$。 -时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是数组 $nums1$ 和 $nums2$ 的长度。 +时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是数组 $\textit{nums1}$ 和 $\textit{nums2}$ 的长度。 @@ -100,12 +99,12 @@ $$ 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] + 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] ``` #### Java @@ -114,18 +113,18 @@ class Solution: 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); + 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) { - 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]); + 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 dp[m][n]; + return f[m][n]; } } ``` @@ -137,15 +136,16 @@ class Solution { public: int maxDotProduct(vector& nums1, vector& nums2) { int m = nums1.size(), n = nums2.size(); - vector> dp(m + 1, vector(n + 1, INT_MIN)); + 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]; - 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); + 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 dp[m][n]; + return f[m][n]; } }; ``` @@ -155,21 +155,39 @@ public: ```go 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 + 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] - 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) + 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 dp[m][n] + return f[m][n] +} +``` + +#### TypeScript + +```ts +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]; } ``` @@ -177,23 +195,20 @@ func maxDotProduct(nums1 []int, nums2 []int) int { ```rust impl Solution { - #[allow(dead_code)] pub fn max_dot_product(nums1: Vec, nums2: Vec) -> 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], - ); + 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); } } - dp[n][m] + f[m][n] } } ``` diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README_EN.md b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README_EN.md index 9adf9790e37f4..f5a5e569010c6 100644 --- a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README_EN.md +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/README_EN.md @@ -64,7 +64,18 @@ Their dot product is -1. -### Solution 1 +### 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 $\textit{nums1}$ and the first $j$ elements of $\textit{nums2}$. Initially, $f[i][j] = -\infty$. + +For $f[i][j]$, we have the following cases: + +1. Do not select $\textit{nums1}[i-1]$ or do not select $\textit{nums2}[j-1]$, i.e., $f[i][j] = \max(f[i-1][j], f[i][j-1])$; +2. Select $\textit{nums1}[i-1]$ and $\textit{nums2}[j-1]$, i.e., $f[i][j] = \max(f[i][j], \max(0, f[i-1][j-1]) + \textit{nums1}[i-1] \times \textit{nums2}[j-1])$. + +The final answer is $f[m][n]$. + +The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of the arrays $\textit{nums1}$ and $\textit{nums2}$, respectively. @@ -74,12 +85,12 @@ Their dot product is -1. 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] + 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] ``` #### Java @@ -88,18 +99,18 @@ class Solution: 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); + 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) { - 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]); + 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 dp[m][n]; + return f[m][n]; } } ``` @@ -111,15 +122,16 @@ class Solution { public: int maxDotProduct(vector& nums1, vector& nums2) { int m = nums1.size(), n = nums2.size(); - vector> dp(m + 1, vector(n + 1, INT_MIN)); + 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]; - 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); + 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 dp[m][n]; + return f[m][n]; } }; ``` @@ -129,21 +141,39 @@ public: ```go 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 + 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] - 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) + 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 dp[m][n] + return f[m][n] +} +``` + +#### TypeScript + +```ts +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]; } ``` @@ -151,23 +181,20 @@ func maxDotProduct(nums1 []int, nums2 []int) int { ```rust impl Solution { - #[allow(dead_code)] pub fn max_dot_product(nums1: Vec, nums2: Vec) -> 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], - ); + 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); } } - dp[n][m] + f[m][n] } } ``` diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.cpp b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.cpp index 47369f4a935b4..c6447fbb6dc93 100644 --- a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.cpp +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.cpp @@ -2,14 +2,15 @@ class Solution { public: int maxDotProduct(vector& nums1, vector& nums2) { int m = nums1.size(), n = nums2.size(); - vector> dp(m + 1, vector(n + 1, INT_MIN)); + 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]; - 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); + 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 dp[m][n]; + return f[m][n]; } -}; \ No newline at end of file +}; diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.go b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.go index e2c0e9a62a864..5236d579359a8 100644 --- a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.go +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.go @@ -1,18 +1,18 @@ 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 + 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] - 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) + 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 dp[m][n] -} \ No newline at end of file + return f[m][n] +} diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.java b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.java index 157c44cc99168..7bb198e1a9509 100644 --- a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.java +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.java @@ -1,17 +1,17 @@ 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); + 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) { - 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]); + 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 dp[m][n]; + return f[m][n]; } -} \ No newline at end of file +} diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.py b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.py index 84386e6289691..496562fed1721 100644 --- a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.py +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.py @@ -1,9 +1,9 @@ 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] + 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] diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.rs b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.rs index 01e12026e66b8..02a6bb18bd05f 100644 --- a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.rs +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.rs @@ -1,20 +1,17 @@ impl Solution { - #[allow(dead_code)] pub fn max_dot_product(nums1: Vec, nums2: Vec) -> i32 { - let n = nums1.len(); - let m = nums2.len(); - let mut dp = vec![vec![i32::MIN; m + 1]; n + 1]; + let m = nums1.len(); + let n = nums2.len(); + let mut f = vec![vec![i32::MIN; n + 1]; m + 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], - ); + 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); } } - dp[n][m] + f[m][n] } } diff --git a/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.ts b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.ts new file mode 100644 index 0000000000000..c8e214478932c --- /dev/null +++ b/solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.ts @@ -0,0 +1,13 @@ +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]; +}