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feat: add solutions to lc problem: No.1458 #3354

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105 changes: 60 additions & 45 deletions solution/1400-1499/1458.Max Dot Product of Two Subsequences/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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。

<!-- tabs:start -->

Expand All @@ -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
Expand All @@ -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];
}
}
```
Expand All @@ -137,15 +136,16 @@ 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));
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];
}
};
```
Expand All @@ -155,45 +155,60 @@ 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];
}
```

#### Rust

```rust
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],
);
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]
}
}
```
Expand Down
105 changes: 66 additions & 39 deletions solution/1400-1499/1458.Max Dot Product of Two Subsequences/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -64,7 +64,18 @@ Their dot product is -1.</pre>

<!-- solution:start -->

### 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.

<!-- tabs:start -->

Expand All @@ -74,12 +85,12 @@ Their dot product is -1.</pre>
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
Expand All @@ -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];
}
}
```
Expand All @@ -111,15 +122,16 @@ 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));
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];
}
};
```
Expand All @@ -129,45 +141,60 @@ 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];
}
```

#### Rust

```rust
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],
);
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]
}
}
```
Expand Down
11 changes: 6 additions & 5 deletions solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2,14 +2,15 @@ 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));
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];
}
};
};
18 changes: 9 additions & 9 deletions solution/1400-1499/1458.Max Dot Product of Two Subsequences/Solution.go
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Original file line number Diff line number Diff line change
@@ -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]
}
return f[m][n]
}
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