diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README.md b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README.md
index 17abe43dc30a1..7388b8f90de61 100644
--- a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README.md
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README.md
@@ -41,7 +41,7 @@ tags:
输入：nums = [1,10,3,4,19]
输出：133
解释：下标三元组 (1, 2, 4) 的值是 (nums[1] - nums[2]) * nums[4] = 133 。
-可以证明不存在值大于 133 的有序下标三元组。
+可以证明不存在值大于 133 的有序下标三元组。
示例 3：
@@ -69,7 +69,11 @@ tags:
### 方法一：维护前缀最大值和最大差值
-我们可以用两个变量 $mx$ 和 $mx\_diff$ 分别维护前缀最大值和最大差值。遍历数组时，更新这两个变量，答案为所有 $mx\_diff \times nums[i]$ 的最大值。
+我们用两个变量 $\textit{mx}$ 和 $\textit{mxDiff}$ 分别维护前缀最大值和最大差值，用一个变量 $\textit{ans}$ 维护答案。初始时，这些变量都为 $0$。
+
+接下来，我们枚举数组的每个元素 $x$ 作为 $\textit{nums}[k]$，首先更新答案 $\textit{ans} = \max(\textit{ans}, \textit{mxDiff} \times x)$，然后我们更新最大差值 $\textit{mxDiff} = \max(\textit{mxDiff}, \textit{mx} - x)$，最后更新前缀最大值 $\textit{mx} = \max(\textit{mx}, x)$。
+
+枚举完所有元素后，返回答案 $\textit{ans}$。
时间复杂度 $O(n)$，其中 $n$ 是数组长度。空间复杂度 $O(1)$。
@@ -81,10 +85,10 @@ tags:
class Solution:
def maximumTripletValue(self, nums: List[int]) -> int:
ans = mx = mx_diff = 0
- for num in nums:
- ans = max(ans, mx_diff * num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx - num)
+ for x in nums:
+ ans = max(ans, mx_diff * x)
+ mx_diff = max(mx_diff, mx - x)
+ mx = max(mx, x)
return ans
```
@@ -93,14 +97,12 @@ class Solution:
```java
class Solution {
public long maximumTripletValue(int[] nums) {
- long max, maxDiff, ans;
- max = 0;
- maxDiff = 0;
- ans = 0;
- for (int num : nums) {
- ans = Math.max(ans, num * maxDiff);
- max = Math.max(max, num);
- maxDiff = Math.max(maxDiff, max - num);
+ long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
@@ -113,12 +115,12 @@ class Solution {
class Solution {
public:
long long maximumTripletValue(vector& nums) {
- long long ans = 0;
- int mx = 0, mx_diff = 0;
- for (int num : nums) {
- ans = max(ans, 1LL * mx_diff * num);
- mx = max(mx, num);
- mx_diff = max(mx_diff, mx - num);
+ long long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = max(ans, mxDiff * x);
+ mxDiff = max(mxDiff, 1LL * mx - x);
+ mx = max(mx, x);
}
return ans;
}
@@ -129,11 +131,11 @@ public:
```go
func maximumTripletValue(nums []int) int64 {
- ans, mx, mx_diff := 0, 0, 0
- for _, num := range nums {
- ans = max(ans, mx_diff*num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx-num)
+ ans, mx, mxDiff := 0, 0, 0
+ for _, x := range nums {
+ ans = max(ans, mxDiff*x)
+ mxDiff = max(mxDiff, mx-x)
+ mx = max(mx, x)
}
return int64(ans)
}
@@ -143,16 +145,36 @@ func maximumTripletValue(nums []int) int64 {
```ts
function maximumTripletValue(nums: number[]): number {
- let [ans, mx, mx_diff] = [0, 0, 0];
- for (const num of nums) {
- ans = Math.max(ans, mx_diff * num);
- mx = Math.max(mx, num);
- mx_diff = Math.max(mx_diff, mx - num);
+ let [ans, mx, mxDiff] = [0, 0, 0];
+ for (const x of nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
```
+#### Rust
+
+```rust
+impl Solution {
+ pub fn maximum_triplet_value(nums: Vec) -> i64 {
+ let mut ans: i64 = 0;
+ let mut mx: i32 = 0;
+ let mut mx_diff: i32 = 0;
+
+ for &x in &nums {
+ ans = ans.max(mx_diff as i64 * x as i64);
+ mx_diff = mx_diff.max(mx - x);
+ mx = mx.max(x);
+ }
+
+ ans
+ }
+}
+```
+
diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README_EN.md b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README_EN.md
index 42059e472f811..33c09398f9fc5 100644
--- a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README_EN.md
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README_EN.md
@@ -31,7 +31,7 @@ tags:
Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
-It can be shown that there are no ordered triplets of indices with a value greater than 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77.
Example 2:
@@ -65,9 +65,13 @@ It can be shown that there are no ordered triplets of indices with a value great
-### Solution 1: Maintain Maximum Prefix Value and Maximum Difference
+### Solution 1: Maintaining Prefix Maximum and Maximum Difference
-We can use two variables $mx$ and $mx\_diff$ to maintain the maximum prefix value and maximum difference, respectively. When traversing the array, we update these two variables, and the answer is the maximum value of all $mx\_diff \times nums[i]$.
+We use two variables $\textit{mx}$ and $\textit{mxDiff}$ to maintain the prefix maximum value and maximum difference, respectively, and use a variable $\textit{ans}$ to maintain the answer. Initially, these variables are all $0$.
+
+Next, we iterate through each element $x$ in the array as $\textit{nums}[k]$. First, we update the answer $\textit{ans} = \max(\textit{ans}, \textit{mxDiff} \times x)$. Then we update the maximum difference $\textit{mxDiff} = \max(\textit{mxDiff}, \textit{mx} - x)$. Finally, we update the prefix maximum value $\textit{mx} = \max(\textit{mx}, x)$.
+
+After iterating through all elements, we return the answer $\textit{ans}$.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
@@ -79,10 +83,10 @@ The time complexity is $O(n)$, where $n$ is the length of the array. The space c
class Solution:
def maximumTripletValue(self, nums: List[int]) -> int:
ans = mx = mx_diff = 0
- for num in nums:
- ans = max(ans, mx_diff * num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx - num)
+ for x in nums:
+ ans = max(ans, mx_diff * x)
+ mx_diff = max(mx_diff, mx - x)
+ mx = max(mx, x)
return ans
```
@@ -91,14 +95,12 @@ class Solution:
```java
class Solution {
public long maximumTripletValue(int[] nums) {
- long max, maxDiff, ans;
- max = 0;
- maxDiff = 0;
- ans = 0;
- for (int num : nums) {
- ans = Math.max(ans, num * maxDiff);
- max = Math.max(max, num);
- maxDiff = Math.max(maxDiff, max - num);
+ long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
@@ -111,12 +113,12 @@ class Solution {
class Solution {
public:
long long maximumTripletValue(vector& nums) {
- long long ans = 0;
- int mx = 0, mx_diff = 0;
- for (int num : nums) {
- ans = max(ans, 1LL * mx_diff * num);
- mx = max(mx, num);
- mx_diff = max(mx_diff, mx - num);
+ long long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = max(ans, mxDiff * x);
+ mxDiff = max(mxDiff, 1LL * mx - x);
+ mx = max(mx, x);
}
return ans;
}
@@ -127,11 +129,11 @@ public:
```go
func maximumTripletValue(nums []int) int64 {
- ans, mx, mx_diff := 0, 0, 0
- for _, num := range nums {
- ans = max(ans, mx_diff*num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx-num)
+ ans, mx, mxDiff := 0, 0, 0
+ for _, x := range nums {
+ ans = max(ans, mxDiff*x)
+ mxDiff = max(mxDiff, mx-x)
+ mx = max(mx, x)
}
return int64(ans)
}
@@ -141,16 +143,36 @@ func maximumTripletValue(nums []int) int64 {
```ts
function maximumTripletValue(nums: number[]): number {
- let [ans, mx, mx_diff] = [0, 0, 0];
- for (const num of nums) {
- ans = Math.max(ans, mx_diff * num);
- mx = Math.max(mx, num);
- mx_diff = Math.max(mx_diff, mx - num);
+ let [ans, mx, mxDiff] = [0, 0, 0];
+ for (const x of nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
```
+#### Rust
+
+```rust
+impl Solution {
+ pub fn maximum_triplet_value(nums: Vec) -> i64 {
+ let mut ans: i64 = 0;
+ let mut mx: i32 = 0;
+ let mut mx_diff: i32 = 0;
+
+ for &x in &nums {
+ ans = ans.max(mx_diff as i64 * x as i64);
+ mx_diff = mx_diff.max(mx - x);
+ mx = mx.max(x);
+ }
+
+ ans
+ }
+}
+```
+
diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.cpp b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.cpp
index fa6cb8efa928e..a26f8ee375a6d 100644
--- a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.cpp
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.cpp
@@ -1,13 +1,13 @@
class Solution {
public:
long long maximumTripletValue(vector& nums) {
- long long ans = 0;
- int mx = 0, mx_diff = 0;
- for (int num : nums) {
- ans = max(ans, 1LL * mx_diff * num);
- mx = max(mx, num);
- mx_diff = max(mx_diff, mx - num);
+ long long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = max(ans, mxDiff * x);
+ mxDiff = max(mxDiff, 1LL * mx - x);
+ mx = max(mx, x);
}
return ans;
}
-};
\ No newline at end of file
+};
diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.go b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.go
index b1f300b08467a..a7bc0db301cc0 100644
--- a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.go
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.go
@@ -1,9 +1,9 @@
func maximumTripletValue(nums []int) int64 {
- ans, mx, mx_diff := 0, 0, 0
- for _, num := range nums {
- ans = max(ans, mx_diff*num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx-num)
+ ans, mx, mxDiff := 0, 0, 0
+ for _, x := range nums {
+ ans = max(ans, mxDiff*x)
+ mxDiff = max(mxDiff, mx-x)
+ mx = max(mx, x)
}
return int64(ans)
-}
\ No newline at end of file
+}
diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.java b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.java
index 2a020a1a66cca..f75a8dd4efb3e 100644
--- a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.java
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.java
@@ -1,14 +1,12 @@
class Solution {
public long maximumTripletValue(int[] nums) {
- long max, maxDiff, ans;
- max = 0;
- maxDiff = 0;
- ans = 0;
- for (int num : nums) {
- ans = Math.max(ans, num * maxDiff);
- max = Math.max(max, num);
- maxDiff = Math.max(maxDiff, max - num);
+ long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
-}
\ No newline at end of file
+}
diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.py b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.py
index a0b5b3e824a3a..4290207f24879 100644
--- a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.py
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.py
@@ -1,8 +1,8 @@
class Solution:
def maximumTripletValue(self, nums: List[int]) -> int:
ans = mx = mx_diff = 0
- for num in nums:
- ans = max(ans, mx_diff * num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx - num)
+ for x in nums:
+ ans = max(ans, mx_diff * x)
+ mx_diff = max(mx_diff, mx - x)
+ mx = max(mx, x)
return ans
diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.rs b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.rs
new file mode 100644
index 0000000000000..cb7f14099f53f
--- /dev/null
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.rs
@@ -0,0 +1,15 @@
+impl Solution {
+ pub fn maximum_triplet_value(nums: Vec) -> i64 {
+ let mut ans: i64 = 0;
+ let mut mx: i32 = 0;
+ let mut mx_diff: i32 = 0;
+
+ for &x in &nums {
+ ans = ans.max(mx_diff as i64 * x as i64);
+ mx_diff = mx_diff.max(mx - x);
+ mx = mx.max(x);
+ }
+
+ ans
+ }
+}
diff --git a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.ts b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.ts
index f70346b89e985..c1a3b5a088e84 100644
--- a/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.ts
+++ b/solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.ts
@@ -1,9 +1,9 @@
function maximumTripletValue(nums: number[]): number {
- let [ans, mx, mx_diff] = [0, 0, 0];
- for (const num of nums) {
- ans = Math.max(ans, mx_diff * num);
- mx = Math.max(mx, num);
- mx_diff = Math.max(mx_diff, mx - num);
+ let [ans, mx, mxDiff] = [0, 0, 0];
+ for (const x of nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README.md b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README.md
index 55b6402ce953e..6f8732b67ac97 100644
--- a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README.md
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README.md
@@ -41,7 +41,7 @@ tags:
输入：nums = [1,10,3,4,19]
输出：133
解释：下标三元组 (1, 2, 4) 的值是 (nums[1] - nums[2]) * nums[4] = 133 。
-可以证明不存在值大于 133 的有序下标三元组。
+可以证明不存在值大于 133 的有序下标三元组。
示例 3：
@@ -69,7 +69,11 @@ tags:
### 方法一：维护前缀最大值和最大差值
-我们可以用两个变量 $mx$ 和 $mx\_diff$ 分别维护前缀最大值和最大差值。遍历数组时，更新这两个变量，答案为所有 $mx\_diff \times nums[i]$ 的最大值。
+我们用两个变量 $\textit{mx}$ 和 $\textit{mxDiff}$ 分别维护前缀最大值和最大差值，用一个变量 $\textit{ans}$ 维护答案。初始时，这些变量都为 $0$。
+
+接下来，我们枚举数组的每个元素 $x$ 作为 $\textit{nums}[k]$，首先更新答案 $\textit{ans} = \max(\textit{ans}, \textit{mxDiff} \times x)$，然后我们更新最大差值 $\textit{mxDiff} = \max(\textit{mxDiff}, \textit{mx} - x)$，最后更新前缀最大值 $\textit{mx} = \max(\textit{mx}, x)$。
+
+枚举完所有元素后，返回答案 $\textit{ans}$。
时间复杂度 $O(n)$，其中 $n$ 是数组长度。空间复杂度 $O(1)$。
@@ -81,10 +85,10 @@ tags:
class Solution:
def maximumTripletValue(self, nums: List[int]) -> int:
ans = mx = mx_diff = 0
- for num in nums:
- ans = max(ans, mx_diff * num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx - num)
+ for x in nums:
+ ans = max(ans, mx_diff * x)
+ mx_diff = max(mx_diff, mx - x)
+ mx = max(mx, x)
return ans
```
@@ -93,14 +97,12 @@ class Solution:
```java
class Solution {
public long maximumTripletValue(int[] nums) {
- long max, maxDiff, ans;
- max = 0;
- maxDiff = 0;
- ans = 0;
- for (int num : nums) {
- ans = Math.max(ans, num * maxDiff);
- max = Math.max(max, num);
- maxDiff = Math.max(maxDiff, max - num);
+ long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
@@ -113,12 +115,12 @@ class Solution {
class Solution {
public:
long long maximumTripletValue(vector& nums) {
- long long ans = 0;
- int mx = 0, mx_diff = 0;
- for (int num : nums) {
- ans = max(ans, 1LL * mx_diff * num);
- mx = max(mx, num);
- mx_diff = max(mx_diff, mx - num);
+ long long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = max(ans, mxDiff * x);
+ mxDiff = max(mxDiff, 1LL * mx - x);
+ mx = max(mx, x);
}
return ans;
}
@@ -129,11 +131,11 @@ public:
```go
func maximumTripletValue(nums []int) int64 {
- ans, mx, mx_diff := 0, 0, 0
- for _, num := range nums {
- ans = max(ans, mx_diff*num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx-num)
+ ans, mx, mxDiff := 0, 0, 0
+ for _, x := range nums {
+ ans = max(ans, mxDiff*x)
+ mxDiff = max(mxDiff, mx-x)
+ mx = max(mx, x)
}
return int64(ans)
}
@@ -143,16 +145,36 @@ func maximumTripletValue(nums []int) int64 {
```ts
function maximumTripletValue(nums: number[]): number {
- let [ans, mx, mx_diff] = [0, 0, 0];
- for (const num of nums) {
- ans = Math.max(ans, mx_diff * num);
- mx = Math.max(mx, num);
- mx_diff = Math.max(mx_diff, mx - num);
+ let [ans, mx, mxDiff] = [0, 0, 0];
+ for (const x of nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
```
+#### Rust
+
+```rust
+impl Solution {
+ pub fn maximum_triplet_value(nums: Vec) -> i64 {
+ let mut ans: i64 = 0;
+ let mut mx: i32 = 0;
+ let mut mx_diff: i32 = 0;
+
+ for &x in &nums {
+ ans = ans.max(mx_diff as i64 * x as i64);
+ mx_diff = mx_diff.max(mx - x);
+ mx = mx.max(x);
+ }
+
+ ans
+ }
+}
+```
+
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README_EN.md b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README_EN.md
index 6ca64b15433a0..24b82d12a4169 100644
--- a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README_EN.md
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README_EN.md
@@ -31,7 +31,7 @@ tags:
Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
-It can be shown that there are no ordered triplets of indices with a value greater than 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77.
Example 2:
@@ -65,9 +65,13 @@ It can be shown that there are no ordered triplets of indices with a value great
-### Solution 1: Maintain Maximum Prefix Value and Maximum Difference
+### Solution 1: Maintaining Prefix Maximum and Maximum Difference
-We can use two variables $mx$ and $mx\_diff$ to maintain the maximum prefix value and maximum difference, respectively. When traversing the array, we update these two variables, and the answer is the maximum value of all $mx\_diff \times nums[i]$.
+We use two variables $\textit{mx}$ and $\textit{mxDiff}$ to maintain the prefix maximum value and maximum difference, respectively, and use a variable $\textit{ans}$ to maintain the answer. Initially, these variables are all $0$.
+
+Next, we iterate through each element $x$ in the array as $\textit{nums}[k]$. First, we update the answer $\textit{ans} = \max(\textit{ans}, \textit{mxDiff} \times x)$. Then we update the maximum difference $\textit{mxDiff} = \max(\textit{mxDiff}, \textit{mx} - x)$. Finally, we update the prefix maximum value $\textit{mx} = \max(\textit{mx}, x)$.
+
+After iterating through all elements, we return the answer $\textit{ans}$.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
@@ -79,10 +83,10 @@ The time complexity is $O(n)$, where $n$ is the length of the array. The space c
class Solution:
def maximumTripletValue(self, nums: List[int]) -> int:
ans = mx = mx_diff = 0
- for num in nums:
- ans = max(ans, mx_diff * num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx - num)
+ for x in nums:
+ ans = max(ans, mx_diff * x)
+ mx_diff = max(mx_diff, mx - x)
+ mx = max(mx, x)
return ans
```
@@ -91,14 +95,12 @@ class Solution:
```java
class Solution {
public long maximumTripletValue(int[] nums) {
- long max, maxDiff, ans;
- max = 0;
- maxDiff = 0;
- ans = 0;
- for (int num : nums) {
- ans = Math.max(ans, num * maxDiff);
- max = Math.max(max, num);
- maxDiff = Math.max(maxDiff, max - num);
+ long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
@@ -111,12 +113,12 @@ class Solution {
class Solution {
public:
long long maximumTripletValue(vector& nums) {
- long long ans = 0;
- int mx = 0, mx_diff = 0;
- for (int num : nums) {
- ans = max(ans, 1LL * mx_diff * num);
- mx = max(mx, num);
- mx_diff = max(mx_diff, mx - num);
+ long long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = max(ans, mxDiff * x);
+ mxDiff = max(mxDiff, 1LL * mx - x);
+ mx = max(mx, x);
}
return ans;
}
@@ -127,11 +129,11 @@ public:
```go
func maximumTripletValue(nums []int) int64 {
- ans, mx, mx_diff := 0, 0, 0
- for _, num := range nums {
- ans = max(ans, mx_diff*num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx-num)
+ ans, mx, mxDiff := 0, 0, 0
+ for _, x := range nums {
+ ans = max(ans, mxDiff*x)
+ mxDiff = max(mxDiff, mx-x)
+ mx = max(mx, x)
}
return int64(ans)
}
@@ -141,16 +143,36 @@ func maximumTripletValue(nums []int) int64 {
```ts
function maximumTripletValue(nums: number[]): number {
- let [ans, mx, mx_diff] = [0, 0, 0];
- for (const num of nums) {
- ans = Math.max(ans, mx_diff * num);
- mx = Math.max(mx, num);
- mx_diff = Math.max(mx_diff, mx - num);
+ let [ans, mx, mxDiff] = [0, 0, 0];
+ for (const x of nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
```
+#### Rust
+
+```rust
+impl Solution {
+ pub fn maximum_triplet_value(nums: Vec) -> i64 {
+ let mut ans: i64 = 0;
+ let mut mx: i32 = 0;
+ let mut mx_diff: i32 = 0;
+
+ for &x in &nums {
+ ans = ans.max(mx_diff as i64 * x as i64);
+ mx_diff = mx_diff.max(mx - x);
+ mx = mx.max(x);
+ }
+
+ ans
+ }
+}
+```
+
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.cpp b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.cpp
index fa6cb8efa928e..a26f8ee375a6d 100644
--- a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.cpp
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.cpp
@@ -1,13 +1,13 @@
class Solution {
public:
long long maximumTripletValue(vector& nums) {
- long long ans = 0;
- int mx = 0, mx_diff = 0;
- for (int num : nums) {
- ans = max(ans, 1LL * mx_diff * num);
- mx = max(mx, num);
- mx_diff = max(mx_diff, mx - num);
+ long long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = max(ans, mxDiff * x);
+ mxDiff = max(mxDiff, 1LL * mx - x);
+ mx = max(mx, x);
}
return ans;
}
-};
\ No newline at end of file
+};
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.go b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.go
index b1f300b08467a..a7bc0db301cc0 100644
--- a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.go
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.go
@@ -1,9 +1,9 @@
func maximumTripletValue(nums []int) int64 {
- ans, mx, mx_diff := 0, 0, 0
- for _, num := range nums {
- ans = max(ans, mx_diff*num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx-num)
+ ans, mx, mxDiff := 0, 0, 0
+ for _, x := range nums {
+ ans = max(ans, mxDiff*x)
+ mxDiff = max(mxDiff, mx-x)
+ mx = max(mx, x)
}
return int64(ans)
-}
\ No newline at end of file
+}
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.java b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.java
index 2a020a1a66cca..f75a8dd4efb3e 100644
--- a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.java
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.java
@@ -1,14 +1,12 @@
class Solution {
public long maximumTripletValue(int[] nums) {
- long max, maxDiff, ans;
- max = 0;
- maxDiff = 0;
- ans = 0;
- for (int num : nums) {
- ans = Math.max(ans, num * maxDiff);
- max = Math.max(max, num);
- maxDiff = Math.max(maxDiff, max - num);
+ long ans = 0, mxDiff = 0;
+ int mx = 0;
+ for (int x : nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}
-}
\ No newline at end of file
+}
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.py b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.py
index a0b5b3e824a3a..4290207f24879 100644
--- a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.py
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.py
@@ -1,8 +1,8 @@
class Solution:
def maximumTripletValue(self, nums: List[int]) -> int:
ans = mx = mx_diff = 0
- for num in nums:
- ans = max(ans, mx_diff * num)
- mx = max(mx, num)
- mx_diff = max(mx_diff, mx - num)
+ for x in nums:
+ ans = max(ans, mx_diff * x)
+ mx_diff = max(mx_diff, mx - x)
+ mx = max(mx, x)
return ans
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.rs b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.rs
new file mode 100644
index 0000000000000..cb7f14099f53f
--- /dev/null
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.rs
@@ -0,0 +1,15 @@
+impl Solution {
+ pub fn maximum_triplet_value(nums: Vec) -> i64 {
+ let mut ans: i64 = 0;
+ let mut mx: i32 = 0;
+ let mut mx_diff: i32 = 0;
+
+ for &x in &nums {
+ ans = ans.max(mx_diff as i64 * x as i64);
+ mx_diff = mx_diff.max(mx - x);
+ mx = mx.max(x);
+ }
+
+ ans
+ }
+}
diff --git a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.ts b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.ts
index f70346b89e985..c1a3b5a088e84 100644
--- a/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.ts
+++ b/solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.ts
@@ -1,9 +1,9 @@
function maximumTripletValue(nums: number[]): number {
- let [ans, mx, mx_diff] = [0, 0, 0];
- for (const num of nums) {
- ans = Math.max(ans, mx_diff * num);
- mx = Math.max(mx, num);
- mx_diff = Math.max(mx_diff, mx - num);
+ let [ans, mx, mxDiff] = [0, 0, 0];
+ for (const x of nums) {
+ ans = Math.max(ans, mxDiff * x);
+ mxDiff = Math.max(mxDiff, mx - x);
+ mx = Math.max(mx, x);
}
return ans;
}