From e913a991f095c844d914ccf81edd1d7fa89ae574 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Tue, 18 Mar 2025 16:17:16 +0800 Subject: [PATCH] feat: add solutions to lc problems: No.2873,2874 * No.2873.Maximum Value of an Ordered Triplet I * No.2874.Maximum Value of an Ordered Triplet II --- .../README.md | 82 +++++++++++------- .../README_EN.md | 84 ++++++++++++------- .../Solution.cpp | 14 ++-- .../Solution.go | 12 +-- .../Solution.java | 16 ++-- .../Solution.py | 8 +- .../Solution.rs | 15 ++++ .../Solution.ts | 10 +-- .../README.md | 82 +++++++++++------- .../README_EN.md | 84 ++++++++++++------- .../Solution.cpp | 14 ++-- .../Solution.go | 12 +-- .../Solution.java | 16 ++-- .../Solution.py | 8 +- .../Solution.rs | 15 ++++ .../Solution.ts | 10 +-- 16 files changed, 298 insertions(+), 184 deletions(-) create mode 100644 solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.rs create mode 100644 solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.rs 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; }