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feat: add solutions to lc problem: No.2419 #3983

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Jan 22, 2025
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feat: add solutions to lc problem: No.2419 #3983

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42 changes: 9 additions & 33 deletions ...on/2000-2099/2011.Final Value of Variable After Performing Operations/README.md
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Expand Up @@ -50,7 +50,7 @@ X++：X 加 1 ，X = 0 + 1 = 1
<pre>
<strong>输入：</strong>operations = ["++X","++X","X++"]
<strong>输出：</strong>3
<strong>解释：</strong>操作按下述步骤执行：
<strong>解释：</strong>操作按下述步骤执行：
最初，X = 0
++X：X 加 1 ，X = 0 + 1 = 1
++X：X 加 1 ，X = 1 + 1 = 2
Expand Down Expand Up @@ -85,11 +85,11 @@ X--：X 减 1 ，X = 1 - 1 = 0

<!-- solution:start -->

### 方法一：模拟
### 方法一：计数

遍历数组 `operations`，对于每个操作 $operations[i]$，如果包含 `'+'`，那么答案加 $1$，否则答案减 $1$。
我们遍历数组 $\textit{operations}$，对于每个操作 $\textit{operations}[i]$，如果包含 `'+'`，那么答案加 $1$，否则答案减 $1$。

时间复杂度为 $O(n)$，其中 $n$ 为数组 `operations` 的长度。空间复杂度 $O(1)$。
时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{operations}$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -122,7 +122,9 @@ class Solution {
public:
int finalValueAfterOperations(vector<string>& operations) {
int ans = 0;
for (auto& s : operations) ans += (s[1] == '+' ? 1 : -1);
for (auto& s : operations) {
ans += s[1] == '+' ? 1 : -1;
}
return ans;
}
};
Expand All @@ -147,11 +149,7 @@ func finalValueAfterOperations(operations []string) (ans int) {

```ts
function finalValueAfterOperations(operations: string[]): number {
let ans = 0;
for (let operation of operations) {
ans += operation.includes('+') ? 1 : -1;
}
return ans;
return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
}
```

Expand All @@ -177,11 +175,7 @@ impl Solution {
* @return {number}
*/
var finalValueAfterOperations = function (operations) {
let ans = 0;
for (const s of operations) {
ans += s[1] === '+' ? 1 : -1;
}
return ans;
return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
};
```

Expand All @@ -201,22 +195,4 @@ int finalValueAfterOperations(char** operations, int operationsSize) {

<!-- solution:end -->

<!-- solution:start -->

### 方法二

<!-- tabs:start -->

#### TypeScript

```ts
function finalValueAfterOperations(operations: string[]): number {
return operations.reduce((r, v) => r + (v[1] === '+' ? 1 : -1), 0);
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
40 changes: 8 additions & 32 deletions ...2000-2099/2011.Final Value of Variable After Performing Operations/README_EN.md
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Expand Up @@ -83,11 +83,11 @@ X--: X is decremented by 1, X = 1 - 1 = 0.

<!-- solution:start -->

### Solution 1: Simulation
### Solution 1: Counting

Traverse the array `operations`. For each operation $operations[i]$, if it contains `'+'`, then the answer increases by $1$, otherwise the answer decreases by $1$.
We traverse the array $\textit{operations}$. For each operation $\textit{operations}[i]$, if it contains `'+'`, we increment the answer by $1$, otherwise, we decrement the answer by $1$.

The time complexity is $O(n)$, where $n$ is the length of the array `operations`. The space complexity is $O(1)$.
The time complexity is $O(n)$, where $n$ is the length of the array $\textit{operations}$. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -120,7 +120,9 @@ class Solution {
public:
int finalValueAfterOperations(vector<string>& operations) {
int ans = 0;
for (auto& s : operations) ans += (s[1] == '+' ? 1 : -1);
for (auto& s : operations) {
ans += s[1] == '+' ? 1 : -1;
}
return ans;
}
};
Expand All @@ -145,11 +147,7 @@ func finalValueAfterOperations(operations []string) (ans int) {

```ts
function finalValueAfterOperations(operations: string[]): number {
let ans = 0;
for (let operation of operations) {
ans += operation.includes('+') ? 1 : -1;
}
return ans;
return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
}
```

Expand All @@ -175,11 +173,7 @@ impl Solution {
* @return {number}
*/
var finalValueAfterOperations = function (operations) {
let ans = 0;
for (const s of operations) {
ans += s[1] === '+' ? 1 : -1;
}
return ans;
return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
};
```

Expand All @@ -199,22 +193,4 @@ int finalValueAfterOperations(char** operations, int operationsSize) {

<!-- solution:end -->

<!-- solution:start -->

### Solution 2

<!-- tabs:start -->

#### TypeScript

```ts
function finalValueAfterOperations(operations: string[]): number {
return operations.reduce((r, v) => r + (v[1] === '+' ? 1 : -1), 0);
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
6 changes: 4 additions & 2 deletions solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.cpp
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Expand Up @@ -2,7 +2,9 @@ class Solution {
public:
int finalValueAfterOperations(vector<string>& operations) {
int ans = 0;
for (auto& s : operations) ans += (s[1] == '+' ? 1 : -1);
for (auto& s : operations) {
ans += s[1] == '+' ? 1 : -1;
}
return ans;
}
};
};
6 changes: 1 addition & 5 deletions solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.js
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Expand Up @@ -3,9 +3,5 @@
* @return {number}
*/
var finalValueAfterOperations = function (operations) {
let ans = 0;
for (const s of operations) {
ans += s[1] === '+' ? 1 : -1;
}
return ans;
return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
};
6 changes: 1 addition & 5 deletions solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,7 +1,3 @@
function finalValueAfterOperations(operations: string[]): number {
let ans = 0;
for (let operation of operations) {
ans += operation.includes('+') ? 1 : -1;
}
return ans;
return operations.reduce((acc, op) => acc + (op[1] === '+' ? 1 : -1), 0);
}
3 changes: 0 additions & 3 deletions solution/2000-2099/2011.Final Value of Variable After Performing Operations/Solution2.ts
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This file was deleted.

82 changes: 49 additions & 33 deletions solution/2400-2499/2419.Longest Subarray With Maximum Bitwise AND/README.md
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Expand Up @@ -52,7 +52,7 @@ tags:
<strong>输入：</strong>nums = [1,2,3,4]
<strong>输出：</strong>1
<strong>解释：</strong>
子数组按位与运算的最大值是 4 。
子数组按位与运算的最大值是 4 。
能得到此结果的最长子数组是 [4]，所以返回 1 。
</pre>

Expand All @@ -77,9 +77,9 @@ tags:

题目可以转换为求最大值在数组中最多连续出现的次数。

先遍历一遍数组，求出最大值，然后再遍历一遍数组，求出最大值连续出现的次数，最后返回这个次数即可。
我们先遍历数组 $\textit{nums}$ 找到最大值 $\textit{mx}$，然后再遍历数组一次，找到最大值连续出现的次数，最后返回这个次数即可。

时间复杂度 $O(n)$。其中 $n$ 为数组的长度。
时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -90,8 +90,8 @@ class Solution:
def longestSubarray(self, nums: List[int]) -> int:
mx = max(nums)
ans = cnt = 0
for v in nums:
if v == mx:
for x in nums:
if x == mx:
cnt += 1
ans = max(ans, cnt)
else:
Expand All @@ -104,15 +104,11 @@ class Solution:
```java
class Solution {
public int longestSubarray(int[] nums) {
int mx = 0;
for (int v : nums) {
mx = Math.max(mx, v);
}
int mx = Arrays.stream(nums).max().getAsInt();
int ans = 0, cnt = 0;
for (int v : nums) {
if (v == mx) {
++cnt;
ans = Math.max(ans, cnt);
for (int x : nums) {
if (x == mx) {
ans = Math.max(ans, ++cnt);
} else {
cnt = 0;
}
Expand All @@ -128,12 +124,11 @@ class Solution {
class Solution {
public:
int longestSubarray(vector<int>& nums) {
int mx = *max_element(nums.begin(), nums.end());
int mx = ranges::max(nums);
int ans = 0, cnt = 0;
for (int v : nums) {
if (v == mx) {
++cnt;
ans = max(ans, cnt);
for (int x : nums) {
if (x == mx) {
ans = max(ans, ++cnt);
} else {
cnt = 0;
}
Expand All @@ -146,18 +141,18 @@ public:
#### Go

```go
func longestSubarray(nums []int) int {
func longestSubarray(nums []int) (ans int) {
mx := slices.Max(nums)
ans, cnt := 0, 0
for _, v := range nums {
if v == mx {
cnt := 0
for _, x := range nums {
if x == mx {
cnt++
ans = max(ans, cnt)
} else {
cnt = 0
}
}
return ans
return
}
```

Expand All @@ -167,38 +162,59 @@ func longestSubarray(nums []int) int {
function longestSubarray(nums: number[]): number {
const mx = Math.max(...nums);
let [ans, cnt] = [0, 0];

for (const x of nums) {
if (x === mx) {
cnt++;
ans = Math.max(ans, cnt);
ans = Math.max(ans, ++cnt);
} else {
cnt = 0;
}
}

return ans;
}
```

#### Rust

```rust
impl Solution {
pub fn longest_subarray(nums: Vec<i32>) -> i32 {
let mx = *nums.iter().max().unwrap();
let mut ans = 0;
let mut cnt = 0;

for &x in nums.iter() {
if x == mx {
cnt += 1;
ans = ans.max(cnt);
} else {
cnt = 0;
}
}

ans
}
}
```

#### JavaScript

```js
function longestSubarray(nums) {
/**
* @param {number[]} nums
* @return {number}
*/
var longestSubarray = function (nums) {
const mx = Math.max(...nums);
let [ans, cnt] = [0, 0];

for (const x of nums) {
if (x === mx) {
cnt++;
ans = Math.max(ans, cnt);
ans = Math.max(ans, ++cnt);
} else {
cnt = 0;
}
}

return ans;
}
};
```

<!-- tabs:end -->
Expand Down
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