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feat: add solutions to lc problem: No.1014 (#3552)
No.1014.Best Sightseeing Pair
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‎solution/1000-1099/1014.Best Sightseeing Pair/README.md‎

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@@ -57,11 +57,11 @@ tags:
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<!-- solution:start -->
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### 方法一：枚举 + 维护前缀最大值
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### 方法一：枚举
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62-
我们可以在 $[1,..n - 1]$ 的范围内枚举 $j$，那么我们要在 $[0,..j - 1]$ 的范围内找到一个 $i$，使得 $values[i] + values[j] + i - j$ 的值最大。我们可以维护一个前缀最大值，即 $values[i] + i$ 的最大值，那么我们只需要在枚举 $j$ 的过程中，不断地更新答案即可。
62+
我们可以从左到右枚举 $j$，同时维护 $j$ 左侧元素中 $values[i] + i$ 的最大值 $mx$，这样对于每个 $j$，最大得分为 $mx + values[j] - j$。我们取所有位置的最大得分的最大值即为答案。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组的长度。
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时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{values}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -70,10 +70,10 @@ tags:
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```python
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class Solution:
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def maxScoreSightseeingPair(self, values: List[int]) -> int:
73-
ans, mx = 0, values[0]
74-
for j in range(1, len(values)):
75-
ans = max(ans, values[j] - j + mx)
76-
mx = max(mx, values[j] + j)
73+
ans = mx = 0
74+
for j, x in enumerate(values):
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ans = max(ans, mx + x - j)
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mx = max(mx, x + j)
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return ans
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```
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@@ -82,9 +82,9 @@ class Solution:
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```java
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class Solution {
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public int maxScoreSightseeingPair(int[] values) {
85-
int ans = 0, mx = values[0];
86-
for (int j = 1; j < values.length; ++j) {
87-
ans = Math.max(ans, values[j] - j + mx);
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int ans = 0, mx = 0;
86+
for (int j = 0; j < values.length; ++j) {
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ans = Math.max(ans, mx + values[j] - j);
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mx = Math.max(mx, values[j] + j);
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}
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return ans;
@@ -98,9 +98,9 @@ class Solution {
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class Solution {
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public:
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int maxScoreSightseeingPair(vector<int>& values) {
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int ans = 0, mx = values[0];
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for (int j = 1; j < values.size(); ++j) {
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ans = max(ans, values[j] - j + mx);
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int ans = 0, mx = 0;
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for (int j = 0; j < values.size(); ++j) {
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ans = max(ans, mx + values[j] - j);
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mx = max(mx, values[j] + j);
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}
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return ans;
@@ -112,9 +112,10 @@ public:
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```go
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func maxScoreSightseeingPair(values []int) (ans int) {
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for j, mx := 1, values[0]; j < len(values); j++ {
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ans = max(ans, values[j]-j+mx)
117-
mx = max(mx, values[j]+j)
115+
mx := 0
116+
for j, x := range values {
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ans = max(ans, mx+x-j)
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mx = max(mx, x+j)
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}
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return
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}
@@ -124,16 +125,31 @@ func maxScoreSightseeingPair(values []int) (ans int) {
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```ts
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function maxScoreSightseeingPair(values: number[]): number {
127-
let ans = 0;
128-
let mx = values[0];
129-
for (let j = 1; j < values.length; ++j) {
130-
ans = Math.max(ans, values[j] - j + mx);
128+
let [ans, mx] = [0, 0];
129+
for (let j = 0; j < values.length; ++j) {
130+
ans = Math.max(ans, mx + values[j] - j);
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mx = Math.max(mx, values[j] + j);
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}
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return ans;
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn max_score_sightseeing_pair(values: Vec<i32>) -> i32 {
142+
let mut ans = 0;
143+
let mut mx = 0;
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for (j, &x) in values.iter().enumerate() {
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ans = ans.max(mx + x - j as i32);
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mx = mx.max(x + j as i32);
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}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1000-1099/1014.Best Sightseeing Pair/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Enumeration
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We can enumerate $j$ from left to right while maintaining the maximum value of $values[i] + i$ for elements to the left of $j$, denoted as $mx$. For each $j$, the maximum score is $mx + values[j] - j$. The answer is the maximum of these maximum scores for all positions.
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The time complexity is $O(n)$, where $n$ is the length of the array $\textit{values}$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def maxScoreSightseeingPair(self, values: List[int]) -> int:
67-
ans, mx = 0, values[0]
68-
for j in range(1, len(values)):
69-
ans = max(ans, values[j] - j + mx)
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mx = max(mx, values[j] + j)
71+
ans = mx = 0
72+
for j, x in enumerate(values):
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ans = max(ans, mx + x - j)
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mx = max(mx, x + j)
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return ans
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```
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```java
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class Solution {
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public int maxScoreSightseeingPair(int[] values) {
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int ans = 0, mx = values[0];
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for (int j = 1; j < values.length; ++j) {
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ans = Math.max(ans, values[j] - j + mx);
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int ans = 0, mx = 0;
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for (int j = 0; j < values.length; ++j) {
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ans = Math.max(ans, mx + values[j] - j);
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mx = Math.max(mx, values[j] + j);
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}
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return ans;
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class Solution {
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public:
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int maxScoreSightseeingPair(vector<int>& values) {
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int ans = 0, mx = values[0];
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for (int j = 1; j < values.size(); ++j) {
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ans = max(ans, values[j] - j + mx);
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int ans = 0, mx = 0;
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for (int j = 0; j < values.size(); ++j) {
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ans = max(ans, mx + values[j] - j);
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mx = max(mx, values[j] + j);
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}
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return ans;
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```go
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func maxScoreSightseeingPair(values []int) (ans int) {
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for j, mx := 1, values[0]; j < len(values); j++ {
110-
ans = max(ans, values[j]-j+mx)
111-
mx = max(mx, values[j]+j)
113+
mx := 0
114+
for j, x := range values {
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ans = max(ans, mx+x-j)
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mx = max(mx, x+j)
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}
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return
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}
@@ -118,16 +123,31 @@ func maxScoreSightseeingPair(values []int) (ans int) {
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119124
```ts
120125
function maxScoreSightseeingPair(values: number[]): number {
121-
let ans = 0;
122-
let mx = values[0];
123-
for (let j = 1; j < values.length; ++j) {
124-
ans = Math.max(ans, values[j] - j + mx);
126+
let [ans, mx] = [0, 0];
127+
for (let j = 0; j < values.length; ++j) {
128+
ans = Math.max(ans, mx + values[j] - j);
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mx = Math.max(mx, values[j] + j);
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}
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return ans;
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}
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```
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#### Rust
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137+
```rust
138+
impl Solution {
139+
pub fn max_score_sightseeing_pair(values: Vec<i32>) -> i32 {
140+
let mut ans = 0;
141+
let mut mx = 0;
142+
for (j, &x) in values.iter().enumerate() {
143+
ans = ans.max(mx + x - j as i32);
144+
mx = mx.max(x + j as i32);
145+
}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1000-1099/1014.Best Sightseeing Pair/Solution.cpp‎

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class Solution {
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public:
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int maxScoreSightseeingPair(vector<int>& values) {
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int ans = 0, mx = values[0];
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for (int j = 1; j < values.size(); ++j) {
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ans = max(ans, values[j] - j + mx);
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int ans = 0, mx = 0;
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for (int j = 0; j < values.size(); ++j) {
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ans = max(ans, mx + values[j] - j);
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mx = max(mx, values[j] + j);
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}
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return ans;
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}
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};
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};

‎solution/1000-1099/1014.Best Sightseeing Pair/Solution.go‎

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func maxScoreSightseeingPair(values []int) (ans int) {
2-
for j, mx := 1, values[0]; j < len(values); j++ {
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ans = max(ans, values[j]-j+mx)
4-
mx = max(mx, values[j]+j)
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mx := 0
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for j, x := range values {
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ans = max(ans, mx+x-j)
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mx = max(mx, x+j)
56
}
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return
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}
8+
}

‎solution/1000-1099/1014.Best Sightseeing Pair/Solution.java‎

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class Solution {
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public int maxScoreSightseeingPair(int[] values) {
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int ans = 0, mx = values[0];
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for (int j = 1; j < values.length; ++j) {
5-
ans = Math.max(ans, values[j] - j + mx);
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int ans = 0, mx = 0;
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for (int j = 0; j < values.length; ++j) {
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ans = Math.max(ans, mx + values[j] - j);
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mx = Math.max(mx, values[j] + j);
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}
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return ans;
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}
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}
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}

‎solution/1000-1099/1014.Best Sightseeing Pair/Solution.py‎

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class Solution:
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def maxScoreSightseeingPair(self, values: List[int]) -> int:
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ans, mx = 0, values[0]
4-
for j in range(1, len(values)):
5-
ans = max(ans, values[j] - j + mx)
6-
mx = max(mx, values[j] + j)
3+
ans = mx = 0
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for j, x in enumerate(values):
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ans = max(ans, mx + x - j)
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mx = max(mx, x + j)
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return ans

‎solution/1000-1099/1014.Best Sightseeing Pair/Solution.rs‎

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impl Solution {
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pub fn max_score_sightseeing_pair(values: Vec<i32>) -> i32 {
3+
let mut ans = 0;
4+
let mut mx = 0;
5+
for (j, &x) in values.iter().enumerate() {
6+
ans = ans.max(mx + x - j as i32);
7+
mx = mx.max(x + j as i32);
8+
}
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ans
10+
}
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}

‎solution/1000-1099/1014.Best Sightseeing Pair/Solution.ts‎

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function maxScoreSightseeingPair(values: number[]): number {
2-
let ans = 0;
3-
let mx = values[0];
4-
for (let j = 1; j < values.length; ++j) {
5-
ans = Math.max(ans, values[j] - j + mx);
2+
let [ans, mx] = [0, 0];
3+
for (let j = 0; j < values.length; ++j) {
4+
ans = Math.max(ans, mx + values[j] - j);
65
mx = Math.max(mx, values[j] + j);
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}
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return ans;

‎solution/1000-1099/1018.Binary Prefix Divisible By 5/README.md‎

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### 方法一：模拟
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遍历数组，每一次遍历都将当前数字加到前面的数字上，然后对 $5$ 取模，如果结果为 $0$，则当前数字可以被 $5$ 整除，答案设置为 `true`，否则为 `false`。
67+
我们用一个变量 $x$ 来表示当前的二进制前缀，然后遍历数组 $nums$，对于每个元素 $v$，我们将 $x$ 左移一位，然后加上 $v$，再对 $5$ 取模，判断是否等于 $0$，如果等于 $0$，则说明当前的二进制前缀可以被 $5$ 整除，我们将 $\textit{true}$ 加入答案数组，否则将 $\textit{false}$ 加入答案数组。
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69-
时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
69+
时间复杂度 $O(n)$，忽略答案数组的空间消耗，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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‎solution/1000-1099/1018.Binary Prefix Divisible By 5/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
63+
### Solution 1: Simulation
64+
65+
We use a variable $x$ to represent the current binary prefix, then traverse the array $nums$. For each element $v$, we left shift $x$ by one bit, then add $v$, and take the result modulo $5$. If the result equals $0$, it means the current binary prefix is divisible by $5$, and we add $\textit{true}$ to the answer array; otherwise, we add $\textit{false}$ to the answer array.
66+
67+
The time complexity is $O(n)$, and ignoring the space consumption of the answer array, the space complexity is $O(1)$.
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<!-- tabs:start -->
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