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Commit 7fb0b03

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feat: add solutions to lc problem: No.3147 (#4775)
1 parent 8585d28 commit 7fb0b03

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‎solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README.md‎

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<!-- solution:start -->
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### 方法一:枚举 + 后缀和
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### 方法一:枚举 + 逆序遍历
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我们可以在 $[n - k, n)$ 的范围内枚举终点,然后从终点开始向前遍历,每次累加间隔为 $k$ 的魔法师的能量值,更新答案。
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时间复杂度 $O(n),ドル其中 $n$ 是数组 `energy` 的长度。空间复杂度 $O(1)$。
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时间复杂度 $O(n),ドル其中 $n$ 是数组 $\textit{energy}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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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 maximum_energy(energy: Vec<i32>, k: i32) -> i32 {
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let n = energy.len();
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let mut ans = i32::MIN;
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for i in n - k as usize..n {
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let mut s = 0;
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let mut j = i as i32;
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while j >= 0 {
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s += energy[j as usize];
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ans = ans.max(s);
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j -= k;
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}
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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/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README_EN.md‎

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<!-- solution:start -->
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### Solution 1: Enumeration + Suffix Sum
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### Solution 1: Enumeration + Reverse Traversal
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We can enumerate the endpoint in the range of $[n - k, n),ドル then start from the endpoint and traverse backwards, adding the energy value of the magician at each $k$ interval, and update the answer.
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We can enumerate the endpoints within the range $[n - k, n),ドル then traverse backwards from each endpoint, accumulating the energy values of wizards at intervals of $k,ドル and update the answer.
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The time complexity is $O(n),ドル where $n$ is the length of the array `energy`. The space complexity is $O(1)$.
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The time complexity is $O(n),ドル where $n$ is the length of array $\textit{energy}$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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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 maximum_energy(energy: Vec<i32>, k: i32) -> i32 {
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let n = energy.len();
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let mut ans = i32::MIN;
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for i in n - k as usize..n {
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let mut s = 0;
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let mut j = i as i32;
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while j >= 0 {
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s += energy[j as usize];
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ans = ans.max(s);
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j -= k;
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}
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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 -->
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impl Solution {
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pub fn maximum_energy(energy: Vec<i32>, k: i32) -> i32 {
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let n = energy.len();
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let mut ans = i32::MIN;
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for i in n - k as usize..n {
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let mut s = 0;
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let mut j = i as i32;
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while j >= 0 {
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s += energy[j as usize];
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ans = ans.max(s);
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j -= k;
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}
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}
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ans
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}
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}

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