Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly
Sign up
Appearance settings

Commit ac0cf6e

Browse files
feat: add solutions to lc problem: No.3495 (#4697)
1 parent 4410451 commit ac0cf6e

File tree

3 files changed

+106
-2
lines changed

3 files changed

+106
-2
lines changed

‎solution/3400-3499/3495.Minimum Operations to Make Array Elements Zero/README.md‎

Lines changed: 40 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -103,7 +103,15 @@ tags:
103103

104104
<!-- solution:start -->
105105

106-
### 方法一
106+
### 方法一:前缀和
107+
108+
根据题目描述,我们不妨假设把一个元素 $x$ 变为 0ドル$ 需要的最小操作次数为 $p,ドル那么 $p$ 是满足 4ドル^p \gt x$ 的最小整数。
109+
110+
我们知道了一个元素的最小操作次数,那么对于一个范围 $[l, r],ドル我们假设 $[l, r]$ 范围内所有元素的**最小操作次数之和**为 $s,ドル**最大操作次数**为元素 $r$ 的操作次数 $mx,ドル那么 $[l, r]$ 范围内所有元素变为 0ドル$ 的最少操作次数为 $\max(\lceil s / 2 \rceil, mx)$。
111+
112+
我们定义一个函数 $f(x),ドル表示范围 $[1, x]$ 内所有元素的最小操作次数之和,那么对于每个查询 $[l, r],ドル我们可以计算出 $s = f(r) - f(l - 1)$ 和 $mx = f(r) - f(r - 1),ドル从而得到答案。
113+
114+
时间复杂度 $O(q \log M),ドル其中 $q$ 是查询次数,而 $M$ 是查询范围的最大值。空间复杂度 $O(1)$。
107115

108116
<!-- tabs:start -->
109117

@@ -243,6 +251,37 @@ function minOperations(queries: number[][]): number {
243251
}
244252
```
245253

254+
#### Rust
255+
256+
```rust
257+
impl Solution {
258+
pub fn min_operations(queries: Vec<Vec<i32>>) -> i64 {
259+
let f = |x: i64| -> i64 {
260+
let mut res: i64 = 0;
261+
let mut p: i64 = 1;
262+
let mut i: i64 = 1;
263+
while p <= x {
264+
let cnt = std::cmp::min(p * 4 - 1, x) - p + 1;
265+
res += cnt * i;
266+
i += 1;
267+
p *= 4;
268+
}
269+
res
270+
};
271+
272+
let mut ans: i64 = 0;
273+
for q in queries {
274+
let l = q[0] as i64;
275+
let r = q[1] as i64;
276+
let s = f(r) - f(l - 1);
277+
let mx = f(r) - f(r - 1);
278+
ans += std::cmp::max((s + 1) / 2, mx);
279+
}
280+
ans
281+
}
282+
}
283+
```
284+
246285
<!-- tabs:end -->
247286

248287
<!-- solution:end -->

‎solution/3400-3499/3495.Minimum Operations to Make Array Elements Zero/README_EN.md‎

Lines changed: 40 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -100,7 +100,15 @@ tags:
100100

101101
<!-- solution:start -->
102102

103-
### Solution 1
103+
### Solution 1: Prefix Sum
104+
105+
According to the problem description, suppose the minimum number of operations required to make an element $x$ become 0ドル$ is $p,ドル where $p$ is the smallest integer such that 4ドル^p > x$.
106+
107+
Once we know the minimum number of operations for each element, for a range $[l, r],ドル let $s$ be the sum of the minimum operations for all elements in $[l, r],ドル and let $mx$ be the maximum number of operations, which is the number of operations for element $r$. Then, the minimum number of operations to make all elements in $[l, r]$ become 0ドル$ is $\max(\lceil s / 2 \rceil, mx)$.
108+
109+
We define a function $f(x)$ to represent the sum of the minimum operations for all elements in the range $[1, x]$. For each query $[l, r],ドル we can compute $s = f(r) - f(l - 1)$ and $mx = f(r) - f(r - 1)$ to obtain the answer.
110+
111+
The time complexity is $O(q \log M),ドル where $q$ is the number of queries and $M$ is the maximum value in the query range. The space complexity is $O(1)$.
104112

105113
<!-- tabs:start -->
106114

@@ -240,6 +248,37 @@ function minOperations(queries: number[][]): number {
240248
}
241249
```
242250

251+
#### Rust
252+
253+
```rust
254+
impl Solution {
255+
pub fn min_operations(queries: Vec<Vec<i32>>) -> i64 {
256+
let f = |x: i64| -> i64 {
257+
let mut res: i64 = 0;
258+
let mut p: i64 = 1;
259+
let mut i: i64 = 1;
260+
while p <= x {
261+
let cnt = std::cmp::min(p * 4 - 1, x) - p + 1;
262+
res += cnt * i;
263+
i += 1;
264+
p *= 4;
265+
}
266+
res
267+
};
268+
269+
let mut ans: i64 = 0;
270+
for q in queries {
271+
let l = q[0] as i64;
272+
let r = q[1] as i64;
273+
let s = f(r) - f(l - 1);
274+
let mx = f(r) - f(r - 1);
275+
ans += std::cmp::max((s + 1) / 2, mx);
276+
}
277+
ans
278+
}
279+
}
280+
```
281+
243282
<!-- tabs:end -->
244283

245284
<!-- solution:end -->
Lines changed: 26 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,26 @@
1+
impl Solution {
2+
pub fn min_operations(queries: Vec<Vec<i32>>) -> i64 {
3+
let f = |x: i64| -> i64 {
4+
let mut res: i64 = 0;
5+
let mut p: i64 = 1;
6+
let mut i: i64 = 1;
7+
while p <= x {
8+
let cnt = std::cmp::min(p * 4 - 1, x) - p + 1;
9+
res += cnt * i;
10+
i += 1;
11+
p *= 4;
12+
}
13+
res
14+
};
15+
16+
let mut ans: i64 = 0;
17+
for q in queries {
18+
let l = q[0] as i64;
19+
let r = q[1] as i64;
20+
let s = f(r) - f(l - 1);
21+
let mx = f(r) - f(r - 1);
22+
ans += std::cmp::max((s + 1) / 2, mx);
23+
}
24+
ans
25+
}
26+
}

0 commit comments

Comments
(0)

AltStyle によって変換されたページ (->オリジナル) /