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feat: update solutions to lc problems: No.1883,3119 #2618

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Apr 19, 2024
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feat: update solutions to lc problems: No.1883,3119 #2618

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42 changes: 18 additions & 24 deletions solution/1800-1899/1883.Minimum Skips to Arrive at Meeting On Time/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -106,6 +106,24 @@ class Solution:
return -1
```

```python
class Solution:
def minSkips(self, dist: List[int], speed: int, hoursBefore: int) -> int:
n = len(dist)
f = [[inf] * (n + 1) for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(dist, 1):
for j in range(i + 1):
if j < i:
f[i][j] = min(f[i][j], ((f[i - 1][j] + x - 1) // speed + 1) * speed)
if j:
f[i][j] = min(f[i][j], f[i - 1][j - 1] + x)
for j in range(n + 1):
if f[n][j] <= hoursBefore * speed:
return j
return -1
```

```java
class Solution {
public int minSkips(int[] dist, int speed, int hoursBefore) {
Expand Down Expand Up @@ -223,28 +241,4 @@ function minSkips(dist: number[], speed: number, hoursBefore: number): number {

<!-- tabs:end -->

### 方法二

<!-- tabs:start -->

```python
class Solution:
def minSkips(self, dist: List[int], speed: int, hoursBefore: int) -> int:
n = len(dist)
f = [[inf] * (n + 1) for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(dist, 1):
for j in range(i + 1):
if j < i:
f[i][j] = min(f[i][j], ((f[i - 1][j] + x - 1) // speed + 1) * speed)
if j:
f[i][j] = min(f[i][j], f[i - 1][j - 1] + x)
for j in range(n + 1):
if f[n][j] <= hoursBefore * speed:
return j
return -1
```

<!-- tabs:end -->

<!-- end -->
58 changes: 33 additions & 25 deletions solution/1800-1899/1883.Minimum Skips to Arrive at Meeting On Time/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -65,7 +65,21 @@ You can skip the first and third rest to arrive in ((7/2 + <u>0</u>) + (3/2 + 0)

## Solutions

### Solution 1
### Solution 1: Dynamic Programming

We define $f[i][j]$ as the shortest time considering the first $i$ roads and exactly skipping $j$ rest times. Initially, $f[0][0]=0$, and the rest $f[i][j]=\infty$.

Since we can choose to skip or not skip the rest time of the $i$-th road, we can list the state transition equation:

$$
f[i][j]=\min\left\{\begin{aligned} \lceil f[i-1][j]+\frac{d_i}{s}\rceil & \text{Do not skip the rest time of the $i$-th road} \\ f[i-1][j-1]+\frac{d_i}{s} & \text{Skip the rest time of the $i$-th road} \end{aligned}\right.
$$

Where $\lceil x\rceil$ represents rounding $x$ up. It should be noted that since we need to ensure that exactly $j$ rest times are skipped, we must have $j\le i$; moreover, if $j=0$, no rest time can be skipped.

Due to the possible precision error brought by floating-point operations and rounding up, we introduce a constant $eps = 10^{-8}$ to represent a very small positive real number. We subtract $eps$ before rounding the floating-point number, and finally, when comparing $f[n][j]$ and $hoursBefore$, we need to add $eps$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$, where $n$ is the number of roads.

<!-- tabs:start -->

Expand All @@ -88,6 +102,24 @@ class Solution:
return -1
```

```python
class Solution:
def minSkips(self, dist: List[int], speed: int, hoursBefore: int) -> int:
n = len(dist)
f = [[inf] * (n + 1) for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(dist, 1):
for j in range(i + 1):
if j < i:
f[i][j] = min(f[i][j], ((f[i - 1][j] + x - 1) // speed + 1) * speed)
if j:
f[i][j] = min(f[i][j], f[i - 1][j - 1] + x)
for j in range(n + 1):
if f[n][j] <= hoursBefore * speed:
return j
return -1
```

```java
class Solution {
public int minSkips(int[] dist, int speed, int hoursBefore) {
Expand Down Expand Up @@ -205,28 +237,4 @@ function minSkips(dist: number[], speed: number, hoursBefore: number): number {

<!-- tabs:end -->

### Solution 2

<!-- tabs:start -->

```python
class Solution:
def minSkips(self, dist: List[int], speed: int, hoursBefore: int) -> int:
n = len(dist)
f = [[inf] * (n + 1) for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(dist, 1):
for j in range(i + 1):
if j < i:
f[i][j] = min(f[i][j], ((f[i - 1][j] + x - 1) // speed + 1) * speed)
if j:
f[i][j] = min(f[i][j], f[i - 1][j - 1] + x)
for j in range(n + 1):
if f[n][j] <= hoursBefore * speed:
return j
return -1
```

<!-- tabs:end -->

<!-- end -->
73 changes: 69 additions & 4 deletions solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -94,6 +94,8 @@ class Solution:
t = min(budget // (k + 1), cnt[k])
ans += t * k
budget -= t * (k + 1)
if budget == 0:
break
cnt[k - 1] += cnt[k] - t
return ans
```
Expand All @@ -114,7 +116,7 @@ class Solution {
}
}
int ans = 0;
for (k = n - 1; k > 0; --k) {
for (k = n - 1; k > 0 && budget > 0; --k) {
int t = Math.min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
Expand Down Expand Up @@ -142,7 +144,7 @@ public:
}
}
int ans = 0;
for (k = n - 1; k; --k) {
for (k = n - 1; k && budget; --k) {
int t = min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
Expand All @@ -167,7 +169,7 @@ func maxPotholes(road string, budget int) (ans int) {
k = 0
}
}
for k = n - 1; k > 0; k-- {
for k = n - 1; k > 0 && budget > 0; k-- {
t := min(budget/(k+1), cnt[k])
ans += t * k
budget -= t * (k + 1)
Expand All @@ -192,7 +194,7 @@ function maxPotholes(road: string, budget: number): number {
}
}
let ans = 0;
for (k = n - 1; k; --k) {
for (k = n - 1; k && budget; --k) {
const t = Math.min(Math.floor(budget / (k + 1)), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
Expand All @@ -202,6 +204,69 @@ function maxPotholes(road: string, budget: number): number {
}
```

```rust
impl Solution {
pub fn max_potholes(road: String, budget: i32) -> i32 {
let mut cs: Vec<char> = road.chars().collect();
cs.push('.');
let n = cs.len();
let mut cnt: Vec<i32> = vec![0; n];
let mut k = 0;

for c in cs.iter() {
if *c == 'x' {
k += 1;
} else if k > 0 {
cnt[k] += 1;
k = 0;
}
}

let mut ans = 0;
let mut budget = budget;

for k in (1..n).rev() {
if budget == 0 {
break;
}
let t = std::cmp::min(budget / ((k as i32) + 1), cnt[k]);
ans += t * (k as i32);
budget -= t * ((k as i32) + 1);
cnt[k - 1] += cnt[k] - t;
}

ans
}
}
```

```cs
public class Solution {
public int MaxPotholes(string road, int budget) {
road += '.';
int n = road.Length;
int[] cnt = new int[n];
int k = 0;
foreach (char c in road) {
if (c == 'x') {
++k;
} else if (k > 0) {
++cnt[k];
k = 0;
}
}
int ans = 0;
for (k = n - 1; k > 0 && budget > 0; --k) {
int t = Math.Min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
cnt[k - 1] += cnt[k] - t;
}
return ans;
}
}
```

<!-- tabs:end -->

<!-- end -->
73 changes: 69 additions & 4 deletions solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -92,6 +92,8 @@ class Solution:
t = min(budget // (k + 1), cnt[k])
ans += t * k
budget -= t * (k + 1)
if budget == 0:
break
cnt[k - 1] += cnt[k] - t
return ans
```
Expand All @@ -112,7 +114,7 @@ class Solution {
}
}
int ans = 0;
for (k = n - 1; k > 0; --k) {
for (k = n - 1; k > 0 && budget > 0; --k) {
int t = Math.min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
Expand Down Expand Up @@ -140,7 +142,7 @@ public:
}
}
int ans = 0;
for (k = n - 1; k; --k) {
for (k = n - 1; k && budget; --k) {
int t = min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
Expand All @@ -165,7 +167,7 @@ func maxPotholes(road string, budget int) (ans int) {
k = 0
}
}
for k = n - 1; k > 0; k-- {
for k = n - 1; k > 0 && budget > 0; k-- {
t := min(budget/(k+1), cnt[k])
ans += t * k
budget -= t * (k + 1)
Expand All @@ -190,7 +192,7 @@ function maxPotholes(road: string, budget: number): number {
}
}
let ans = 0;
for (k = n - 1; k; --k) {
for (k = n - 1; k && budget; --k) {
const t = Math.min(Math.floor(budget / (k + 1)), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
Expand All @@ -200,6 +202,69 @@ function maxPotholes(road: string, budget: number): number {
}
```

```rust
impl Solution {
pub fn max_potholes(road: String, budget: i32) -> i32 {
let mut cs: Vec<char> = road.chars().collect();
cs.push('.');
let n = cs.len();
let mut cnt: Vec<i32> = vec![0; n];
let mut k = 0;

for c in cs.iter() {
if *c == 'x' {
k += 1;
} else if k > 0 {
cnt[k] += 1;
k = 0;
}
}

let mut ans = 0;
let mut budget = budget;

for k in (1..n).rev() {
if budget == 0 {
break;
}
let t = std::cmp::min(budget / ((k as i32) + 1), cnt[k]);
ans += t * (k as i32);
budget -= t * ((k as i32) + 1);
cnt[k - 1] += cnt[k] - t;
}

ans
}
}
```

```cs
public class Solution {
public int MaxPotholes(string road, int budget) {
road += '.';
int n = road.Length;
int[] cnt = new int[n];
int k = 0;
foreach (char c in road) {
if (c == 'x') {
++k;
} else if (k > 0) {
++cnt[k];
k = 0;
}
}
int ans = 0;
for (k = n - 1; k > 0 && budget > 0; --k) {
int t = Math.Min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
cnt[k - 1] += cnt[k] - t;
}
return ans;
}
}
```

<!-- tabs:end -->

<!-- end -->
2 changes: 1 addition & 1 deletion solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -14,7 +14,7 @@ class Solution {
}
}
int ans = 0;
for (k = n - 1; k; --k) {
for (k = n - 1; k && budget; --k) {
int t = min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
Expand Down
24 changes: 24 additions & 0 deletions solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/Solution.cs
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Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
public class Solution {
public int MaxPotholes(string road, int budget) {
road += '.';
int n = road.Length;
int[] cnt = new int[n];
int k = 0;
foreach (char c in road) {
if (c == 'x') {
++k;
} else if (k > 0) {
++cnt[k];
k = 0;
}
}
int ans = 0;
for (k = n - 1; k > 0 && budget > 0; --k) {
int t = Math.Min(budget / (k + 1), cnt[k]);
ans += t * k;
budget -= t * (k + 1);
cnt[k - 1] += cnt[k] - t;
}
return ans;
}
}
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