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

* No.1883.Minimum Skips to Arrive at Meeting On Time
* No.3119.Maximum Number of Potholes That Can Be Fixed

Author: yanglbme

solution/1800-1899/1883.Minimum Skips to Arrive at Meeting On Time/README.md

@@ -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) {
@@ -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 -->

solution/1800-1899/1883.Minimum Skips to Arrive at Meeting On Time/README_EN.md

@@ -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 -->
 
@@ -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) {
@@ -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 -->

solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/README.md

@@ -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
 ```
@@ -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);
@@ -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);
@@ -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)
@@ -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);
@@ -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 -->

solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/README_EN.md

@@ -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
 ```
@@ -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);
@@ -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);
@@ -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)
@@ -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);
@@ -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 -->

solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/Solution.cpp

@@ -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);

solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/Solution.cs

@@ -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;
+    }
+}