feat: add solutions to lc problems: No.2244,2589 (#2788)

Author: yanglbme

solution/2200-2299/2244.Minimum Rounds to Complete All Tasks/README.md

@@ -142,6 +142,28 @@ function minimumRounds(tasks: number[]): number {
 }
 ```
 
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn minimum_rounds(tasks: Vec<i32>) -> i32 {
+        let mut cnt = HashMap::new();
+        for &t in tasks.iter() {
+            let count = cnt.entry(t).or_insert(0);
+            *count += 1;
+        }
+        let mut ans = 0;
+        for &v in cnt.values() {
+            if v == 1 {
+                return -1;
+            }
+            ans += v / 3 + (if v % 3 == 0 { 0 } else { 1 });
+        }
+        ans
+    }
+}
+```
+
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 <!-- end -->

solution/2200-2299/2244.Minimum Rounds to Complete All Tasks/README_EN.md

@@ -42,7 +42,13 @@ It can be shown that all the tasks cannot be completed in fewer than 4 rounds, s
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Hash Table
+
+We use a hash table to count the number of tasks for each difficulty level. Then we traverse the hash table. For each difficulty level, if the number of tasks is $1$, then it is impossible to complete all tasks, so we return $-1$. Otherwise, we calculate the number of rounds needed to complete tasks of this difficulty level and add it to the answer.
+
+Finally, we return the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the `tasks` array.
 
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@@ -134,6 +140,28 @@ function minimumRounds(tasks: number[]): number {
 }
 ```
 
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn minimum_rounds(tasks: Vec<i32>) -> i32 {
+        let mut cnt = HashMap::new();
+        for &t in tasks.iter() {
+            let count = cnt.entry(t).or_insert(0);
+            *count += 1;
+        }
+        let mut ans = 0;
+        for &v in cnt.values() {
+            if v == 1 {
+                return -1;
+            }
+            ans += v / 3 + (if v % 3 == 0 { 0 } else { 1 });
+        }
+        ans
+    }
+}
+```
+
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 <!-- end -->

solution/2200-2299/2244.Minimum Rounds to Complete All Tasks/Solution.rs

@@ -0,0 +1,19 @@
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn minimum_rounds(tasks: Vec<i32>) -> i32 {
+        let mut cnt = HashMap::new();
+        for &t in tasks.iter() {
+            let count = cnt.entry(t).or_insert(0);
+            *count += 1;
+        }
+        let mut ans = 0;
+        for &v in cnt.values() {
+            if v == 1 {
+                return -1;
+            }
+            ans += v / 3 + (if v % 3 == 0 { 0 } else { 1 });
+        }
+        ans
+    }
+}

solution/2200-2299/2244.Minimum Rounds to Complete All Tasks/Soluttion.ts

@@ -58,7 +58,31 @@ There are no cornered paths in the grid that result in a product with a trailing
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Prefix Sum + Enumerate Turning Point
+
+Firstly, we need to understand that for a product, the number of trailing zeros depends on the smaller count of $2$ and $5$ in its factors. Also, each corner path should cover as many numbers as possible, so it must start from a boundary, reach a turning point, and then reach another boundary.
+
+Therefore, we can create four two-dimensional arrays $r2$, $c2$, $r5$, $c5$ to record the counts of $2$ and $5$ in each row and column. Where:
+
+-   `r2[i][j]` represents the count of $2$ from the first column to the $j$-th column in the $i$-th row;
+-   `c2[i][j]` represents the count of $2$ from the first row to the $i$-th row in the $j$-th column;
+-   `r5[i][j]` represents the count of $5$ from the first column to the $j$-th column in the $i$-th row;
+-   `c5[i][j]` represents the count of $5$ from the first row to the $i$-th row in the $j$-th column.
+
+Next, we traverse the two-dimensional array `grid`. For each number, we calculate its counts of $2$ and $5$, and then update the four two-dimensional arrays.
+
+Then, we enumerate the turning point $(i, j)$. For each turning point, we calculate four values:
+
+-   `a` represents the smaller count of $2$ and $5$ in the path that moves right from $(i, 1)$ to $(i, j)$, then turns and moves up to $(1, j)$;
+-   `b` represents the smaller count of $2$ and $5$ in the path that moves right from $(i, 1)$ to $(i, j)$, then turns and moves down to $(m, j)$;
+-   `c` represents the smaller count of $2$ and $5$ in the path that moves left from $(i, n)$ to $(i, j)$, then turns and moves up to $(1, j)$;
+-   `d` represents the smaller count of $2$ and $5$ in the path that moves left from $(i, n)$ to $(i, j)$, then turns and moves down to $(m, j)$.
+
+Each time we enumerate, we take the maximum of these four values, and then update the answer.
+
+Finally, we return the answer.
+
+The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the `grid` array, respectively.
 
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solution/2200-2299/2245.Maximum Trailing Zeros in a Cornered Path/README_EN.md

@@ -44,7 +44,15 @@ It can be proven that there is no longer path that satisfies the conditions.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Tree-shaped DP
+
+First, we construct an adjacency list $g$ based on the array $parent$, where $g[i]$ represents all child nodes of node $i$.
+
+Then we start DFS from the root node. For each node $i$, we traverse each child node $j$ in $g[i]$. If $s[i] \neq s[j]$, then we can start from node $i$, pass through node $j$, and reach a leaf node. The length of this path is $x = 1 + \text{dfs}(j)$. We use $mx$ to record the longest path length starting from node $i$. At the same time, we update the answer $ans = \max(ans, mx + x)$ during the traversal process.
+
+Finally, we return $ans + 1$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes.
 
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solution/2200-2299/2246.Longest Path With Different Adjacent Characters/README_EN.md

@@ -51,7 +51,25 @@ Note that the trip 4 -> 1 -> 0 -> 1 is not allowed because you visit th
 
 ## Solutions
 
-### Solution 1
+### Solution 1: State Compression Dynamic Programming
+
+We notice that the problem requires exactly $k$ roads to be passed, and each city can only be visited once. The number of cities is $n$, so we can pass at most $n - 1$ roads. Therefore, if $k \ge n$, we cannot meet the requirements of the problem, and we can directly return $-1$.
+
+In addition, we can also find that the number of cities $n$ does not exceed $15$, which suggests that we can consider using the method of state compression dynamic programming to solve this problem. We use a binary number of length $n$ to represent the cities that have been passed, where the $i$-th bit is $1$ indicates that the $i$-th city has been passed, and $0$ indicates that the $i$-th city has not been passed yet.
+
+We use $f[i][j]$ to represent the maximum travel cost when the cities that have been passed are $i$ and the last city passed is $j$. Initially, $f[2^i][i]=0$, and the rest $f[i][j]=-\infty$.
+
+Consider how $f[i][j]$ transitions. For $f[i]$, we enumerate all cities $j$. If the $j$-th bit of $i$ is $1$, then we can reach city $j$ from other city $h$ through the road, at this time the value of $f[i][j]$ is the maximum value of $f[i][h]+cost(h, j)$, where $cost(h, j)$ represents the travel cost from city $h$ to city $j$. Therefore, we can get the state transition equation:
+
+$$
+f[i][j]=\max_{h \in \text{city}}\{f[i \backslash j][h]+cost(h, j)\}
+$$
+
+where $i \backslash j$ represents changing the $j$-th bit of $i$ to $0$.
+
+After calculating $f[i][j]$, we judge whether the number of cities passed is $k+1$, that is, whether the number of $1$s in the binary representation of $i$ is $k+1$. If so, we update the answer as $ans = \max(ans, f[i][j])$.
+
+The time complexity is $O(2^n \times n^2)$, and the space complexity is $O(2^n \times n)$, where $n$ represents the number of cities.
 
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solution/2200-2299/2247.Maximum Cost of Trip With K Highways/README_EN.md

@@ -38,7 +38,11 @@ There does not exist any integer present both in nums[0] and nums[1], so we retu
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Counting
+
+Traverse the array `nums`. For each sub-array `arr`, count the occurrence of each number in `arr`. Then traverse the count array, count the numbers that appear as many times as the length of the array `nums`, which are the answers.
+
+The time complexity is $O(N)$, and the space complexity is $O(1000)$. Where $N$ is the total number of numbers in the array `nums`.
 
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solution/2200-2299/2248.Intersection of Multiple Arrays/README_EN.md

@@ -154,7 +154,7 @@ func findMinimumTime(tasks [][]int) (ans int) {
 ```ts
 function findMinimumTime(tasks: number[][]): number {
     tasks.sort((a, b) => a[1] - b[1]);
-    const vis = new Array(2010).fill(0);
+    const vis: number[] = Array(2010).fill(0);
     let ans = 0;
     for (let [start, end, duration] of tasks) {
         for (let i = start; i <= end; ++i) {
@@ -171,6 +171,35 @@ function findMinimumTime(tasks: number[][]): number {
 }
 ```
 
+```rust
+impl Solution {
+    pub fn find_minimum_time(tasks: Vec<Vec<i32>>) -> i32 {
+        let mut tasks = tasks;
+        tasks.sort_by(|a, b| a[1].cmp(&b[1]));
+        let mut vis = vec![0; 2010];
+        let mut ans = 0;
+
+        for task in tasks {
+            let start = task[0] as usize;
+            let end = task[1] as usize;
+            let mut duration = task[2] - vis[start..=end].iter().sum::<i32>();
+            let mut i = end;
+
+            while i >= start && duration > 0 {
+                if vis[i] == 0 {
+                    duration -= 1;
+                    vis[i] = 1;
+                    ans += 1;
+                }
+                i -= 1;
+            }
+        }
+
+        ans
+    }
+}
+```
+
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solution/2500-2599/2589.Minimum Time to Complete All Tasks/README.md

@@ -152,7 +152,7 @@ func findMinimumTime(tasks [][]int) (ans int) {
 ```ts
 function findMinimumTime(tasks: number[][]): number {
     tasks.sort((a, b) => a[1] - b[1]);
-    const vis = new Array(2010).fill(0);
+    const vis: number[] = Array(2010).fill(0);
     let ans = 0;
     for (let [start, end, duration] of tasks) {
         for (let i = start; i <= end; ++i) {
@@ -169,6 +169,35 @@ function findMinimumTime(tasks: number[][]): number {
 }
 ```
 
+```rust
+impl Solution {
+    pub fn find_minimum_time(tasks: Vec<Vec<i32>>) -> i32 {
+        let mut tasks = tasks;
+        tasks.sort_by(|a, b| a[1].cmp(&b[1]));
+        let mut vis = vec![0; 2010];
+        let mut ans = 0;
+
+        for task in tasks {
+            let start = task[0] as usize;
+            let end = task[1] as usize;
+            let mut duration = task[2] - vis[start..=end].iter().sum::<i32>();
+            let mut i = end;
+
+            while i >= start && duration > 0 {
+                if vis[i] == 0 {
+                    duration -= 1;
+                    vis[i] = 1;
+                    ans += 1;
+                }
+                i -= 1;
+            }
+        }
+
+        ans
+    }
+}
+```
+
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solution/2500-2599/2589.Minimum Time to Complete All Tasks/README_EN.md

@@ -0,0 +1,26 @@
+impl Solution {
+    pub fn find_minimum_time(tasks: Vec<Vec<i32>>) -> i32 {
+        let mut tasks = tasks;
+        tasks.sort_by(|a, b| a[1].cmp(&b[1]));
+        let mut vis = vec![0; 2010];
+        let mut ans = 0;
+
+        for task in tasks {
+            let start = task[0] as usize;
+            let end = task[1] as usize;
+            let mut duration = task[2] - vis[start..=end].iter().sum::<i32>();
+            let mut i = end;
+
+            while i >= start && duration > 0 {
+                if vis[i] == 0 {
+                    duration -= 1;
+                    vis[i] = 1;
+                    ans += 1;
+                }
+                i -= 1;
+            }
+        }
+
+        ans
+    }
+}

solution/2500-2599/2589.Minimum Time to Complete All Tasks/Solution.rs

@@ -1,6 +1,6 @@
 function findMinimumTime(tasks: number[][]): number {
     tasks.sort((a, b) => a[1] - b[1]);
-    const vis = new Array(2010).fill(0);
+    const vis: number[] = Array(2010).fill(0);
     let ans = 0;
     for (let [start, end, duration] of tasks) {
         for (let i = start; i <= end; ++i) {