feat: update solutions to lc problems: No.2678~2683 (#3121)

Author: yanglbme

solution/2600-2699/2678.Number of Senior Citizens/README.md

@@ -135,52 +135,7 @@ func countSeniors(details []string) (ans int) {
 
 ```ts
 function countSeniors(details: string[]): number {
-    let ans = 0;
-    for (const x of details) {
-        const age = parseInt(x.slice(11, 13));
-        if (age > 60) {
-            ++ans;
-        }
-    }
-    return ans;
-}
-```
-
-#### Rust
-
-```rust
-impl Solution {
-    pub fn count_seniors(details: Vec<String>) -> i32 {
-        let mut ans = 0;
-
-        for s in details.iter() {
-            if let Ok(age) = s[11..13].parse::<i32>() {
-                if age > 60 {
-                    ans += 1;
-                }
-            }
-        }
-
-        ans
-    }
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### TypeScript
-
-```ts
-function countSeniors(details: string[]): number {
-    return details.filter(v => parseInt(v.slice(11, 13)) > 60).length;
+    return details.filter(x => +x.slice(11, 13) > 60).length;
 }
 ```
 

solution/2600-2699/2678.Number of Senior Citizens/README_EN.md

@@ -133,52 +133,7 @@ func countSeniors(details []string) (ans int) {
 
 ```ts
 function countSeniors(details: string[]): number {
-    let ans = 0;
-    for (const x of details) {
-        const age = parseInt(x.slice(11, 13));
-        if (age > 60) {
-            ++ans;
-        }
-    }
-    return ans;
-}
-```
-
-#### Rust
-
-```rust
-impl Solution {
-    pub fn count_seniors(details: Vec<String>) -> i32 {
-        let mut ans = 0;
-
-        for s in details.iter() {
-            if let Ok(age) = s[11..13].parse::<i32>() {
-                if age > 60 {
-                    ans += 1;
-                }
-            }
-        }
-
-        ans
-    }
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### TypeScript
-
-```ts
-function countSeniors(details: string[]): number {
-    return details.filter(v => parseInt(v.slice(11, 13)) > 60).length;
+    return details.filter(x => +x.slice(11, 13) > 60).length;
 }
 ```
 

solution/2600-2699/2678.Number of Senior Citizens/Solution.rs

@@ -1,15 +1,9 @@
 impl Solution {
     pub fn count_seniors(details: Vec<String>) -> i32 {
-        let mut ans = 0;
-
-        for s in details.iter() {
-            if let Ok(age) = s[11..13].parse::<i32>() {
-                if age > 60 {
-                    ans += 1;
-                }
-            }
-        }
-
-        ans
+        details
+            .iter()
+            .filter_map(|s| s[11..13].parse::<i32>().ok())
+            .filter(|&age| age > 60)
+            .count() as i32
     }
 }

solution/2600-2699/2678.Number of Senior Citizens/Solution.ts

@@ -1,10 +1,3 @@
 function countSeniors(details: string[]): number {
-    let ans = 0;
-    for (const x of details) {
-        const age = parseInt(x.slice(11, 13));
-        if (age > 60) {
-            ++ans;
-        }
-    }
-    return ans;
+    return details.filter(x => +x.slice(11, 13) > 60).length;
 }

solution/2600-2699/2678.Number of Senior Citizens/Solution2.rs

@@ -168,22 +168,18 @@ function matrixSum(nums: number[][]): number {
 
 ```rust
 impl Solution {
-    pub fn matrix_sum(nums: Vec<Vec<i32>>) -> i32 {
-        let mut nums = nums;
-        for row in nums.iter_mut() {
+    pub fn matrix_sum(mut nums: Vec<Vec<i32>>) -> i32 {
+        for row in &mut nums {
             row.sort();
         }
-        let transposed: Vec<Vec<i32>> = (0..nums[0].len())
-            .map(|i| {
-                nums.iter()
-                    .map(|row| row[i])
-                    .collect()
-            })
-            .collect();
-
-        transposed
-            .iter()
-            .map(|row| row.iter().max().unwrap())
+        (0..nums[0].len())
+            .map(|col|
+                nums
+                    .iter()
+                    .map(|row| row[col])
+                    .max()
+                    .unwrap()
+            )
             .sum()
     }
 }
@@ -193,38 +189,4 @@ impl Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Rust
-
-```rust
-impl Solution {
-    pub fn matrix_sum(nums: Vec<Vec<i32>>) -> i32 {
-        let mut nums = nums.clone();
-        for row in nums.iter_mut() {
-            row.sort();
-        }
-
-        let mut ans = 0;
-        for j in 0..nums[0].len() {
-            let mut mx = 0;
-            for row in &nums {
-                mx = mx.max(row[j]);
-            }
-            ans += mx;
-        }
-
-        ans
-    }
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2600-2699/2678.Number of Senior Citizens/Solution2.ts

@@ -159,22 +159,18 @@ function matrixSum(nums: number[][]): number {
 
 ```rust
 impl Solution {
-    pub fn matrix_sum(nums: Vec<Vec<i32>>) -> i32 {
-        let mut nums = nums;
-        for row in nums.iter_mut() {
+    pub fn matrix_sum(mut nums: Vec<Vec<i32>>) -> i32 {
+        for row in &mut nums {
             row.sort();
         }
-        let transposed: Vec<Vec<i32>> = (0..nums[0].len())
-            .map(|i| {
-                nums.iter()
-                    .map(|row| row[i])
-                    .collect()
-            })
-            .collect();
-
-        transposed
-            .iter()
-            .map(|row| row.iter().max().unwrap())
+        (0..nums[0].len())
+            .map(|col|
+                nums
+                    .iter()
+                    .map(|row| row[col])
+                    .max()
+                    .unwrap()
+            )
             .sum()
     }
 }
@@ -184,38 +180,4 @@ impl Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Rust
-
-```rust
-impl Solution {
-    pub fn matrix_sum(nums: Vec<Vec<i32>>) -> i32 {
-        let mut nums = nums.clone();
-        for row in nums.iter_mut() {
-            row.sort();
-        }
-
-        let mut ans = 0;
-        for j in 0..nums[0].len() {
-            let mut mx = 0;
-            for row in &nums {
-                mx = mx.max(row[j]);
-            }
-            ans += mx;
-        }
-
-        ans
-    }
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2600-2699/2679.Sum in a Matrix/README.md

@@ -1,20 +1,16 @@
 impl Solution {
-    pub fn matrix_sum(nums: Vec<Vec<i32>>) -> i32 {
-        let mut nums = nums;
-        for row in nums.iter_mut() {
+    pub fn matrix_sum(mut nums: Vec<Vec<i32>>) -> i32 {
+        for row in &mut nums {
             row.sort();
         }
-        let transposed: Vec<Vec<i32>> = (0..nums[0].len())
-            .map(|i| {
-                nums.iter()
-                    .map(|row| row[i])
-                    .collect()
-            })
-            .collect();
-
-        transposed
-            .iter()
-            .map(|row| row.iter().max().unwrap())
+        (0..nums[0].len())
+            .map(|col|
+                nums
+                    .iter()
+                    .map(|row| row[col])
+                    .max()
+                    .unwrap()
+            )
             .sum()
     }
 }

solution/2600-2699/2679.Sum in a Matrix/README_EN.md

@@ -70,7 +70,23 @@ The sum of powers of all groups is 8 + 1 + 64 + 4 + 32 + 16 + 16 = 141.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Sorting + Mathematics
+
+We notice that the problem involves the maximum and minimum values of a subsequence, and the order of elements in the array does not affect the final result. Therefore, we can sort the array first.
+
+Next, we enumerate each element as the minimum value of the subsequence. Let's denote each element of the array as $a_1, a_2, \cdots, a_n$. The contribution of the subsequence with $a_i$ as the minimum value is:
+
+$$
+a_i \times (a_{i}^{2} + a_{i+1}^2 + 2 \times a_{i+2}^2 + 4 \times a_{i+3}^2 + \cdots + 2^{n-i-1} \times a_n^2)
+$$
+
+We notice that each $a_i$ will be multiplied by $a_i^2$, which we can directly add to the answer. For the remaining part, we can maintain it with a variable $p$, initially set to $0$.
+
+Then, we enumerate $a_i$ from right to left. Each time, we add $a_i \times p$ to the answer, and then set $p = p \times 2 + a_i^2$.
+
+After enumerating all $a_i$, return the answer.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.
 
 <!-- tabs:start -->
 

solution/2600-2699/2679.Sum in a Matrix/Solution.rs

@@ -76,7 +76,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+We use an array `vis` to record whether each friend has received the ball, initially, all friends have not received the ball. Then, we simulate the game process according to the rules described in the problem statement until a friend receives the ball for the second time.
+
+In the simulation process, we use two variables $i$ and $p$ to represent the current friend holding the ball and the current passing step length, respectively. Initially, $i=0, p=1$, indicating the first friend receives the ball. Each time the ball is passed, we update $i$ to $(i+p \times k) \bmod n$, representing the next friend's number to receive the ball, and then update $p$ to $p+1$, representing the step length for the next pass. The game ends when a friend receives the ball for the second time.
+
+Finally, we iterate through the array `vis` and add the numbers of friends who have not received the ball to the answer array.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of friends.
 
 <!-- tabs:start -->