feat: update lc problems

solution/0600-0699/0624.Maximum Distance in Arrays/README.md

@@ -62,11 +62,13 @@ tags:
 
 ### 方法一：维护最大值和最小值
 
-我们注意到，最大距离一定是两个数组中的一个最大值和另一个最小值之间的距离。因此，我们可以维护两个变量，分别表示当前数组中的最大值和最小值，然后遍历数组，更新最大距离，同时更新最大值和最小值。
+我们注意到，最大距离一定是两个数组中的一个最大值和另一个最小值之间的距离。因此，我们可以维护两个变量 $\textit{mi}$ 和 $\textit{mx}$，分别表示已经遍历过的数组中的最小值和最大值。初始时 $\textit{mi}$ 和 $\textit{mx}$ 分别为第一个数组的第一个元素和最后一个元素。
+
+接下来，我们从第二个数组开始遍历，对于每个数组，我们首先计算当前数组的第一个元素和 $\textit{mx}$ 之间的距离，以及当前数组的最后一个元素和 $\textit{mi}$ 之间的距离，然后更新最大距离。同时，我们更新 $\textit{mi}$ 和 $\textit{mx}$ 为当前数组的第一个元素和最后一个元素。
 
 遍历结束后，即可得到最大距离。
 
-时间复杂度 $O(m)$，空间复杂度 $O(1)$。其中 $m$ 为数组的个数。
+时间复杂度 $O(m)$，其中 $m$ 为数组的个数。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -152,66 +154,42 @@ func abs(x int) int {
 
 ```ts
 function maxDistance(arrays: number[][]): number {
-    const n = arrays.length;
-    let res = 0;
-    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
-
-    for (let i = 0; i < n; i++) {
-        const a = arrays[i];
-        res = Math.max(Math.max(a.at(-1)! - min, max - a[0]), res);
-        min = Math.min(min, a[0]);
-        max = Math.max(max, a.at(-1)!);
+    let ans = 0;
+    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)!];
+    for (let i = 1; i < arrays.length; ++i) {
+        const arr = arrays[i];
+        const a = Math.abs(arr[0] - mx);
+        const b = Math.abs(arr.at(-1)! - mi);
+        ans = Math.max(ans, a, b);
+        mi = Math.min(mi, arr[0]);
+        mx = Math.max(mx, arr.at(-1)!);
     }
-
-    return res;
+    return ans;
 }
 ```
 
-#### JavaScript
+#### Rust
 
-```js
-/**
- * @param {number[][]} arrays
- * @return {number}
- */
-var maxDistance = function (arrays) {
-    const n = arrays.length;
-    let res = 0;
-    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
-
-    for (let i = 0; i < n; i++) {
-        const a = arrays[i];
-        res = Math.max(Math.max(a.at(-1) - min, max - a[0]), res);
-        min = Math.min(min, a[0]);
-        max = Math.max(max, a.at(-1));
-    }
+```rust
+impl Solution {
+    pub fn max_distance(arrays: Vec<Vec<i32>>) -> i32 {
+        let mut ans = 0;
+        let mut mi = arrays[0][0];
+        let mut mx = arrays[0][arrays[0].len() - 1];
 
-    return res;
-};
-```
+        for i in 1..arrays.len() {
+            let arr = &arrays[i];
+            let a = (arr[0] - mx).abs();
+            let b = (arr[arr.len() - 1] - mi).abs();
+            ans = ans.max(a).max(b);
 
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
-
-### 方法二：一行
-
-<!-- tabs:start -->
-
-#### TypeScript
+            mi = mi.min(arr[0]);
+            mx = mx.max(arr[arr.len() - 1]);
+        }
 
-```ts
-const maxDistance = (arrays: number[][]): number =>
-    arrays.reduce(
-        ([res, min, max], a) => [
-            Math.max(Math.max(a.at(-1)! - min, max - a[0]), res),
-            Math.min(min, a[0]),
-            Math.max(max, a.at(-1)!),
-        ],
-        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
-    )[0];
+        ans
+    }
+}
 ```
 
 #### JavaScript
@@ -221,15 +199,19 @@ const maxDistance = (arrays: number[][]): number =>
  * @param {number[][]} arrays
  * @return {number}
  */
-var maxDistance = arrays =>
-    arrays.reduce(
-        ([res, min, max], a) => [
-            Math.max(Math.max(a.at(-1) - min, max - a[0]), res),
-            Math.min(min, a[0]),
-            Math.max(max, a.at(-1)),
-        ],
-        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
-    )[0];
+var maxDistance = function (arrays) {
+    let ans = 0;
+    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)];
+    for (let i = 1; i < arrays.length; ++i) {
+        const arr = arrays[i];
+        const a = Math.abs(arr[0] - mx);
+        const b = Math.abs(arr.at(-1) - mi);
+        ans = Math.max(ans, a, b);
+        mi = Math.min(mi, arr[0]);
+        mx = Math.max(mx, arr.at(-1));
+    }
+    return ans;
+};
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0624.Maximum Distance in Arrays/README_EN.md

@@ -57,7 +57,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Maintain Maximum and Minimum Values
+
+We notice that the maximum distance must be the distance between the maximum value in one array and the minimum value in another array. Therefore, we can maintain two variables $\textit{mi}$ and $\textit{mx}$, representing the minimum and maximum values of the arrays we have traversed. Initially, $\textit{mi}$ and $\textit{mx}$ are the first and last elements of the first array, respectively.
+
+Next, we traverse from the second array. For each array, we first calculate the distance between the first element of the current array and $\textit{mx}$, and the distance between the last element of the current array and $\textit{mi}$. Then, we update the maximum distance. At the same time, we update $\textit{mi}$ and $\textit{mx}$ to be the first and last elements of the current array.
+
+After traversing all arrays, we get the maximum distance.
+
+The time complexity is $O(m)$, where $m$ is the number of arrays. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -143,66 +151,42 @@ func abs(x int) int {
 
 ```ts
 function maxDistance(arrays: number[][]): number {
-    const n = arrays.length;
-    let res = 0;
-    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
-
-    for (let i = 0; i < n; i++) {
-        const a = arrays[i];
-        res = Math.max(Math.max(a.at(-1)! - min, max - a[0]), res);
-        min = Math.min(min, a[0]);
-        max = Math.max(max, a.at(-1)!);
+    let ans = 0;
+    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)!];
+    for (let i = 1; i < arrays.length; ++i) {
+        const arr = arrays[i];
+        const a = Math.abs(arr[0] - mx);
+        const b = Math.abs(arr.at(-1)! - mi);
+        ans = Math.max(ans, a, b);
+        mi = Math.min(mi, arr[0]);
+        mx = Math.max(mx, arr.at(-1)!);
     }
-
-    return res;
+    return ans;
 }
 ```
 
-#### JavaScript
-
-```js
-/**
- * @param {number[][]} arrays
- * @return {number}
- */
-var maxDistance = function (arrays) {
-    const n = arrays.length;
-    let res = 0;
-    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
-
-    for (let i = 0; i < n; i++) {
-        const a = arrays[i];
-        res = Math.max(Math.max(a.at(-1) - min, max - a[0]), res);
-        min = Math.min(min, a[0]);
-        max = Math.max(max, a.at(-1));
-    }
-
-    return res;
-};
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
+#### Rust
 
-### Solution 2: One-line solution
+```rust
+impl Solution {
+    pub fn max_distance(arrays: Vec<Vec<i32>>) -> i32 {
+        let mut ans = 0;
+        let mut mi = arrays[0][0];
+        let mut mx = arrays[0][arrays[0].len() - 1];
 
-<!-- tabs:start -->
+        for i in 1..arrays.len() {
+            let arr = &arrays[i];
+            let a = (arr[0] - mx).abs();
+            let b = (arr[arr.len() - 1] - mi).abs();
+            ans = ans.max(a).max(b);
 
-#### TypeScript
+            mi = mi.min(arr[0]);
+            mx = mx.max(arr[arr.len() - 1]);
+        }
 
-```ts
-const maxDistance = (arrays: number[][]): number =>
-    arrays.reduce(
-        ([res, min, max], a) => [
-            Math.max(Math.max(a.at(-1)! - min, max - a[0]), res),
-            Math.min(min, a[0]),
-            Math.max(max, a.at(-1)!),
-        ],
-        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
-    )[0];
+        ans
+    }
+}
 ```
 
 #### JavaScript
@@ -212,15 +196,19 @@ const maxDistance = (arrays: number[][]): number =>
  * @param {number[][]} arrays
  * @return {number}
  */
-var maxDistance = arrays =>
-    arrays.reduce(
-        ([res, min, max], a) => [
-            Math.max(Math.max(a.at(-1) - min, max - a[0]), res),
-            Math.min(min, a[0]),
-            Math.max(max, a.at(-1)),
-        ],
-        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
-    )[0];
+var maxDistance = function (arrays) {
+    let ans = 0;
+    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)];
+    for (let i = 1; i < arrays.length; ++i) {
+        const arr = arrays[i];
+        const a = Math.abs(arr[0] - mx);
+        const b = Math.abs(arr.at(-1) - mi);
+        ans = Math.max(ans, a, b);
+        mi = Math.min(mi, arr[0]);
+        mx = Math.max(mx, arr.at(-1));
+    }
+    return ans;
+};
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0624.Maximum Distance in Arrays/Solution.js

@@ -3,16 +3,15 @@
  * @return {number}
  */
 var maxDistance = function (arrays) {
-    const n = arrays.length;
-    let res = 0;
-    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
-
-    for (let i = 0; i < n; i++) {
-        const a = arrays[i];
-        res = Math.max(Math.max(a.at(-1) - min, max - a[0]), res);
-        min = Math.min(min, a[0]);
-        max = Math.max(max, a.at(-1));
+    let ans = 0;
+    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)];
+    for (let i = 1; i < arrays.length; ++i) {
+        const arr = arrays[i];
+        const a = Math.abs(arr[0] - mx);
+        const b = Math.abs(arr.at(-1) - mi);
+        ans = Math.max(ans, a, b);
+        mi = Math.min(mi, arr[0]);
+        mx = Math.max(mx, arr.at(-1));
     }
-
-    return res;
+    return ans;
 };

solution/0600-0699/0624.Maximum Distance in Arrays/Solution.rs

@@ -0,0 +1,19 @@
+impl Solution {
+    pub fn max_distance(arrays: Vec<Vec<i32>>) -> i32 {
+        let mut ans = 0;
+        let mut mi = arrays[0][0];
+        let mut mx = arrays[0][arrays[0].len() - 1];
+
+        for i in 1..arrays.len() {
+            let arr = &arrays[i];
+            let a = (arr[0] - mx).abs();
+            let b = (arr[arr.len() - 1] - mi).abs();
+            ans = ans.max(a).max(b);
+
+            mi = mi.min(arr[0]);
+            mx = mx.max(arr[arr.len() - 1]);
+        }
+
+        ans
+    }
+}

solution/0600-0699/0624.Maximum Distance in Arrays/Solution.ts

@@ -1,14 +1,13 @@
 function maxDistance(arrays: number[][]): number {
-    const n = arrays.length;
-    let res = 0;
-    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
-
-    for (let i = 0; i < n; i++) {
-        const a = arrays[i];
-        res = Math.max(Math.max(a.at(-1)! - min, max - a[0]), res);
-        min = Math.min(min, a[0]);
-        max = Math.max(max, a.at(-1)!);
+    let ans = 0;
+    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)!];
+    for (let i = 1; i < arrays.length; ++i) {
+        const arr = arrays[i];
+        const a = Math.abs(arr[0] - mx);
+        const b = Math.abs(arr.at(-1)! - mi);
+        ans = Math.max(ans, a, b);
+        mi = Math.min(mi, arr[0]);
+        mx = Math.max(mx, arr.at(-1)!);
     }
-
-    return res;
+    return ans;
 }