diff --git a/solution/0500-0599/0598.Range Addition II/README.md b/solution/0500-0599/0598.Range Addition II/README.md
index 4031844398e52..44fd37fcb568b 100644
--- a/solution/0500-0599/0598.Range Addition II/README.md	
+++ b/solution/0500-0599/0598.Range Addition II/README.md	
@@ -67,7 +67,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：脑筋急转弯
+
+我们注意到，所有操作子矩阵的交集就是最终的最大整数所在的子矩阵，并且每个操作子矩阵都是从左上角 $(0, 0)$ 开始的，因此，我们遍历所有操作子矩阵，求出行数和列数的最小值，最后返回这两个值的乘积即可。
+
+注意，如果操作数组为空，那么矩阵中的最大整数个数就是 $m \times n$。
+
+时间复杂度 $O(k)$，其中 $k$ 是操作数组 $\textit{ops}$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -123,6 +129,50 @@ func maxCount(m int, n int, ops [][]int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxCount(m: number, n: number, ops: number[][]): number {
+    for (const [a, b] of ops) {
+        m = Math.min(m, a);
+        n = Math.min(n, b);
+    }
+    return m * n;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn max_count(mut m: i32, mut n: i32, ops: Vec<Vec<i32>>) -> i32 {
+        for op in ops {
+            m = m.min(op[0]);
+            n = n.min(op[1]);
+        }
+        m * n
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number} m
+ * @param {number} n
+ * @param {number[][]} ops
+ * @return {number}
+ */
+var maxCount = function (m, n, ops) {
+    for (const [a, b] of ops) {
+        m = Math.min(m, a);
+        n = Math.min(n, b);
+    }
+    return m * n;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/0500-0599/0598.Range Addition II/README_EN.md b/solution/0500-0599/0598.Range Addition II/README_EN.md
index fd5e5237b9517..7ea839d741336 100644
--- a/solution/0500-0599/0598.Range Addition II/README_EN.md	
+++ b/solution/0500-0599/0598.Range Addition II/README_EN.md	
@@ -61,7 +61,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Brain Teaser
+
+We notice that the intersection of all operation submatrices is the submatrix where the final maximum integer is located, and each operation submatrix starts from the top-left corner $(0, 0)$. Therefore, we traverse all operation submatrices to find the minimum number of rows and columns. Finally, we return the product of these two values.
+
+Note that if the operation array is empty, the number of maximum integers in the matrix is $m \times n$.
+
+The time complexity is $O(k)$, where $k$ is the length of the operation array $\textit{ops}$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -96,7 +102,7 @@ class Solution {
 class Solution {
 public:
     int maxCount(int m, int n, vector<vector<int>>& ops) {
-        for (auto op : ops) {
+        for (const auto& op : ops) {
             m = min(m, op[0]);
             n = min(n, op[1]);
         }
@@ -117,6 +123,50 @@ func maxCount(m int, n int, ops [][]int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxCount(m: number, n: number, ops: number[][]): number {
+    for (const [a, b] of ops) {
+        m = Math.min(m, a);
+        n = Math.min(n, b);
+    }
+    return m * n;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn max_count(mut m: i32, mut n: i32, ops: Vec<Vec<i32>>) -> i32 {
+        for op in ops {
+            m = m.min(op[0]);
+            n = n.min(op[1]);
+        }
+        m * n
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number} m
+ * @param {number} n
+ * @param {number[][]} ops
+ * @return {number}
+ */
+var maxCount = function (m, n, ops) {
+    for (const [a, b] of ops) {
+        m = Math.min(m, a);
+        n = Math.min(n, b);
+    }
+    return m * n;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/0500-0599/0598.Range Addition II/Solution.cpp b/solution/0500-0599/0598.Range Addition II/Solution.cpp
index e78047cff9b35..4ed70e314d9ba 100644
--- a/solution/0500-0599/0598.Range Addition II/Solution.cpp	
+++ b/solution/0500-0599/0598.Range Addition II/Solution.cpp	
@@ -1,10 +1,10 @@
 class Solution {
 public:
     int maxCount(int m, int n, vector<vector<int>>& ops) {
-        for (auto op : ops) {
+        for (const auto& op : ops) {
             m = min(m, op[0]);
             n = min(n, op[1]);
         }
         return m * n;
     }
-};
\ No newline at end of file
+};
diff --git a/solution/0500-0599/0598.Range Addition II/Solution.js b/solution/0500-0599/0598.Range Addition II/Solution.js
new file mode 100644
index 0000000000000..cbb4cb3a71203
--- /dev/null
+++ b/solution/0500-0599/0598.Range Addition II/Solution.js	
@@ -0,0 +1,13 @@
+/**
+ * @param {number} m
+ * @param {number} n
+ * @param {number[][]} ops
+ * @return {number}
+ */
+var maxCount = function (m, n, ops) {
+    for (const [a, b] of ops) {
+        m = Math.min(m, a);
+        n = Math.min(n, b);
+    }
+    return m * n;
+};
diff --git a/solution/0500-0599/0598.Range Addition II/Solution.rs b/solution/0500-0599/0598.Range Addition II/Solution.rs
new file mode 100644
index 0000000000000..152e6066f7dcc
--- /dev/null
+++ b/solution/0500-0599/0598.Range Addition II/Solution.rs	
@@ -0,0 +1,9 @@
+impl Solution {
+    pub fn max_count(mut m: i32, mut n: i32, ops: Vec<Vec<i32>>) -> i32 {
+        for op in ops {
+            m = m.min(op[0]);
+            n = n.min(op[1]);
+        }
+        m * n
+    }
+}
diff --git a/solution/0500-0599/0598.Range Addition II/Solution.ts b/solution/0500-0599/0598.Range Addition II/Solution.ts
new file mode 100644
index 0000000000000..3fe338236f681
--- /dev/null
+++ b/solution/0500-0599/0598.Range Addition II/Solution.ts	
@@ -0,0 +1,7 @@
+function maxCount(m: number, n: number, ops: number[][]): number {
+    for (const [a, b] of ops) {
+        m = Math.min(m, a);
+        n = Math.min(n, b);
+    }
+    return m * n;
+}