diff --git a/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md b/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md
index e7a831416..ec1d771d6 100644
--- a/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md
+++ b/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md
@@ -1,324 +1,323 @@
----
-id: reverse-pairs
-title: Reverse Pairs
-sidebar_label: 493-Reverse-Pairs
-tags:
-  - Array
-  - Binary Search
-  - Divide and conquer
-  - Binary Indexed Tree
-  - Segment Tree
-  - Merge Sort
-  - Ordered Set
-description: The problem is to reverse the pairs.
-sidebar_position: 2667
----
-
-## Problem Statement 
-
-### Problem Description
-
-Given an integer array `nums`, return the number of reverse pairs in the array.
-
-A reverse pair is a pair `(i, j)` where:
-
-`0 <= i < j < nums.length` and
-`nums[i] > 2 * nums[j]`.
-
-### Examples
-
-**Example 1:**
-
-```
-Input: nums = [1,3,2,3,1]
-Output: 2
-Explanation: The reverse pairs are:
-(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1
-(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1
-```
-
-**Example 2:**
-
-```
-Input: nums = [2,4,3,5,1]
-Output: 3
-Explanation: The reverse pairs are:
-(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1
-(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1
-(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1
-```
-
-### Constraints
-
-- `1 <= nums.length <= 5 * 10^4`
-
-## Solution of Given Problem
-
-### Intuition and Approach
-
-The intuition behind this solution is to use a modified Merge Sort algorithm to count the number of reverse pairs in the array.
-
-
-### Approaches
-
-- Divide the array into smaller subarrays until each subarray has only one element.
-- Merge the subarrays back together, counting the number of reverse pairs between each pair of subarrays.
-- The merge step is done in a way that ensures the count of reverse pairs is accurate.
-
-
-#### Codes in Different Languages
-
-<Tabs>
-  <TabItem value="JavaScript" label="JavaScript" default>
-  <SolutionAuthor name="@Ishitamukherjee2004"/>
-   ```javascript
-    function mergeSortedArrays(nums, left, mid, right, temp) {
-  let i = left, j = mid + 1, k = 0;
-  while (i <= mid && j <= right) {
-    if (nums[i] <= nums[j]) temp[k++] = nums[i++];
-    else temp[k++] = nums[j++];
-  }
-  while (i <= mid) temp[k++] = nums[i++];
-  while (j <= right) temp[k++] = nums[j++];
-}
-
-function merge(nums, left, mid, right) {
-  let count = 0;
-  let j = mid + 1;
-  for (let i = left; i <= mid; i++) {
-    while (j <= right && nums[i] > 2 * nums[j]) j++;
-    count += j - mid - 1;
-  }
-  let temp = new Array(right - left + 1);
-  mergeSortedArrays(nums, left, mid, right, temp);
-  for (let i = left; i <= right; i++) nums[i] = temp[i - left];
-  return count;
-}
-function mergeSort(nums, left, right) {
-  if (left >= right) return 0;
-  let mid = left + (right - left) / 2;
-  let count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
-  count += merge(nums, left, mid, right);
-  return count;
-}
-
-class Solution {
-  reversePairs(nums) {
-    return mergeSort(nums, 0, nums.length - 1);
-  }
-}
-
-
-    ```
-
-  </TabItem>
-  <TabItem value="TypeScript" label="TypeScript">
-  <SolutionAuthor name="@Ishitamukherjee2004"/> 
-   ```typescript
-    function mergeSortedArrays(nums: number[], left: number, mid: number, right: number, temp: number[]) {
-  let i = left, j = mid + 1, k = 0;
-  while (i <= mid && j <= right) {
-    if (nums[i] <= nums[j]) temp[k++] = nums[i++];
-    else temp[k++] = nums[j++];
-  }
-  while (i <= mid) temp[k++] = nums[i++];
-  while (j <= right) temp[k++] = nums[j++];
-}
-
-function merge(nums: number[], left: number, mid: number, right: number) {
-  let count = 0;
-  let j = mid + 1;
-  for (let i = left; i <= mid; i++) {
-    while (j <= right && nums[i] > 2 * nums[j]) j++;
-    count += j - mid - 1;
-  }
-  let temp = new Array(right - left + 1);
-  mergeSortedArrays(nums, left, mid, right, temp);
-  for (let i = left; i <= right; i++) nums[i] = temp[i - left];
-  return count;
-}
-function mergeSort(nums: number[], left: number, right: number) {
-  if (left >= right) return 0;
-  let mid = left + (right - left) / 2;
-  let count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
-  count += merge(nums, left, mid, right);
-  return count;
-}
-
-class Solution {
-  reversePairs(nums: number[]) {
-    return mergeSort(nums, 0, nums.length - 1);
-  }
-}
-
-    ```
-
-  </TabItem>
-  <TabItem value="Python" label="Python"> 
-  <SolutionAuthor name="@Ishitamukherjee2004"/>
-   ```python
-    def merge_sorted_arrays(nums, left, mid, right, temp):
-    i, j, k = left, mid + 1, 0
-    while i <= mid and j <= right:
-        if nums[i] <= nums[j]:
-            temp[k] = nums[i]
-            i += 1
-        else:
-            temp[k] = nums[j]
-            j += 1
-        k += 1
-    while i <= mid:
-        temp[k] = nums[i]
-        i += 1
-        k += 1
-    while j <= right:
-        temp[k] = nums[j]
-        j += 1
-        k += 1
-
-def merge(nums, left, mid, right):
-    count = 0
-    j = mid + 1
-    for i in range(left, mid + 1):
-        while j <= right and nums[i] > 2 * nums[j]:
-            j += 1
-        count += j - mid - 1
-    temp = [0] * (right - left + 1)
-    merge_sorted_arrays(nums, left, mid, right, temp)
-    for i in range(left, right + 1):
-        nums[i] = temp[i - left]
-    return count
-
-def merge_sort(nums, left, right):
-    if left >= right:
-        return 0
-    mid = left + (right - left) // 2
-    count = merge_sort(nums, left, mid) + merge_sort(nums, mid + 1, right)
-    count += merge(nums, left, mid, right)
-    return count
-
-class Solution:
-    def reversePairs(self, nums):
-        return merge_sort(nums, 0, len(nums) - 1)
-
-    ```
-
-  </TabItem>
-  <TabItem value="Java" label="Java">
-  <SolutionAuthor name="@Ishitamukherjee2004"/>
-   ```java
-    public class Solution {
-    public int reversePairs(int[] nums) {
-        return mergeSort(nums, 0, nums.length - 1);
-    }
-
-    public int mergeSort(int[] nums, int left, int right) {
-        if (left >= right) {
-            return 0;
-        }
-        int mid = left + (right - left) / 2;
-        int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
-        count += merge(nums, left, mid, right);
-        return count;
-    }
-     public int merge(int[] nums, int left, int mid, int right) {
-        int count = 0;
-        int j = mid + 1;
-        for (int i = left; i <= mid; i++) {
-            while (j <= right && nums[i] > 2 * nums[j]) {
-                j++;
-            }
-            count += j - mid - 1;
-        }
-        int[] temp = new int[right - left + 1];
-        mergeSortedArrays(nums, left, mid, right, temp);
-        for (int i = left; i <= right; i++) {
-            nums[i] = temp[i - left];
-        }
-        return count;
-    }
-
-    public void
-    mergeSortedArrays(int[] nums, int left, int mid, int right, int[] temp) {
-        int i = left, j = mid + 1, k = 0;
-        while (i <= mid && j <= right) {
-            if (nums[i] <= nums[j]) {
-                temp[k++] = nums[i++];
-            } else {
-                temp[k++] = nums[j++];
-            }
-        }
-        while (i <= mid) {
-            temp[k++] = nums[i++];
-        }
-        while (j <= right) {
-            temp[k++] = nums[j++];
-        }
-    }
-}
-
-    ```
-  </TabItem>
-  <TabItem value="cpp" label="C++">
-  <SolutionAuthor name="@Ishitamukherjee2004"/>
-   ```cpp
-    void mergeSortedArrays(vector<int>& nums, int left, int mid, int right, vector<int>& temp) {
-    int i = left, j = mid + 1, k = 0;
-    while (i <= mid && j <= right) {
-        if (nums[i] <= nums[j]) temp[k++] = nums[i++];
-        else temp[k++] = nums[j++];
-    }
-    while (i <= mid) temp[k++] = nums[i++];
-    while (j <= right) temp[k++] = nums[j++];
-}
-int merge(vector<int>& nums, int left, int mid, int right) {
-    int count = 0;
-    int j = mid + 1;
-    for (int i = left; i <= mid; i++) {
-        while (j <= right && nums[i] > 2 * nums[j]) j++;
-        count += j - mid - 1;
-    }
-    vector<int> temp(right - left + 1);
-    mergeSortedArrays(nums, left, mid, right, temp);
-    for (int i = left; i <= right; i++) nums[i] = temp[i - left];
-    return count;
-}
-int mergeSort(vector<int>& nums, int left, int right) {
-    if (left >= right) return 0;
-    int mid = left + (right - left) / 2;
-    int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
-    count += merge(nums, left, mid, right);
-    return count;
-}
-
-
-class Solution {
-public:
-    int reversePairs(vector<int>& nums) {
-       return mergeSort(nums, 0, nums.size() - 1);
-
-
-    }
-};
-    ```
-  </TabItem>  
-</Tabs>
-
-### Complexity Analysis
-
-- **Time Complexity**: $$O(n*log(n))$$, where n is the length of the input array. This is because the solution uses a modified Merge Sort algorithm, which has a time complexity of O(n log n).
-
-
-- **Space Complexity**: $$O(n)$$, where n is the length of the input array. This is because the solution uses a temporary array to store the merged sorted arrays.
-
-
-
----
-
-<h2>Authors:</h2>
-
-<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
-{['Ishitamukherjee2004', 'ImmidiSivani'].map(username => (
- <Author key={username} username={username} />
-))}
-</div>
\ No newline at end of file
+---
+id: reverse-pairs
+title: Reverse Pairs
+sidebar_label: 493-Reverse-Pairs
+tags:
+  - Array
+  - Binary Search
+  - Divide and conquer
+  - Binary Indexed Tree
+  - Segment Tree
+  - Merge Sort
+  - Ordered Set
+description: The problem is to reverse the pairs.
+sidebar_position: 2667
+---
+
+## Problem Statement 
+
+### Problem Description
+
+Given an integer array `nums`, return the number of reverse pairs in the array.
+
+A reverse pair is a pair `(i, j)` where:
+
+`0 <= i < j < nums.length` and
+`nums[i] > 2 * nums[j]`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: nums = [1,3,2,3,1]
+Output: 2
+Explanation: The reverse pairs are:
+(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,4,3,5,1]
+Output: 3
+Explanation: The reverse pairs are:
+(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1
+(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1
+```
+
+### Constraints
+
+- `1 <= nums.length <= 5 * 10^4`
+
+## Solution of Given Problem
+
+### Intuition and Approach
+
+The intuition behind this solution is to use a modified Merge Sort algorithm to count the number of reverse pairs in the array.
+
+
+### Approaches
+
+- Divide the array into smaller subarrays until each subarray has only one element.
+- Merge the subarrays back together, counting the number of reverse pairs between each pair of subarrays.
+- The merge step is done in a way that ensures the count of reverse pairs is accurate.
+
+
+#### Codes in Different Languages
+
+<Tabs>
+  <TabItem value="JavaScript" label="JavaScript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```javascript
+    function mergeSortedArrays(nums, left, mid, right, temp) {
+  let i = left, j = mid + 1, k = 0;
+  while (i <= mid && j <= right) {
+    if (nums[i] <= nums[j]) temp[k++] = nums[i++];
+    else temp[k++] = nums[j++];
+  }
+  while (i <= mid) temp[k++] = nums[i++];
+  while (j <= right) temp[k++] = nums[j++];
+}
+
+function merge(nums, left, mid, right) {
+  let count = 0;
+  let j = mid + 1;
+  for (let i = left; i <= mid; i++) {
+    while (j <= right && nums[i] > 2 * nums[j]) j++;
+    count += j - mid - 1;
+  }
+  let temp = new Array(right - left + 1);
+  mergeSortedArrays(nums, left, mid, right, temp);
+  for (let i = left; i <= right; i++) nums[i] = temp[i - left];
+  return count;
+}
+function mergeSort(nums, left, right) {
+  if (left >= right) return 0;
+  let mid = left + (right - left) / 2;
+  let count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+  count += merge(nums, left, mid, right);
+  return count;
+}
+
+class Solution {
+  reversePairs(nums) {
+    return mergeSort(nums, 0, nums.length - 1);
+  }
+}
+
+
+    ```
+
+  </TabItem>
+  <TabItem value="TypeScript" label="TypeScript">
+  <SolutionAuthor name="@Ishitamukherjee2004"/> 
+   ```typescript
+    function mergeSortedArrays(nums: number[], left: number, mid: number, right: number, temp: number[]) {
+  let i = left, j = mid + 1, k = 0;
+  while (i <= mid && j <= right) {
+    if (nums[i] <= nums[j]) temp[k++] = nums[i++];
+    else temp[k++] = nums[j++];
+  }
+  while (i <= mid) temp[k++] = nums[i++];
+  while (j <= right) temp[k++] = nums[j++];
+}
+
+function merge(nums: number[], left: number, mid: number, right: number) {
+  let count = 0;
+  let j = mid + 1;
+  for (let i = left; i <= mid; i++) {
+    while (j <= right && nums[i] > 2 * nums[j]) j++;
+    count += j - mid - 1;
+  }
+  let temp = new Array(right - left + 1);
+  mergeSortedArrays(nums, left, mid, right, temp);
+  for (let i = left; i <= right; i++) nums[i] = temp[i - left];
+  return count;
+}
+function mergeSort(nums: number[], left: number, right: number) {
+  if (left >= right) return 0;
+  let mid = left + (right - left) / 2;
+  let count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+  count += merge(nums, left, mid, right);
+  return count;
+}
+
+class Solution {
+  reversePairs(nums: number[]) {
+    return mergeSort(nums, 0, nums.length - 1);
+  }
+}
+
+    ```
+
+  </TabItem>
+  <TabItem value="Python" label="Python"> 
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```python
+    def merge_sorted_arrays(nums, left, mid, right, temp):
+    i, j, k = left, mid + 1, 0
+    while i <= mid and j <= right:
+        if nums[i] <= nums[j]:
+            temp[k] = nums[i]
+            i += 1
+        else:
+            temp[k] = nums[j]
+            j += 1
+        k += 1
+    while i <= mid:
+        temp[k] = nums[i]
+        i += 1
+        k += 1
+    while j <= right:
+        temp[k] = nums[j]
+        j += 1
+        k += 1
+
+def merge(nums, left, mid, right):
+    count = 0
+    j = mid + 1
+    for i in range(left, mid + 1):
+        while j <= right and nums[i] > 2 * nums[j]:
+            j += 1
+        count += j - mid - 1
+    temp = [0] * (right - left + 1)
+    merge_sorted_arrays(nums, left, mid, right, temp)
+    for i in range(left, right + 1):
+        nums[i] = temp[i - left]
+    return count
+
+def merge_sort(nums, left, right):
+    if left >= right:
+        return 0
+    mid = left + (right - left) // 2
+    count = merge_sort(nums, left, mid) + merge_sort(nums, mid + 1, right)
+    count += merge(nums, left, mid, right)
+    return count
+
+class Solution:
+    def reversePairs(self, nums):
+        return merge_sort(nums, 0, len(nums) - 1)
+
+    ```
+
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```java
+    public class Solution {
+    public int reversePairs(int[] nums) {
+        return mergeSort(nums, 0, nums.length - 1);
+    }
+
+    public int mergeSort(int[] nums, int left, int right) {
+        if (left >= right) {
+            return 0;
+        }
+        int mid = left + (right - left) / 2;
+        int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+        count += merge(nums, left, mid, right);
+        return count;
+    }
+     public int merge(int[] nums, int left, int mid, int right) {
+        int count = 0;
+        int j = mid + 1;
+        for (int i = left; i <= mid; i++) {
+            while (j <= right && nums[i] > 2 * nums[j]) {
+                j++;
+            }
+            count += j - mid - 1;
+        }
+        int[] temp = new int[right - left + 1];
+        mergeSortedArrays(nums, left, mid, right, temp);
+        for (int i = left; i <= right; i++) {
+            nums[i] = temp[i - left];
+        }
+        return count;
+    }
+
+    public void
+    mergeSortedArrays(int[] nums, int left, int mid, int right, int[] temp) {
+        int i = left, j = mid + 1, k = 0;
+        while (i <= mid && j <= right) {
+            if (nums[i] <= nums[j]) {
+                temp[k++] = nums[i++];
+            } else {
+                temp[k++] = nums[j++];
+            }
+        }
+        while (i <= mid) {
+            temp[k++] = nums[i++];
+        }
+        while (j <= right) {
+            temp[k++] = nums[j++];
+        }
+    }
+}
+
+    ```
+  </TabItem>
+  <TabItem value="cpp" label="C++">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```cpp
+    void mergeSortedArrays(vector<int>& nums, int left, int mid, int right, vector<int>& temp) {
+    int i = left, j = mid + 1, k = 0;
+    while (i <= mid && j <= right) {
+        if (nums[i] <= nums[j]) temp[k++] = nums[i++];
+        else temp[k++] = nums[j++];
+    }
+    while (i <= mid) temp[k++] = nums[i++];
+    while (j <= right) temp[k++] = nums[j++];
+}
+int merge(vector<int>& nums, int left, int mid, int right) {
+    int count = 0;
+    int j = mid + 1;
+    for (int i = left; i <= mid; i++) {
+        while (j <= right && nums[i] > 2 * nums[j]) j++;
+        count += j - mid - 1;
+    }
+    vector<int> temp(right - left + 1);
+    mergeSortedArrays(nums, left, mid, right, temp);
+    for (int i = left; i <= right; i++) nums[i] = temp[i - left];
+    return count;
+}
+int mergeSort(vector<int>& nums, int left, int right) {
+    if (left >= right) return 0;
+    int mid = left + (right - left) / 2;
+    int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+    count += merge(nums, left, mid, right);
+    return count;
+}
+
+
+class Solution {
+public:
+    int reversePairs(vector<int>& nums) {
+       return mergeSort(nums, 0, nums.size() - 1);
+
+
+    }
+};
+    ```
+  </TabItem>  
+</Tabs>
+
+### Complexity Analysis
+
+- **Time Complexity**: $$O(n*log(n))$$, where n is the length of the input array. This is because the solution uses a modified Merge Sort algorithm, which has a time complexity of O(n log n).
+
+
+- **Space Complexity**: $$O(n)$$, where n is the length of the input array. This is because the solution uses a temporary array to store the merged sorted arrays.
+
+
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['Ishitamukherjee2004'].map(username => (
+ <Author key={username} username={username} />
+))} </div>
diff --git a/dsa-solutions/lc-solutions/3000-3099/3163-string-compression-III.md b/dsa-solutions/lc-solutions/3000-3099/3163-string-compression-III.md
new file mode 100644
index 000000000..4525848dd
--- /dev/null
+++ b/dsa-solutions/lc-solutions/3000-3099/3163-string-compression-III.md
@@ -0,0 +1,241 @@
+---
+id: string-compression
+title: String Compression Solution
+sidebar_label: String Compression
+tags:
+  - String Compression
+  - Brute Force
+  - Optimized
+  - LeetCode
+  - Java
+  - Python
+  - C++
+description: "This is a solution to the String Compression problem on LeetCode."
+sidebar_position: 2
+---
+
+In this tutorial, we will solve the String Compression problem using two different approaches: brute force and optimized. We will provide the implementation of the solution in C++, Java, and Python.
+
+## Problem Description
+
+Given a string `word`, compress it using the following algorithm:
+
+Begin with an empty string `comp`. While `word` is not empty, use the following operation:
+Remove a maximum length prefix of `word` made of a single character `c` repeating at most 9 times.
+Append the length of the prefix followed by `c` to `comp`.
+
+Return the string `comp`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: word = "abcde"
+Output: "1a1b1c1d1e"
+```
+
+**Example 2:**
+
+```
+Input: word = "aaaaaaaaaaaaaabb"
+Output: "9a5a2b"
+```
+
+### Constraints
+
+- `1 <= word.length <= 2 * 105`
+- `word` consists only of lowercase English letters.
+
+---
+
+## Solution for String Compression Problem
+
+### Intuition and Approach
+
+The problem can be solved using a brute force approach or an optimized approach. The brute force approach directly iterates through the string and constructs the result, while the optimized approach efficiently handles consecutive characters.
+
+<Tabs>
+<tabItem value="Brute Force" label="Brute Force">
+
+### Approach 1: Brute Force (Naive)
+
+The brute force approach iterates through each character of the string, counts consecutive characters up to 9, and appends the count followed by the character to the result string.
+
+#### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    string compressString(string word) {
+        string comp;
+        int i = 0;
+        while (i < word.length()) {
+            char c = word[i];
+            int count = 0;
+            while (i < word.length() && word[i] == c && count < 9) {
+                count++;
+                i++;
+            }
+            comp += to_string(count) + c;
+        }
+        return comp;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public String compressString(String word) {
+        StringBuilder comp = new StringBuilder();
+        int i = 0;
+        while (i < word.length()) {
+            char c = word.charAt(i);
+            int count = 0;
+            while (i < word.length() && word.charAt(i) == c && count < 9) {
+                count++;
+                i++;
+            }
+            comp.append(count).append(c);
+        }
+        return comp.toString();
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def compressString(self, word: str) -> str:
+        comp = []
+        i = 0
+        while i < len(word):
+            c = word[i]
+            count = 0
+            while i < len(word) and word[i] == c and count < 9:
+                count += 1
+                i += 1
+            comp.append(f"{count}{c}")
+        return ''.join(comp)
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $O(n)$
+- Space Complexity: $O(n)$
+- Where `n` is the length of `word`.
+- The time complexity is $O(n)$ because we iterate through the string once.
+- The space complexity is $O(n)$ because we store the result in a new string.
+
+</tabItem>
+<tabItem value="Optimized" label="Optimized">
+
+### Approach 2: Optimized Approach
+
+The optimized approach is similar to the brute force but handles the prefix length more efficiently, ensuring it counts up to 9 characters in each iteration.
+
+#### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    string compressString(string word) {
+        string comp;
+        int i = 0;
+        while (i < word.length()) {
+            char c = word[i];
+            int count = 1;
+            while (i + count < word.length() && word[i + count] == c && count < 9) {
+                count++;
+            }
+            comp += to_string(count) + c;
+            i += count;
+        }
+        return comp;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public String compressString(String word) {
+        StringBuilder comp = new StringBuilder();
+        int i = 0;
+        while (i < word.length()) {
+            char c = word.charAt(i);
+            int count = 1;
+            while (i + count < word.length() && word.charAt(i + count) == c && count < 9) {
+                count++;
+            }
+            comp.append(count).append(c);
+            i += count;
+        }
+        return comp.toString();
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def compressString(self, word: str) -> str:
+        comp = []
+        i = 0
+        while i < len(word):
+            c = word[i]
+            count = 1
+            while i + count < len(word) and word[i + count] == c and count < 9:
+                count += 1
+            comp.append(f"{count}{c}")
+            i += count
+        return ''.join(comp)
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $O(n)$
+- Space Complexity: $O(n)$
+- Where `n` is the length of `word`.
+- The time complexity is $O(n)$ because we iterate through the string once.
+- The space complexity is $O(n)$ because we store the result in a new string.
+
+</tabItem>
+</Tabs>
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['ImmidiSivani'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>