[Docs]: Want to Add Folder which will Contain Different questions rel…

…ated to topic And their Solution #351 (#374)

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* Added solution to 3 problems and categorised them

* renamed

* nodemodules removed

docs/Practice_Question/Array/Battleships-in-a-Board.md

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+---
+id: battleships-in-a-board
+title: Battleships in a Board
+sidebar_label: 0419-Battleships-in-a-Board
+tags:
+- Array
+- Depth-First Search
+- Matrix
+description: "Given an `m x n` board where each cell is a battleship 'X' or empty '.', count the number of battleships on the board."
+---
+
+## Problem
+
+Given an `m x n` board where each cell is a battleship `'X'` or empty `'.'`, return the number of battleships on the board.
+
+Battleships can only be placed horizontally or vertically on the board. In other words, they can only occupy a contiguous line of cells horizontally or vertically. Each battleship must be separated by at least one empty cell (either horizontally or vertically) from any other battleship.
+
+### Examples
+
+**Example 1:**
+
+**Input:** `board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]` 
+**Output:** `2`
+
+**Example 2:**
+
+**Input:** `board = [["."]]` 
+**Output:** `0`
+
+### Constraints
+
+- `m == board.length`
+- `n == board[i].length`
+- `1 <= m, n <= 200`
+- `board[i][j]` is either `'X'` or `'.'`.
+
+### Follow-up
+
+Could you do it in one-pass, using only `O(1)` extra memory and without modifying the value of the board?
+
+---
+
+## Approach
+
+To count the number of battleships in one pass and without extra memory, we can traverse the board and count only the top-left cell of each battleship. A cell qualifies as the top-left cell if there are no battleship cells above it or to the left of it.
+
+### Steps:
+
+1. Initialize a counter to keep track of the number of battleships.
+2. Traverse each cell in the board.
+3. For each cell, check if it is a battleship (`'X'`):
+ - If it is, check if there is no battleship cell above it and no battleship cell to the left of it.
+ - If both conditions are met, increment the battleship counter.
+4. Return the counter after the traversal.
+
+### C++ Solution
+
+```cpp
+class Solution {
+public:
+ int countBattleships(vector<vector<char>>& board) {
+ int count = 0;
+ for (int i = 0; i < board.size(); i++) {
+ for (int j = 0; j < board[0].size(); j++) {
+ if (board[i][j] == 'X') {
+ if (i > 0 && board[i - 1][j] == 'X') continue;
+ if (j > 0 && board[i][j - 1] == 'X') continue;
+ count++;
+ }
+ }
+ }
+ return count;
+ }
+};
+```
+### Complexity Analysis
+**Time Complexity:** O(m * n)
+>Reason: We traverse each cell of the board once.
+
+**Space Complexity:** O(1)
+>Reason: We do not use any extra space that scales with the input size.
+
+### References
+**LeetCode Problem:** Battleships in a Board
+
+**Solution Link:** Battleships in a Board Solution on LeetCode

docs/Practice_Question/String/Longest-Uncommon-Subsequence-II.md

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+---
+id: longest-uncommon-subsequence-ii
+title: Longest Uncommon Subsequence II
+sidebar_label: 0522-Longest-Uncommon-Subsequence-II
+tags:
+- String
+- Sorting
+description: "Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1."
+---
+
+## Problem
+
+Given an array of strings `strs`, return the length of the longest uncommon subsequence between them. An uncommon subsequence is a subsequence that is not common to any other string in the array.
+
+A **subsequence** of a string `s` is a string that can be obtained after deleting any number of characters from `s`.
+
+### Examples
+
+**Example 1:**
+
+**Input:** `strs = ["aba","cdc","eae"]` 
+**Output:** `3`
+
+**Example 2:**
+
+**Input:** `strs = ["aaa","aaa","aa"]` 
+**Output:** `-1`
+
+### Constraints
+
+- `2 <= strs.length <= 50`
+- `1 <= strs[i].length <= 10`
+- `strs[i]` consists of lowercase English letters.
+
+---
+
+## Approach
+
+To find the longest uncommon subsequence, we need to consider the following:
+
+1. If a string is unique in the list (i.e., it does not appear more than once), we need to check if it is not a subsequence of any other string. 
+2. If a string is not unique (i.e., it appears more than once), it cannot be an uncommon subsequence.
+3. We can start by checking the strings in descending order of their lengths. This way, we can return the first string that meets the criteria as soon as we find it.
+
+### Steps:
+
+1. Sort the strings by length in descending order.
+2. Check each string:
+ - If the string appears only once in the list.
+ - If it is not a subsequence of any other string longer than it.
+3. If such a string is found, return its length.
+4. If no such string is found, return `-1`.
+
+### Helper Function
+
+A helper function `is_subsequence` can be used to determine if one string is a subsequence of another.
+
+### C++ Solution
+```cpp
+#include <vector>
+#include <string>
+#include <algorithm>
+using namespace std;
+
+class Solution {
+public:
+ int findLUSlength(vector<string>& strs) {
+ sort(strs.begin(), strs.end(), [](const string &a, const string &b) {
+ return b.size() < a.size();
+ });
+
+  for (int i = 0; i < strs.size(); i++) {
+  bool isSubsequence = false;
+  for (int j = 0; j < strs.size(); j++) {
+  if (i != j && isSubsequence(strs[i], strs[j])) {
+  isSubsequence = true;
+  break;
+  }
+ }
+ if (!isSubsequence) {
+  return strs[i].size();
+  }
+  }
+  return -1;
+ }
+
+private:
+ bool isSubsequence(const string &a, const string &b) {
+ int i = 0, j = 0;
+ while (i < a.size() && j < b.size()) {
+ if (a[i] == b[j]) {
+ i++;
+ }
+ j++;
+ }
+ return i == a.size();
+ }
+};
+```
+### Complexity Analysis
+**Time Complexity:** O(n^2 * l)
+>Reason: Sorting the list takes O(n log n). Checking if a string is a subsequence of another string takes O(l) time, and this is done for each pair of strings, leading to O(n^2 * l) in total.
+
+**Space Complexity:** O(1)
+>Reason: The space complexity is constant as we are not using any extra space proportional to the input size, other than the space required for sorting.
+
+### References
+**LeetCode Problem:** Longest Uncommon Subsequence II
+
+**Solution Link:** LeetCode Solution
+
+**Wikipedia:** Subsequence

docs/Practice_Question/Tree/Convert-BST-to-Greater-Tree.md

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+---
+id: convert-bst-to-greater-tree
+title: Convert BST to Greater Tree
+sidebar_label: 0538-Convert-BST-to-Greater-Tree
+tags:
+- Tree
+- Depth-First Search
+- Binary Search Tree
+- Binary Tree
+description: "Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST."
+---
+
+## Problem
+
+Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.
+
+### Examples
+
+**Example 1:**
+
+**Input:** `root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]` 
+**Output:** `[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]`
+
+**Example 2:**
+
+**Input:** `root = [0,null,1]` 
+**Output:** `[1,null,1]`
+
+### Constraints
+
+- The number of nodes in the tree is in the range `[0, 10^4]`.
+- `-10^4 <= Node.val <= 10^4`
+- All the values in the tree are unique.
+- `root` is guaranteed to be a valid binary search tree.
+
+---
+
+## Approach
+
+To convert a BST to a Greater Tree, we need to accumulate the sum of all nodes that are greater than the current node. This can be efficiently done using a reverse in-order traversal (right-root-left), which processes nodes from the largest to the smallest.
+
+### Steps:
+
+1. Initialize a variable to keep track of the running sum.
+2. Perform a reverse in-order traversal of the tree.
+3. During the traversal, update the value of each node to include the running sum.
+4. Update the running sum with the value of the current node.
+
+### C++ Solution
+
+```cpp
+#include <iostream>
+using namespace std;
+
+struct TreeNode {
+ int val;
+ TreeNode *left;
+ TreeNode *right;
+ TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+};
+
+class Solution {
+private:
+ int sum = 0;
+public:
+ TreeNode* convertBST(TreeNode* root) {
+ if (root != nullptr) {
+ convertBST(root->right);
+ sum += root->val;
+ root->val = sum;
+ convertBST(root->left);
+ }
+ return root;
+ }
+};
+```
+### Complexity Analysis
+**Time Complexity:** O(n)
+>Reason: Each node is visited once during the traversal.
+
+**Space Complexity:** O(h)
+>Reason: The space complexity is determined by the recursion stack, which in the worst case (unbalanced tree) is O(n), but on average (balanced tree) is O(log n).
+
+### References
+**LeetCode Problem:** Convert BST to Greater Tree
+
+**Solution Link:** LeetCode Solution
+
+**Wikipedia:** Binary Search Tree