diff --git a/Domains/CompetitiveProgramming/Programs/C++/Binary Tree Maximum Path Sum.cpp b/Domains/CompetitiveProgramming/Programs/C++/Binary Tree Maximum Path Sum.cpp
new file mode 100644
index 00000000..902558df
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+++ b/Domains/CompetitiveProgramming/Programs/C++/Binary Tree Maximum Path Sum.cpp	
@@ -0,0 +1,82 @@
+/*
+Problem: Binary Tree Maximum Path Sum
+Platform: LeetCode
+Problem Code: 124
+Difficulty: Hard
+Link: https://leetcode.com/problems/binary-tree-maximum-path-sum/
+
+A path in a binary tree is a sequence of nodes where each pair of adjacent nodes 
+has an edge connecting them. A node can appear in the path at most once. The path 
+does not necessarily need to pass through the root.
+
+The path sum of a path is the sum of the node's values in the path.
+
+Given the root of a binary tree, return the maximum path sum of any non-empty path.
+
+Example 1:
+Input: root = [1,2,3]
+Output: 6
+Explanation: The optimal path is 2 → 1 → 3, which gives a path sum of 6.
+
+Example 2:
+Input: root = [-10,9,20,null,null,15,7]
+Output: 42
+Explanation: The optimal path is 15 → 20 → 7, which gives a path sum of 42.
+
+Approach:
+1. Use post-order DFS traversal to compute the maximum path sum passing through each node.
+2. For every node, calculate:
+   - `l_sum`: maximum sum path in the left subtree.
+   - `r_sum`: maximum sum path in the right subtree.
+3. The best path through the current node could include:
+   - only the node value,
+   - the node and left subtree,
+   - the node and right subtree,
+   - or both subtrees through the node.
+4. Update a global `max_sum` with the maximum path found so far.
+5. Return to the parent the maximum single-path sum (either left or right) plus current node value.
+
+Time Complexity: O(n) — each node is visited once.
+Space Complexity: O(h) — recursion stack, where h is the height of the tree.
+
+Contributor: shwetakul2005
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// Definition for a binary tree node.
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode() : val(0), left(nullptr), right(nullptr) {}
+    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+};
+
+class Solution {
+public:
+    int calc_sum(TreeNode* root, int& max_sum) {
+        if (root == NULL) return 0;
+
+        int l_sum = calc_sum(root->left, max_sum);
+        int r_sum = calc_sum(root->right, max_sum);
+
+        int c_sum = root->val;
+        if (l_sum > 0) c_sum += l_sum;
+        if (r_sum > 0) c_sum += r_sum;
+
+        // Update global maximum
+        max_sum = max(max_sum, c_sum);
+
+        // Return maximum path sum for parent usage
+        return max(0, max(l_sum, r_sum)) + root->val;
+    }
+
+    int maxPathSum(TreeNode* root) {
+        int max_sum = root->val;
+        calc_sum(root, max_sum);
+        return max_sum;
+    }
+};
diff --git a/Domains/CompetitiveProgramming/Programs/C++/Insert into a Binary Search Tree.cpp b/Domains/CompetitiveProgramming/Programs/C++/Insert into a Binary Search Tree.cpp
new file mode 100644
index 00000000..db3a5100
--- /dev/null
+++ b/Domains/CompetitiveProgramming/Programs/C++/Insert into a Binary Search Tree.cpp	
@@ -0,0 +1,57 @@
+/*
+Problem: Insert into a Binary Search Tree
+Platform: LeetCode
+Problem Code: 701
+Difficulty: Medium
+Link: https://leetcode.com/problems/insert-into-a-binary-search-tree/
+
+Given the root of a binary search tree (BST) and a value to insert into the tree, 
+insert the value into the BST. Return the root of the BST after the insertion. 
+It is guaranteed that the new value does not exist in the original BST.
+
+Notice that there may exist multiple valid ways for the insertion, as long as the tree 
+remains a BST after insertion. You can return any of them.
+
+Example 1:
+Input: root = [4,2,7,1,3], val = 5
+Output: [4,2,7,1,3,5]
+
+Example 2:
+Input: root = [40,20,60,10,30,50,70], val = 25
+Output: [40,20,60,10,30,50,70,null,null,25]
+
+Approach:
+1. Recursively traverse the tree to find the correct position for the new value.
+2. If the value is less than the current node, go to the left subtree.
+3. If the value is greater, go to the right subtree.
+4. When a NULL position is found, create a new node with the given value.
+5. Return the root of the updated BST.
+
+Time Complexity: O(h), where h is the height of the BST.
+Space Complexity: O(h) due to recursion stack.
+
+Contributor: shwetakul2005
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// Definition for a binary tree node.
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode() : val(0), left(nullptr), right(nullptr) {}
+    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+};
+
+class Solution {
+public:
+    TreeNode* insertIntoBST(TreeNode* root, int val) {
+        if (root == NULL) return new TreeNode(val);
+        if (val < root->val) root->left = insertIntoBST(root->left, val);
+        else root->right = insertIntoBST(root->right, val);
+        return root;
+    }
+};
diff --git a/Domains/CompetitiveProgramming/Programs/C++/Merge k Sorted Lists.cpp b/Domains/CompetitiveProgramming/Programs/C++/Merge k Sorted Lists.cpp
new file mode 100644
index 00000000..317bdcb3
--- /dev/null
+++ b/Domains/CompetitiveProgramming/Programs/C++/Merge k Sorted Lists.cpp	
@@ -0,0 +1,77 @@
+/*
+Problem: Merge k Sorted Lists
+Platform: LeetCode
+Problem Code: 23
+Difficulty: Hard
+Link: https://leetcode.com/problems/merge-k-sorted-lists/
+
+You are given an array of k linked-lists, each linked-list is sorted in ascending order.
+Merge all the linked-lists into one sorted linked-list and return it.
+
+Example 1:
+Input: lists = [[1,4,5],[1,3,4],[2,6]]
+Output: [1,1,2,3,4,4,5,6]
+
+Example 2:
+Input: lists = []
+Output: []
+
+Example 3:
+Input: lists = [[]]
+Output: []
+
+Approach:
+1. Traverse through each linked list and push all node values into a min-heap (priority queue).
+2. Extract the smallest element one by one from the heap to form a new sorted linked list.
+3. This approach ensures a sorted result, though it uses extra memory for all node values.
+
+Time Complexity: O(N log N) where N is the total number of nodes (since we push and pop each node in the heap).
+Space Complexity: O(N) for storing all values in the priority queue.
+
+Contributor: shwetakul2005
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// Definition for singly-linked list.
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode() : val(0), next(nullptr) {}
+    ListNode(int x) : val(x), next(nullptr) {}
+    ListNode(int x, ListNode *next) : val(x), next(next) {}
+};
+
+class Solution {
+public:
+    ListNode* mergeKLists(vector<ListNode*>& lists) {
+        int n = lists.size();
+        priority_queue<int, vector<int>, greater<int>> pq; // min-heap
+
+        // Step 1: Push all node values into heap
+        for (int i = 0; i < n; i++) {
+            ListNode* temp = lists[i];
+            while (temp != nullptr) {
+                pq.push(temp->val);
+                temp = temp->next;
+            }
+        }
+
+        // Step 2: Handle empty case
+        if (pq.empty()) return nullptr;
+
+        // Step 3: Construct the new sorted list
+        ListNode* res = new ListNode(pq.top());
+        pq.pop();
+        ListNode* temp = res;
+
+        while (!pq.empty()) {
+            temp->next = new ListNode(pq.top());
+            temp = temp->next;
+            pq.pop();
+        }
+
+        return res;
+    }
+};
diff --git a/Domains/CompetitiveProgramming/Programs/C++/Split Array Largest Sum.cpp b/Domains/CompetitiveProgramming/Programs/C++/Split Array Largest Sum.cpp
new file mode 100644
index 00000000..72b2a859
--- /dev/null
+++ b/Domains/CompetitiveProgramming/Programs/C++/Split Array Largest Sum.cpp	
@@ -0,0 +1,91 @@
+/*
+Problem: Split Array Largest Sum
+Platform: LeetCode
+Problem Code: 410
+Difficulty: Medium
+Link: https://leetcode.com/problems/split-array-largest-sum/
+
+Given an array `arr` of non-negative integers and an integer `m`, split the array into `m` non-empty continuous subarrays. 
+Write an algorithm to minimize the largest sum among these `m` subarrays.
+
+Example 1:
+Input: arr = [7,2,5,10,8], m = 2
+Output: 18
+Explanation: There are four ways to split arr into two subarrays. The best way is [7,2,5] and [10,8], 
+which has the largest sum = max(7+2+5,10+8) = 18.
+
+Example 2:
+Input: arr = [1,2,3,4,5], m = 2
+Output: 9
+
+Approach:
+1. Use binary search on the range of possible sums: 
+   - `low = max element of array` (minimum possible largest sum)
+   - `high = sum of array` (maximum possible largest sum)
+2. For a candidate sum `mid`, check if it is possible to split the array into ≤ m subarrays such that no subarray sum exceeds `mid`.
+3. If possible, try a smaller `mid` (move `high`), else increase `mid` (move `low`).
+4. Continue until `low > high`, and the minimum feasible `mid` is the answer.
+
+Time Complexity: O(n log(sum-max)), where sum-max = total sum of array minus max element.
+Space Complexity: O(1) extra space.
+
+Contributor: shwetakul2005
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    // Check if we can split array into ≤ m subarrays with max sum ≤ mid
+    bool ispos(vector<int>& v, int mid, int m) {
+        int n = v.size();
+        int s = 0, cnt = 0;
+
+        for (int i = 0; i < n; i++) {
+            s += v[i];
+            if (s == mid) {
+                cnt++;
+                s = 0;
+            } 
+            else if (s > mid) {
+                cnt++;
+                s = v[i];
+                if (i == n - 1) {
+                    cnt++;
+                }
+            } 
+            else if (s < mid && i == n - 1) {
+                cnt++;
+            }
+        }
+
+        return cnt <= m;
+    }
+
+    int splitArray(vector<int>& arr, int m) {
+        int n = arr.size();
+        if (n < m) return -1;
+
+        int tot_sum = 0, maxi = INT_MIN;
+        for (int i = 0; i < n; i++) {
+            tot_sum += arr[i];
+            maxi = max(maxi, arr[i]);
+        }
+
+        int low = maxi, high = tot_sum;
+        int ans = tot_sum;
+
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            if (ispos(arr, mid, m)) {
+                ans = mid;
+                high = mid - 1;
+            } else {
+                low = mid + 1;
+            }
+        }
+
+        return ans;
+    }
+};
\ No newline at end of file
diff --git a/Domains/CompetitiveProgramming/Programs/C++/Trapping Rain Water.cpp b/Domains/CompetitiveProgramming/Programs/C++/Trapping Rain Water.cpp
new file mode 100644
index 00000000..451cfad7
--- /dev/null
+++ b/Domains/CompetitiveProgramming/Programs/C++/Trapping Rain Water.cpp	
@@ -0,0 +1,69 @@
+/*
+Problem: Trapping Rain Water
+Platform: LeetCode
+Problem Code: 42
+Difficulty: Hard
+Link: https://leetcode.com/problems/trapping-rain-water/
+
+Given n non-negative integers representing an elevation map where the width of each bar is 1, 
+compute how much water it can trap after raining.
+
+Example 1:
+Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
+Output: 6
+Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. 
+In this case, 6 units of rain water (blue section) are being trapped.
+
+Example 2:
+Input: height = [4,2,0,3,2,5]
+Output: 9
+
+Approach:
+1. Use two pointers — one from the left and one from the right.
+2. Maintain two variables `leftMax` and `rightMax` to store the maximum height encountered so far from each side.
+3. Move the pointer with the smaller height inward.
+   - If the current height is less than the corresponding max, water can be trapped = max - height.
+   - Otherwise, update the max height.
+4. Continue until both pointers meet.
+5. This approach eliminates the need for extra arrays and efficiently computes the trapped water.
+
+Time Complexity: O(n)
+Space Complexity: O(1)
+
+Contributor: shwetakul2005
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    int trap(vector<int>& height) {
+        int n = height.size();
+        if (n == 0) return 0;
+
+        int left = 0, right = n - 1;
+        int leftMax = 0, rightMax = 0;
+        int water = 0;
+
+        while (left < right) {
+            if (height[left] < height[right]) {
+                // Update left max
+                if (height[left] >= leftMax)
+                    leftMax = height[left];
+                else
+                    water += leftMax - height[left];
+                left++;
+            } 
+            else {
+                // Update right max
+                if (height[right] >= rightMax)
+                    rightMax = height[right];
+                else
+                    water += rightMax - height[right];
+                right--;
+            }
+        }
+        return water;
+    }
+};
diff --git a/Domains/CompetitiveProgramming/Programs/C++/Vertical Order Traversal of a Binary Tree.cpp b/Domains/CompetitiveProgramming/Programs/C++/Vertical Order Traversal of a Binary Tree.cpp
new file mode 100644
index 00000000..f1b12b7d
--- /dev/null
+++ b/Domains/CompetitiveProgramming/Programs/C++/Vertical Order Traversal of a Binary Tree.cpp	
@@ -0,0 +1,91 @@
+/*
+Problem: Vertical Order Traversal of a Binary Tree
+Platform: LeetCode
+Problem Code: 987
+Difficulty: Medium
+Link: https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/
+
+Given the root of a binary tree, calculate the vertical order traversal of the binary tree.
+
+For each node at position (row, col):
+- The root is at (0, 0).
+- Left child is at (row+1, col-1)
+- Right child is at (row+1, col+1)
+
+Return a list of non-empty reports of nodes for each column sorted from left to right. 
+- Nodes in the same row and column are sorted by value.
+
+Example 1:
+Input: root = [3,9,20,null,null,15,7]
+Output: [[9],[3,15],[20],[7]]
+
+Example 2:
+Input: root = [1,2,3,4,5,6,7]
+Output: [[4],[2],[1,5,6],[3],[7]]
+
+Approach:
+1. Perform a DFS traversal of the tree while recording (col, row, value) for each node.
+2. Store all nodes in a vector of tuples: (col, row, value).
+3. Sort the vector:
+   - first by column (left to right)
+   - then by row (top to bottom)
+   - then by node value (ascending) if row and column are the same
+4. Group the sorted nodes by column to get the final vertical order.
+
+Time Complexity: O(n log n) — due to sorting all n nodes.
+Space Complexity: O(n) — for storing node tuples.
+
+Contributor: shwetakul2005
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// Definition for a binary tree node.
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode() : val(0), left(nullptr), right(nullptr) {}
+    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+};
+
+class Solution {
+public:
+    // DFS traversal storing (col, row, value)
+    void traversal(TreeNode* root, vector<tuple<int,int,int>>& st, int col=0, int row=0){
+        if(root == NULL) return;
+
+        st.push_back({col, row, root->val});
+        traversal(root->left, st, col-1, row+1);
+        traversal(root->right, st, col+1, row+1);
+    }
+
+    vector<vector<int>> verticalTraversal(TreeNode* root) {
+        vector<tuple<int,int,int>> st; // store (col, row, value)
+        traversal(root, st);
+
+        // Sort by column, then row, then value
+        sort(st.begin(), st.end(), [](const auto& a, const auto& b) {
+            if (get<0>(a) != get<0>(b)) return get<0>(a) < get<0>(b);
+            if (get<1>(a) != get<1>(b)) return get<1>(a) < get<1>(b);
+            return get<2>(a) < get<2>(b);
+        });
+
+        // Group by column
+        vector<vector<int>> ans;
+        vector<int> res;
+        int k = get<0>(st[0]);
+        for (int i = 0; i < st.size(); i++) {
+            if (get<0>(st[i]) != k) {
+                ans.push_back(res);
+                res.clear();
+                k = get<0>(st[i]);
+            }
+            res.push_back(get<2>(st[i]));
+        }
+        if (!res.empty()) ans.push_back(res);
+        return ans;
+    }
+};