From 618b4390c21d5d45e836781df68921ac57700130 Mon Sep 17 00:00:00 2001
From: SOHAM-GADEKAR <sohamg.3456@gmail.com>
Date: Mon, 20 Oct 2025 12:08:44 +0530
Subject: [PATCH] Added Solution of LeetCode Problem No. 430

---
 .../Programs/C++/LeetCode/Q430. cpp           | 140 ++++++++++++++++++
 1 file changed, 140 insertions(+)
 create mode 100644 Domains/CompetitiveProgramming/Programs/C++/LeetCode/Q430. cpp

diff --git a/Domains/CompetitiveProgramming/Programs/C++/LeetCode/Q430. cpp b/Domains/CompetitiveProgramming/Programs/C++/LeetCode/Q430. cpp
new file mode 100644
index 00000000..d49e1625
--- /dev/null
+++ b/Domains/CompetitiveProgramming/Programs/C++/LeetCode/Q430. cpp	
@@ -0,0 +1,140 @@
+/*
+===============================================================
+430. Flatten a Multilevel Doubly Linked List
+===============================================================
+
+Difficulty: Medium
+Tags: Linked List, Depth-First Search, Doubly Linked List
+
+---------------------------------------------------------------
+Problem Statement:
+---------------------------------------------------------------
+You are given a doubly linked list that contains nodes with:
+- a 'next' pointer,
+- a 'prev' pointer,
+- and an additional 'child' pointer.
+
+The 'child' pointer may or may not point to another doubly linked list.
+These child lists may have their own children, and so on, creating a 
+multilevel data structure.
+
+Given the head of the first level of the list, flatten the list so that 
+all the nodes appear in a single-level, doubly linked list.
+
+Let 'curr' be a node with a child list. 
+The nodes in the child list should appear *after curr* and *before curr->next*
+in the flattened list.
+
+Return the head of the flattened list. 
+All 'child' pointers in the flattened list must be set to NULL.
+
+---------------------------------------------------------------
+Example 1:
+---------------------------------------------------------------
+Input:  head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
+Output: [1,2,3,7,8,11,12,9,10,4,5,6]
+
+Explanation:
+The multilevel linked list in the input is shown below:
+1---2---3---4---5---6--NULL
+        |
+        7---8---9---10--NULL
+            |
+            11--12--NULL
+
+After flattening, the list becomes:
+1---2---3---7---8---11---12---9---10---4---5---6--NULL
+
+---------------------------------------------------------------
+Example 2:
+---------------------------------------------------------------
+Input:  head = [1,2,null,3]
+Output: [1,3,2]
+
+Explanation:
+The multilevel linked list is:
+1---2--NULL
+|
+3--NULL
+After flattening, it becomes:
+1---3---2--NULL
+
+---------------------------------------------------------------
+Example 3:
+---------------------------------------------------------------
+Input:  head = []
+Output: []
+
+Explanation:
+The list may be empty.
+
+---------------------------------------------------------------
+Constraints:
+---------------------------------------------------------------
+- The number of nodes will not exceed 1000.
+- 1 <= Node.val <= 10^5
+
+---------------------------------------------------------------
+Representation Notes:
+---------------------------------------------------------------
+The multilevel linked list from Example 1 is represented as:
+Level 1: [1,2,3,4,5,6,null]
+Level 2: [7,8,9,10,null]
+Level 3: [11,12,null]
+
+Merged representation (serialization):
+[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
+===============================================================
+*/
+
+/*
+// Definition for a Node.
+class Node {
+public:
+    int val;
+    Node* prev;
+    Node* next;
+    Node* child;
+};
+*/
+
+class Q430{
+public:
+    Node* flatten(Node* head) {
+        // Pointer to traverse the linked list
+        Node* curr = head;
+
+        // Traverse through all nodes in the list
+        while (curr) {
+
+            // If the current node has a child, we need to flatten that child list
+            if (curr->child) {
+
+                // Step 1: Find the tail of the child list
+                Node* temp = curr->child;
+                while (temp->next) 
+                    temp = temp->next;
+
+                // Step 2: Connect the tail of the child list to curr->next
+                temp->next = curr->next;
+
+                // If there is a next node, update its prev pointer
+                if (curr->next) 
+                    curr->next->prev = temp;
+
+                // Step 3: Connect the child list to the main list
+                curr->next = curr->child;
+                curr->child->prev = curr;
+
+                // Step 4: Remove the child pointer (since it's now flattened)
+                curr->child = nullptr;
+            }
+
+            // Move to the next node in the list
+            curr = curr->next;
+        }
+
+        // Return the head of the flattened list
+        return head;
+    }
+};