🚀 11-Jul-2020

Author: sureshmangs

competitive programming/leetcode/2020-July-Challenge/Day-10-Flatten a Multilevel Doubly Linked List.cpp

@@ -0,0 +1,118 @@
+You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.
+
+Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.
+
+
+
+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 as follows:
+
+
+
+After flattening the multilevel linked list it becomes:
+
+
+Example 2:
+
+Input: head = [1,2,null,3]
+Output: [1,3,2]
+Explanation:
+
+The input multilevel linked list is as follows:
+
+  1---2---NULL
+  |
+  3---NULL
+Example 3:
+
+Input: head = []
+Output: []
+
+
+How multilevel linked list is represented in test case:
+
+We use the multilevel linked list from Example 1 above:
+
+ 1---2---3---4---5---6--NULL
+         |
+         7---8---9---10--NULL
+             |
+             11--12--NULL
+The serialization of each level is as follows:
+
+[1,2,3,4,5,6,null]
+[7,8,9,10,null]
+[11,12,null]
+To serialize all levels together we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:
+
+[1,2,3,4,5,6,null]
+[null,null,7,8,9,10,null]
+[null,11,12,null]
+Merging the serialization of each level and removing trailing nulls we obtain:
+
+[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
+
+
+Constraints:
+
+Number of Nodes will not exceed 1000.
+1 <= Node.val <= 10^5
+
+
+
+
+
+
+
+
+
+
+
+/*
+// Definition for a Node.
+class Node {
+public:
+    int val;
+    Node* prev;
+    Node* next;
+    Node* child;
+};
+*/
+
+class Solution {
+public:
+    Node* flatten(Node* head) {
+        Node *curr=head, *tail=head;
+        stack<Node*> s;
+        while(curr!=NULL){
+            if(curr->child!=NULL){
+                if(curr->next!=NULL){
+                    s.push(curr->next);
+                    curr->next->prev=NULL;
+                }
+                Node* child=curr->child;
+                curr->next=child;
+                child->prev=curr;
+                curr->child=NULL;
+            }
+            tail=curr;
+            curr=curr->next;
+        }
+        while(!s.empty()){
+            curr=s.top();
+            s.pop();
+            tail->next=curr;
+            curr->prev=tail;
+            while(curr!=NULL){
+                tail=curr;
+                curr=curr->next;
+            }
+        }
+        return head;
+    }
+};

competitive programming/leetcode/430. Flatten a Multilevel Doubly Linked List.cpp

@@ -0,0 +1,118 @@
+You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.
+
+Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.
+
+
+
+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 as follows:
+
+
+
+After flattening the multilevel linked list it becomes:
+
+
+Example 2:
+
+Input: head = [1,2,null,3]
+Output: [1,3,2]
+Explanation:
+
+The input multilevel linked list is as follows:
+
+  1---2---NULL
+  |
+  3---NULL
+Example 3:
+
+Input: head = []
+Output: []
+
+
+How multilevel linked list is represented in test case:
+
+We use the multilevel linked list from Example 1 above:
+
+ 1---2---3---4---5---6--NULL
+         |
+         7---8---9---10--NULL
+             |
+             11--12--NULL
+The serialization of each level is as follows:
+
+[1,2,3,4,5,6,null]
+[7,8,9,10,null]
+[11,12,null]
+To serialize all levels together we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:
+
+[1,2,3,4,5,6,null]
+[null,null,7,8,9,10,null]
+[null,11,12,null]
+Merging the serialization of each level and removing trailing nulls we obtain:
+
+[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
+
+
+Constraints:
+
+Number of Nodes will not exceed 1000.
+1 <= Node.val <= 10^5
+
+
+
+
+
+
+
+
+
+
+
+/*
+// Definition for a Node.
+class Node {
+public:
+    int val;
+    Node* prev;
+    Node* next;
+    Node* child;
+};
+*/
+
+class Solution {
+public:
+    Node* flatten(Node* head) {
+        Node *curr=head, *tail=head;
+        stack<Node*> s;
+        while(curr!=NULL){
+            if(curr->child!=NULL){
+                if(curr->next!=NULL){
+                    s.push(curr->next);
+                    curr->next->prev=NULL;
+                }
+                Node* child=curr->child;
+                curr->next=child;
+                child->prev=curr;
+                curr->child=NULL;
+            }
+            tail=curr;
+            curr=curr->next;
+        }
+        while(!s.empty()){
+            curr=s.top();
+            s.pop();
+            tail->next=curr;
+            curr->prev=tail;
+            while(curr!=NULL){
+                tail=curr;
+                curr=curr->next;
+            }
+        }
+        return head;
+    }
+};