Equal Beauty CodeChef SnackDown 2021 Round 1A Question The beauty of an (non-empty) array of integers is defined as the difference between its largest and smallest element. For example, the beauty of the array [2,3,4,4,6] is 6−2=4. An array A is said to be good if it is possible to partition the elements of A into two non-empty arrays B1 and B2 such that B1 and B2 have the same beauty. Each element of array A should be in exactly one array: either in B1 or in B2. For example, the array [6,2,4,4,4] is good because its elements can be partitioned into two arrays B1=[6,4,4] and B2=[2,4], where both B1 and B2 have the same beauty (6−4=4−2=2). You are given an array A of length N. In one move you can: Select an index i (1≤i≤N) and either increase Ai by 1 or decrease Ai by 1. Find the minimum number of moves required to make the array A good. Input Format The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follow. Each test
Construct Binary Search Tree from Preorder Traversal LeetCode Solution
Question
Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.
It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.
A binary search tree is a binary tree where for every node, any descendant of Node.left
has a value strictly less than Node.val
, and any descendant of Node.right
has a value strictly greater than Node.val
.
A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left
, then traverses Node.right
.
Example 1:
Input: preorder = [8,5,1,7,10,12] Output: [8,5,10,1,7,null,12]
Example 2:
Input: preorder = [1,3] Output: [1,null,3]
Constraints:
1 <= preorder.length <= 100
1 <= preorder[i] <= 108
- All the values of
preorder
are unique.
Explanation
Constructing Binary Search Tree(BST) from its traversal is a very common question. Here it is given from preorder traversal this could also be done with postorder traversal. We know that for the Binary Search Tree the inorder traversal is always sorted, so using this approach we can easily find the binary search tree. Consider that values less than the root value are present in the left subtree and values greater than the root node is present in the right subtree. This logic is true for every node of the Binary Search Tree. Thus its easy to solve now just code it and print the BST. I recommend to give a try first if you are stuck then utilize the solution given below in Java, C++ and Python.
More information about LeetCode Daily Challenge
More Information About LeetCode Daily Challenge
LeetCode is a great coding platform which provides daily problems to solve and maintain a streak of coding. This platform helps in improvement of coding skills and attempting the daily challenge makes our practice of coding regular. With its bunch of questions available it is one of the best coding platforms where you can have self improvement. LeetCode also holds weekly and biweekly contests which is a great place to implement you knowledge and coding skills which you learnt through out the week by performing the daily challenge. It also provides badges if a coder solves all the daily problems of a month. For some lucky winners T-shirts are also given and that too for free. If you get the pro version you can get more from LeetCode like questions that are asked in interviews, detailed explanation of each chapter and each concept and much more. LeetCode challenges participants with a problem from our carefully curated collection of interview problems every 24 hours.
All users with all levels of coding background are welcome to join!
@PREMIUM USERS will have an extra carefully curated premium challenge weekly and earn extra LeetCoin rewards. The premium challenge will be available together with the first October challenge of the week and closed together with the November challenge of the week.
Starting from 2021, users who completed all Daily LeetCoding Challenge non-premium problems for the month (with or without using Time Travel Tickets) will win a badge to recognize their consistency! Your badges will be displayed on your profile page. Completing each non-premium daily challenge. (+10 LeetCoins)
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Lucky Draw: Those who complete all 31 non-premium daily challenges will be automatically entered into a Lucky Draw, where LeetCode staff will randomly select 3 lucky participants to each receive one LeetCode Polo Shirt on top of their rewards! Do you have experience trying to keep a streak but failed because you missed one of the challenges?
Hoping you can travel back to the missed deadline to complete the challenge but that's not possible......or is it?
With the new Time Travel Ticket , it's possible! In order to time travel, you will need to redeem a Time Travel Ticket with your LeetCoins. You can redeem up to 3 tickets per month and the tickets are only usable through the end of the month in which they are redeemed but you can use the tickets to make up any invalid or late submission of the current year. To join, just start solving the daily problem on the calendar of the problem page. You can also find the daily problem listed on the top of the problem page. No registration is required. The daily problem will be updated on every 12 a.m. Coordinated Universal Time (UTC) and you will have 24 hours to solve that challenge. Thar's all you needed to know about Daily LeetCode Challenge.
Program Code
Java
class Solution {
public TreeNode bstFromPreorder(int[] preorder) {
TreeNode root=new TreeNode(preorder[0]);
for (int i=1; i<preorder.length; i++) {
root=addNode(root,preorder[i]);
}
return root;
}
static TreeNode addNode(TreeNode root, int val) {
if (root==null) {
TreeNode node=new TreeNode(val);
return node;
}
if (val<root.val) {
root.left=addNode(root.left,val);
} else {
root.right=addNode(root.right,val);
}
return root;
}
}
C++
class Solution {
static void insert(TreeNode** pnode, int n)
{
if(*pnode == nullptr) {
*pnode = new TreeNode(n);
return;
}
if(n < (*pnode)->val)
insert(&(*pnode)->left, n);
else
insert(&(*pnode)->right, n);
}
public:
TreeNode* bstFromPreorder(vector<int>& preorder) {
TreeNode* root = nullptr;
for(int n : preorder)
insert(&root, n);
return root;
}
};
Python
import bisect
class Solution:
def bstFromPreorder(self, preorder: List[int]) -> Optional[TreeNode]:
if not preorder:
return None
val = preorder.pop(0)
root = TreeNode(val=val)
idx = bisect.bisect_left(preorder, val)
root.left = self.bstFromPreorder(preorder[:idx])
root.right = self.bstFromPreorder(preorder[idx:])
return root
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