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Equal Beauty CodeChef SnackDown 2021 Round 1A

 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
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Implement Trie Prefix Tree LeetCode Solution

 Implement Trie Prefix Tree LeetCode Solution

Implement Trie Prefix Tree LeetCode Solution


Question

A trie (pronounced as "try") or prefix tree is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.

Implement the Trie class:

  • Trie() Initializes the trie object.
  • void insert(String word) Inserts the string word into the trie.
  • boolean search(String word) Returns true if the string word is in the trie (i.e., was inserted before), and false otherwise.
  • boolean startsWith(String prefix) Returns true if there is a previously inserted string word that has the prefix prefix, and false otherwise.

 

Example 1:

Input
["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
[[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
Output
[null, null, true, false, true, null, true]

Explanation
Trie trie = new Trie();
trie.insert("apple");
trie.search("apple");   // return True
trie.search("app");     // return False
trie.startsWith("app"); // return True
trie.insert("app");
trie.search("app");     // return True

 

Constraints:

  • 1 <= word.length, prefix.length <= 2000
  • word and prefix consist only of lowercase English letters.
  • At most 3 * 104 calls in total will be made to insert, search, and startsWith.

Explanation

This problem is quite hard to understand since to solve this problem you need to know thoroughly about trees and their implementation. Trie is a very important data structure that is one of the most efficient methods for searching. Trie is a tree structure that has n number of nodes and has a flag value taht checks whether the word is end or not and another array that contains the tree structure. We are now constructing the functions insert, search, startwith. Now the main thing is we insert values and search for a specified value and traverse the entire tree. If no child is present for a specified node print null otherwise if child is present then check whether it is same as the specified node or not print true or false according to the check. That's all now you need to code it. I recommend first to give a try to the problem and then if you are stuck then visit the solution given below in Java, C++, Python.    

Program Code

Java

class Trie {
    
    class Node {
        Node [] childs;
        boolean isEnd;
        
        Node(){
            childs = new Node[26];
            isEnd = false;
        }
    }
    
    final private Node root;
    
    public Trie() {
        root = new Node();
    }
    
 
    public void insert(String word) {
        Node curr = root;
        
        for(int i = 0;i<word.length();i++){
            char ch = word.charAt(i);
            
            if(curr.childs[ch - 'a'] == null){
                curr.childs[ch - 'a'] = new Node();
            }
            curr = curr.childs[ch - 'a'];
        }
        
        curr.isEnd = true;
    }
    
 
    public boolean search(String word) {
        Node curr = root;
        
        for(int i = 0;i<word.length();i++){
            char ch = word.charAt(i);
            
            if(curr.childs[ch - 'a'] == null) return false;
            curr = curr.childs[ch - 'a'];
        }
        return curr.isEnd;
    }
    
  
    public boolean startsWith(String prefix) {
          Node curr = root;
        
        for(int i = 0;i<prefix.length();i++){
            char ch = prefix.charAt(i);
            
            if(curr.childs[ch - 'a'] == null) return false;
            curr = curr.childs[ch - 'a'];
        }
        
        return true;
    }
}

C++

class TrieNode
{
public: 
    bool isEnd;
    vector<TrieNode*> children;
    TrieNode()
    {
        isEnd = false;
        for(int i = 0; i < 26; i++)
            children.push_back(nullptr);
    } 
};

class Trie 
{
public:
    TrieNode* root;
    Trie()
    {
        root = new TrieNode();
    }

    
    void insert(string word) 
    {
        TrieNode* ptr = root;
        for(char& c : word){
            if(!(ptr -> children[c - 'a']))
                ptr -> children[c-'a'] = new TrieNode();
            ptr = ptr -> children[c-'a'];
        }
        ptr -> isEnd = true;        // mark the end of a word
    }
    
    bool search(string word) 
    {
        TrieNode* ptr = root;
        for(char&c : word){
            if(!(ptr -> children[c-'a']))
                return false;
            ptr = ptr -> children[c-'a'];
        }
        return ptr -> isEnd;
    }
    
    bool startsWith(string prefix) 
    {
        TrieNode* ptr = root;
        for(char& c : prefix){
            if(!(ptr -> children[c-'a']))
                return false;
            ptr = ptr -> children[c-'a'];
        }
        return true;
    }
};

Python

import collections

class Trie:
    def __init__(self):
        self.children = collections.defaultdict(Trie)
        self.end = False
        
    def insert(self, word: str) -> None:
        curr = self
        for char in word:
            curr = curr.children[char]
        curr.end = True 
        
    def search(self, word: str) -> bool:
        curr = self
        for char in word:
            if char not in curr.children: return False
            curr = curr.children[char]
        return curr.end

    def startsWith(self, prefix: str) -> bool:
        curr = self
        for char in prefix:
            if char not in curr.children: return False
            curr = curr.children[char]
        return True

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)
Completing 25 to 30 non-premium daily challenges without using Time Travel Ticket will be eligible for an additional 25 LeetCoins. (Total = 275 to 325 LeetCoins)
Completing all 31 non-premium daily challenges without using Time Travel Ticket will be eligible for an additional 50 LeetCoins, plus a chance to win a secret prize ( Total = 385 LeetCoins + *Lucky Draw)!
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.

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