Algorithm


Problem Name: Data Structures - Contacts

Problem Link: https://www.hackerrank.com/challenges/contacts/problem?isFullScreen=true

In this HackerRank in Data Structures - Contacts solutions

We're going to make our own Contacts application! The application must perform two types of operations:

  1. add name, where name is a string denoting a contact name. This must store name as a new contact in the application.
  2. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting with partial and print the count on a new line.

Given n sequential add and find operations, perform each operation in order.

Example
Operations are requested as follows:

 

add ed
add eddie
add edward
find ed
add edwina
find edw
find a

Three add operations include the names 'ed', 'eddie', and 'edward'. Next, find ed matches all 3 names. Note that it matches and counts the entire name 'ed'. Next, add 'edwina' and then find 'edw'. 2 names match: 'edward' and 'edwina'. In the final operation, there are 0 names that start with 'a'. Return the array [3,2,0]

Function Description

 

Complete the contacts function below.

 

contacts has the following parameters:

 

  • string queries[n]: the operations to perform

 

Returns

 

  • int[]: the results of each find operation

Input Format

The first line contains a single integer, n the number of operations to perform (the size of queries[])

Each of the following n lines contains a string queries[i]

Constraints

  • 1 <= n <= 10**5
  • 1 <= length of name <= 21
  • 1 <= length of partial <= 21
  • name and partial contain lowercase English letters only.
  • The input does not have any duplicate name for the operation.

Sample Input

STDIN           Function
-----           --------
4               queries[] size n = 4
add hack        queries = ['add hack', 'add hackerrank', 'find hac', 'find hak']
add hackerrank
find hac
find hak

Sample Output

2
0

Explanation

  1. Add a contact named hack.
  2. Add a contact named hackerrank.
  3. Find the number of contact names beginning with hac. Both name start with hac, add 2 to the return array.
  4. Find the number of contact names beginning with hak. neither name starts with hak, add 0 to the return array.

 

 

 

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming


#include <bits/stdc++.h>
using namespace std;
typedef long long ll; 
unordered_map < string,int> s;
int main(int argc, char** argv) {
    int n; cin>>n;
    while(n--){
        string t,x; cin>>t>>x;
        if(t=="add"){
            for(int i = 1; i <= x.length(); i++){
                s[x.substr(0,i>]++;
            }
        }else{
            cout << s[x] << endl;
        }
    }
    return 0;
}
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#3 Code Example with Javascript Programming

Code - Javascript Programming


'use strict';

const fs = require('fs');

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

process.stdin.on('data', function(inputStdin) {
    inputString += inputStdin;
});

process.stdin.on('end', function() {
    inputString = inputString.split('\n');

    main();
});

function readLine() {
    return inputString[currentLine++];
}

/*
 * Complete the 'contacts' function below.
 *
 * The function is expected to return an INTEGER_ARRAY.
 * The function accepts 2D_STRING_ARRAY queries as parameter.
 */

function contacts(queries) {
    class TrieNode {
        children = {}
        isWord = false
        count = 0 // -- we count while adding to be fast enough
    }
    class Trie {
        root
        
        constructor() {
            this.root = new TrieNode()
        }

        insert(word) {
            let node = this.root
            let last = node
            for (const l of word) {
                if (!node.children[l]) {
                    node.children[l] = new TrieNode();
                }
                last = node
                node = node.children[l]
                node.count++ // counting while adding
            }
            last.isWord = true;
        }
        findNode(word) {
            let node = this.root
            let last = node
            for (const l of word) {
                if (!node.children[l])
                    return false
                last = node
                node = node.children[l]
            }
            return node
        }
        // count(node) { -- this is not fast enough
        //     if (!node) return 0
            
        //     let count = 0
        //     if (node.isWord) count++
        //     for (const l in node.children)
        //         count += this.count(node.children[l])
            
        //     return count
        // }
    }
    
    let trie = new Trie()
    let ret = []
    for (const [command, word] of queries) {
        if (command === 'add') {
            trie.insert(word)
        } else if (command === 'find') {
            let node = trie.findNode(word)
            //let count = trie.count(node) -- this is not fast enough
            let count = node.count || 0
            ret.push(count)
        }
    }
    return ret
}

function main() {
    const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

    const queriesRows = parseInt(readLine().trim(), 10);

    let queries = Array(queriesRows);

    for (let i = 0; i  <  queriesRows; i++) {
        queries[i] = readLine().replace(/\s+$/g, '').split(' ');
    }

    const result = contacts(queries);

    ws.write(result.join('\n') + '\n');

    ws.end();
}
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#4 Code Example with Python Programming

Code - Python Programming


#!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'contacts' function below.
#
# The function is expected to return an INTEGER_ARRAY.
# The function accepts 2D_STRING_ARRAY queries as parameter.
#

def contacts(queries):
    mpp, res = {}, []
  
    for cmd,word in queries :
        if cmd == 'add' :
            for i in range(len(word)) :
                mpp[word[:i+1]] = mpp.get(word[:i+1],0)+1
        
        else :
            ans = 0
            ans += mpp.get(word,0)
            res.append(ans)
            
    return res

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    queries_rows = int(input().strip())

    queries = []

    for _ in range(queries_rows):
        queries.append(input().rstrip().split())

    result = contacts(queries)

    fptr.write('\n'.join(map(str, result)))
    fptr.write('\n')

    fptr.close()
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#4 Code Example with Java Programming

Code - Java Programming


import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'contacts' function below.
     *
     * The function is expected to return an INTEGER_ARRAY.
     * The function accepts 2D_STRING_ARRAY queries as parameter.
     */

   public static List < Integer> contacts(List results = new ArrayList();
        Map namePartials = new HashMap();
        
        for (List < String> list : queries) {
            String operation = list.get(0);
            String text = list.get(1);
            
            if (operation.equals("add")) {
                for (int i = 1; i  < = text.length(); ++i) {
                    String partial = text.substring(0, i);
                    Integer count = namePartials.get(partial);
                    if (count == null) {
                        count = 0;
                    }
                    count++;
                    namePartials.put(partial, count);
                }               
            } else if (operation.equals("find")) {
                Integer count = namePartials.get(text);
                if (count == null) {
                    count = 0;
                }
                results.add(count);
            } else {
                throw new RuntimeException("operation " + operation + " is not supported!");
            }
        }
    
        return results;
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int queriesRows = Integer.parseInt(bufferedReader.readLine().trim());

        List < List();

        IntStream.range(0, queriesRows).forEach(i -> {
            try {
                queries.add(
                    Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
                        .collect(toList())
                );
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        });

        List < Integer> result = Result.contacts(queries);

        bufferedWriter.write(
            result.stream()
                .map(Object::toString)
                .collect(joining("\n"))
            + "\n"
        );

        bufferedReader.close();
        bufferedWriter.close();
    }
}
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