HackerRank Sparse Arrays problem solution

In this HackerRank Sparse Arrays problem, we need to develop a program in which for each gives string we need to determine how many times the query present in the string, and then we need to return an array as a result.

Problem solution in Python programming.

n = int(input())
hashmap = {}

for i in range(n):
    string = input()
    hashmap[string] = 1 if string not in hashmap else hashmap[string] + 1

q = int(input())

for j in range(q):
    string = input()
    print(0 if string not in hashmap else hashmap[string])

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        HashMap<String,Integer> hm = new HashMap<>();
        for(int i = 0; i < N; ++i) {
            String input = sc.next();
            if(hm.get(input) == null) hm.put(input,1);
            else {
                int val = hm.get(input);
                hm.put(input,++val);
            }
        }

        int Q = sc.nextInt();
        while(Q-->0) {
            String query = sc.next();
            if(hm.get(query) == null) System.out.println(0);
            else System.out.println(hm.get(query));
        }
    }
}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    map<string,int> count;
    int n;
    cin>>n;
    while(n--){
        string s;
        cin>>s;
        count[s]++;
    }
    cin>>n;
    while(n--){
        string s;
        cin>>s;
        //cout<<s<<endl;
        cout<<count[s]<<endl;
    }
    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int N, Q; 
    char *N_array[1000], *Q_array[1000];
    scanf("%d", &N);
    for (int N_i = 0; N_i < N; N_i++) {
        char s[21];
        scanf("%s", s); 
        N_array[N_i] = malloc(21);
        strcpy(N_array[N_i], s);
    }
    scanf("%d", &Q);
    
    for (int Q_i = 0; Q_i < Q; Q_i++) {
        int occurs = 0, result;
        char s[21];
        scanf("%s", s);
        Q_array[Q_i] = malloc(21);
        strcpy(Q_array[Q_i], s);
        for (int N_i2 = 0; N_i2 < N; N_i2++) {
            result = strcmp(Q_array[Q_i], N_array[N_i2]);
            if (result == 0) occurs++;
        }
        printf("%dn", occurs);
    }
    
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    const split = input.split('n');

    const N = parseInt(split[0], 10);    
    var strings = split.slice(1, N + 1);
    
    const Q = parseInt(split[N+1], 10);
    const queries = split.slice(N+2);
    
    const counts = [];
    const countMap = new Map();
    
    queries.forEach(query => {
        if (countMap.has(query)) {
            console.log(countMap.get(query))         
        } else {
            const prevCount = strings.length;
            strings = strings.filter(string => string !== query);
            const nextCount = strings.length;
            const count = prevCount - nextCount;
            countMap.set(query, count);
            console.log(count)          
        }
    });
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});