Algorithm

Problem Name: Data Structures - Sparse Arrays

Problem Link: https://www.hackerrank.com/challenges/sparse-arrays/problem?isFullScreen=true

In this HackerRank in Data Structures - Sparse Arrays solutions

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results.

 

 

Example

stringList = ['ab',ab', abc']

queries = ['ab',abc',bc']

There are 2 instances of 'ab' , 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, result = [2,1,0]

Function Description

 

Complete the function matchingStrings in the editor below. The function must return an array of integers representing the frequency of occurrence of each query string in stringList.

 

matchingStrings has the following parameters:

 

  • string stringList[n] - an array of strings to search
  • string queries[q] - an array of query strings

 

Returns

 

  • int[q]: an array of results for each query

Input Format

The first line contains and integer n, the size of stringList[].

 

Each of the next n lines contains a string stringList[i].

The next line contains q, the size of queries[].

Each of the next q lines contains a string queries[i].

Constraints

1 <= n <= 1000

1 <= q <= 1000

1 <= stringList[i],queries[i] <= 20

Sample Input 1

CopyDownload
Array: stringList
aba

baba

aba

xzxb



 






Array: queries




aba

xzxb
ab
4
aba
baba
aba
xzxb
3
aba
xzxb
ab

Sample Output 1

2
1
0

 

 

 

Code Examples

#1 Code Example with C Programming

Code - 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("%d\n", occurs);
    }
    
    return 0;
}
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#2 Code Example with C++ Programming

Code - 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;
}
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#3 Code Example with Java Programming

Code - 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));
        }
    }
}
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#4 Code Example with Javascript Programming

Code - 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);
});
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#5 Code Example with Python Programming

Code - 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])
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