[해커랭크] Repeated String

[해커랭크] Repeated String

https://www.hackerrank.com/challenges/repeated-string/problem?isFullScreen=true

Repeated String | HackerRank

 

 

 

There is a string, , of lowercase English letters that is repeated infinitely many times. Given an integer, , find and print the number of letter a's in the first  letters of the infinite string.

Example

The substring we consider is , the first  characters of the infinite string. There are  occurrences of a in the substring.

Function Description

Complete the repeatedString function in the editor below.

repeatedString has the following parameter(s):

• s: a string to repeat
• n: the number of characters to consider

Returns

• int: the frequency of a in the substring

Input Format

The first line contains a single string, .
The second line contains an integer, .

Constraints

•  
•  
• For  of the test cases, .

Sample Input

Sample Input 0

aba
10

Sample Output 0

7

Explanation 0
The first  letters of the infinite string are abaabaabaa. Because there are  a's, we return .

Sample Input 1

a
1000000000000

Sample Output 1

1000000000000

Explanation 1
Because all of the first  letters of the infinite string are a, we return .

 

 

 

 

그냥 전체 문자열에서 'a' 개수 × 반복 횟수 + 나머지만 계산하면 됨

단순히 반복 구조 + 나머지 처리로 해결 가능해서 DP가 필요 없음

 

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 'repeatedString' function below.

     *

     * The function is expected to return a LONG_INTEGER.

     * The function accepts following parameters:

     *  1. STRING s

     *  2. LONG_INTEGER n

     */

    public static long repeatedString(String s, long n) {

        long aCount = 0;

        for (int i = 0; i < s.length(); i++) {

            if (s.charAt(i) == 'a') aCount++;

        }

        long repeatCount = n / s.length();

        long result = repeatCount * aCount;

        int reaminCount = (int) (n % s.length());

        for (int i = 0; i < reaminCount; i++) {

            if (s.charAt(i) == 'a') result++;

        }

        return result;

    }

}

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")));

        String s = bufferedReader.readLine();

        long n = Long.parseLong(bufferedReader.readLine().trim());

        long result = Result.repeatedString(s, n);

        bufferedWriter.write(String.valueOf(result));

        bufferedWriter.newLine();

        bufferedReader.close();

        bufferedWriter.close();

    }

}