[코딜리티] codility lesson 4 Counting Elements - MaxCounters 100%

문제.

Programming language:                 C                        C++                        C#                        Go                        Java                        JavaScript                        Lua                        Objective-C                        Pascal                        PHP                        Perl                        Python                        Ruby                        Scala                        Swift 2                        Swift 3                        VB.NET                    

You are given N counters, initially set to 0, and you have two possible operations on them:

• increase(X) − counter X is increased by 1,
• max counter − all counters are set to the maximum value of any counter.

A non-empty zero-indexed array A of M integers is given. This array represents consecutive operations:

• if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
• if A[K] = N + 1 then operation K is max counter.

For example, given integer N = 5 and array A such that:

A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4

the values of the counters after each consecutive operation will be:

(0, 0, 1, 0, 0) (0, 0, 1, 1, 0) (0, 0, 1, 2, 0) (2, 2, 2, 2, 2) (3, 2, 2, 2, 2) (3, 2, 2, 3, 2) (3, 2, 2, 4, 2)

The goal is to calculate the value of every counter after all operations.

Assume that the following declarations are given:

struct Results { int * C; int L; };

Write a function:

struct Results solution(int N, int A[], int M);

that, given an integer N and a non-empty zero-indexed array A consisting of M integers, returns a sequence of integers representing the values of the counters.

The sequence should be returned as:

• a structure Results (in C), or
• a vector of integers (in C++), or
• a record Results (in Pascal), or
• an array of integers (in any other programming language).

For example, given:

A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4

the function should return [3, 2, 2, 4, 2], as explained above.

Assume that:

• N and M are integers within the range [1..100,000];
• each element of array A is an integer within the range [1..N + 1].

Complexity:

• expected worst-case time complexity is O(N+M);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Copyright 2009–2017 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

풀이.

12345678910111213141516171819202122232425

// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
    public int[] solution(int N, int[] A) {
        // write your code in Java SE 8
        int[] counter = new int[N];
        int max = 0;
        int check = 0;
        

        for(int i = 0 ; i < A.length ; i++){
            if(A[i] == N+1) max = check;
            else{
                if(counter[A[i]-1] < max)  counter[A[i]-1] = max;
                counter[A[i]-1]++;
                if(check < counter[A[i]-1]) check = counter[A[i]-1];
            }
        }

        for(int i = 0 ; i < counter.length ; i++) if(counter[i] < max)  counter[i] = max;
        return counter;
    }
}

설명.

N개의 카운터기를 만들고 A 배열에서 카운터기 번호가 나오면 카운터를 하나 증가시킨다.
만약 N+1 의 카운터기 번호가 나오면 모든 카운터의 수를 가장높은 카운터기 번호로 만든가.

함정. 배열초기화를 시도하면 타임아웃에 걸림.