[Codility] Lesson4 - MaxCounters

Task discription

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 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.

Write a function:

class Solution { public int[] solution(int N, int[] A); }

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

Result array should be returned as an array of integers.

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.

Write an efficient algorithm for the following assumptions:

• 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].

Source Code

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

vector<int> solution(int N, vector<int> &A) {

    vector<int> v(N);
    int max = 0, tmpMax = 0;

    fill(v.begin(), v.end(), 0);

    for(int i=0;i<A.size();i++) {
        if(A[i] == N+1) {
            max = tmpMax;
        }
        else {
            // max보다 작으면 max+1
            if(v[A[i]-1] < max) {
                v[A[i]-1] = max+1;
            }
            // max 이상이면 +1
		    else {
                v[A[i]-1]++;
            }

            // 최댓값 갱신
            if(v[A[i]-1] > tmpMax) {
                tmpMax = v[A[i]-1];
            }
        }
    }

    for(int i=0;i<N;i++){
        // max보다 작은 값들 max로 채움
	    if(v[i] < max) {
            v[i] = max;
        }
	}

    return v;
}

Wrong Answer

max counter일 때(N+1) max로 모든 값을 갱신해주는 방식으로 풀었더니 Performance 부분에서 오답 처리가 되었다.

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

vector<int> solution(int N, vector<int> &A) {

    vector<int> v(N);
    int max = 0;
    fill(v.begin(), v.end(), 0);
    for(int i=0;i<A.size();i++) {
        int a = A[i];
        if(a == N+1) {
            fill(v.begin(), v.end(), max);
        }
        else {
            v[a-1] += 1;
            if(v[a-1] > max) {
            	max = v[a-1];
            }
        }
    }

    return v;
}