Codility----MaxCounters

Task description

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.

Write a function:

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

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.

Solution

Programming language used: Java

Total time used: 23 minutes

Code: 01:01:37 UTC, java, final, score: 100

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// 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 size = A.length;
        int max = 0, maxtmp = 0;
        int []res = new int[N];
        boolean allmax = false;
        for(int i=0; i<size; i++){
            if(A[i] <= N) {
                if(res[A[i]-1] > max) {
                    res[A[i]-1] +=1;
                } else {
                    res[A[i]-1] = max + 1;
                }
                if(res[A[i]-1] > maxtmp) {
                    maxtmp = res[A[i]-1];
                }
            }else if(A[i] == N+1) {
                max = maxtmp;
            }
        }
        for(int j=0; j<N; j++){
            if(res[j] < max) {
                res[j] = max;
            }
        }
        return res;
    }
}


https://codility.com/demo/results/trainingF499M5-C2U/