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As per the instructions are given in MaxCounters-Codility,

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.

I have written this code

public int[] maxCount(int[]A,int N) {
 int[] I = new int[N];
 for (int i = 0; i < A.length; i++) {
 try {
 I[A[i] - 1]++;
 } catch (Exception e) {
 Arrays.sort(I);
 Arrays.fill(I, I[I.length - 1]);
 }
 }
 return I;
}

It gives correct answers for all test cases. Any Idea to do this with time complexity O(N).Its currently on O(N*M).

As per the instructions are given in MaxCounters-Codility,

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.

I have written this code

public int[] maxCount(int[]A,int N) {
 int[] I = new int[N];
 for (int i = 0; i < A.length; i++) {
 try {
 I[A[i] - 1]++;
 } catch (Exception e) {
 Arrays.sort(I);
 Arrays.fill(I, I[I.length - 1]);
 }
 }
 return I;
}

It gives correct answers for all test cases. Any Idea to do this with time complexity O(N). Its currently on O(N*M).

As per the instructions are given in MaxCounters-Codility, I have written the code

public int[] maxCount(int[]A,int N) {
 int[] I = new int[N];
 for (int i = 0; i < A.length; i++) {
 try {
 I[A[i] - 1]++;
 } catch (Exception e) {
 Arrays.sort(I);
 Arrays.fill(I, I[I.length - 1]);
 }
 }
 return I;
}

It gives correct answers for all test cases. Any Idea to do this with time complexity O(N).Its currently on O(N*M).

As per the instructions are given in MaxCounters-Codility,

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.

I have written this code

public int[] maxCount(int[]A,int N) {
 int[] I = new int[N];
 for (int i = 0; i < A.length; i++) {
 try {
 I[A[i] - 1]++;
 } catch (Exception e) {
 Arrays.sort(I);
 Arrays.fill(I, I[I.length - 1]);
 }
 }
 return I;
}

It gives correct answers for all test cases. Any Idea to do this with time complexity O(N).Its currently on O(N*M).