Codility - Lesson 4 Counting elements - 4. MaxCounters

Source Link:
https://codility.com/programmers/lessons/4-counting_elements/max_counters/

Question:

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. 

My Solution:

Notes:
1. key point: maintain the max value
int max = 0;
2. key point: maintain the current_min (very important!!!)
// so, we can move "the 2nd for-loop" outside "the 1st for-loop"  
// by maintaining "min" 
3. new integer array
int[] my_array = new int[N]; 
4.
for(int i=0; i<A.length; i++){
if( A[i] >= 1 && A[i] <= N){ // normal case
...
}
else if( A[i] == N+1){ // special case 
...
}
}
5.  move the 2nd for-loop outside the 1st for-loop  
// update some elements who have not been updated yet
for(int j=0; j<my_array.length; j++){
if(my_array[j] < min){
my_array[j] = min; // update it to "min"
}
}