medium
Calculate the values of counters after applying all alternating operations: increase counter by 1; set value of all counters to current maximum.
100%
Correctness
100%
Performance
100%

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.

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) (not counting the storage required for input arguments).
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Solution
Programming language used Java
Total time used 2 minutes
Effective time used 2 minutes
Notes
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