Tech It Yourself
tháng 1 19, 2020
0
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:
vector<int> solution(int N, vector<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 a vector 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].
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:
vector<int> solution(int N, vector<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 a vector 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].
vector<int> solution(int N, vector<int> &A) { // write your code in C++11 (g++ 4.8.2) vector<int> res(N, 0); int base = 0; int maximum = 0; for (int i = 0; i < int(A.size()); i++) { if (A[i] != N + 1) { res[A[i] - 1] = max(base, res[A[i] - 1]) + 1; maximum = max(maximum, res[A[i] - 1]); } else { base = maximum; } } for (int i = 0; i < N; i++) { if (res[i] < base) { res[i] = base; } } return res; }
