I've been doing some Codility challenges lately and the MaxCounters challenge got me stumped:

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.

Below is my solution in Python. I can't reach 100% on it even though I am pretty confident it is an O(n+m) complexity solution. Codility fails it and says its an O(n*m) complexity solution.

def solution(N, A):
 counter = [0] * N
 max_val = 0
 for v in A:
 if v <= N and v >= 1:
 if counter[v-1] < max_val+1:
 counter[v-1] = max_val+1
 else:
 counter[v-1] += 1
 else:
 max_val = max(counter)
 for i in range(len(counter)):
 if counter[i] < max_val:
 counter[i] = max_val
 return counter

I've been doing some Codility challenges lately and the MaxCounters challenge got me stumped:

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.

[...] The goal is to calculate the value of every counter after all operations.

Below is my solution in Python. I can't reach 100% on it even though I am pretty confident it is an O(n+m) complexity solution. Codility fails it and says its an O(n*m) complexity solution.

def solution(N, A):
 counter = [0] * N
 max_val = 0
 for v in A:
 if v <= N and v >= 1:
 if counter[v-1] < max_val+1:
 counter[v-1] = max_val+1
 else:
 counter[v-1] += 1
 else:
 max_val = max(counter)
 for i in range(len(counter)):
 if counter[i] < max_val:
 counter[i] = max_val
 return counter

Been doing some codility challenges lately and this last one got me stunned.. i cant reach 100% on it even though i am pretty confident it is an O(n+m) complexity solution, codility fails it and says its an O(n*m) complexity solution.

this is the challenge in 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 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.

And below my solution in python.

def solution(N, A):
 counter = [0] * N
 max_val = 0
 for v in A:
 if v <= N and v >= 1:
 if counter[v-1] < max_val+1:
 counter[v-1] = max_val+1
 else:
 counter[v-1] += 1
 else:
 max_val = max(counter)
 for i in range(len(counter)):
 if counter[i] < max_val:
 counter[i] = max_val
 return counter

I've been doing some Codility challenges lately and the MaxCounters challenge got me stumped:

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.

Below is my solution in Python. I can't reach 100% on it even though I am pretty confident it is an O(n+m) complexity solution. Codility fails it and says its an O(n*m) complexity solution.

def solution(N, A):
 counter = [0] * N
 max_val = 0
 for v in A:
 if v <= N and v >= 1:
 if counter[v-1] < max_val+1:
 counter[v-1] = max_val+1
 else:
 counter[v-1] += 1
 else:
 max_val = max(counter)
 for i in range(len(counter)):
 if counter[i] < max_val:
 counter[i] = max_val
 return counter

Been doing some codility challenges lately and this last one got me stunned.. i cant reach 100% on it even though i am pretty confident it is an O(n+m) complexity solution, codility fails it and says its an O(n*m) complexity solution.

this is the challenge in question: "https://app.codility.com/programmers/lessons/4-counting_elements/max_counters/"

And below my solution in python. Appreciate any help!

def solution(N, A):
counter = [0] * N
max_val = 0
for v in A:
 if v <= N and v >= 1:
 if counter[v-1] < max_val+1:
 counter[v-1] = max_val+1
 else:
 counter[v-1] += 1
 else:
 max_val = max(counter)
for i in range(len(counter)):
 if counter[i] < max_val:
 counter[i] = max_val
return counter

Been doing some codility challenges lately and this last one got me stunned.. i cant reach 100% on it even though i am pretty confident it is an O(n+m) complexity solution, codility fails it and says its an O(n*m) complexity solution.

this is the challenge in 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 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.

And below my solution in python.

def solution(N, A):
counter = [0] * N
max_val = 0
for v in A:
 if v <= N and v >= 1:
 if counter[v-1] < max_val+1:
 counter[v-1] = max_val+1
 else:
 counter[v-1] += 1
 else:
 max_val = max(counter)
for i in range(len(counter)):
 if counter[i] < max_val:
 counter[i] = max_val
return counter