MaxCounters solution

Question

I am doing this Codility problem

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.

I have ended up with this code -

def solution(N, array): counters = [0]*N maximum = 0 for a in array: # O(m) as array has m integers if a >=1 and a <= N: counters[a-1] += 1 maximum = max(counters[a-1], maximum) if a == N+1: counters = [maximum]*N return counters 

I am failing two test cases because of timeout errors. I timed the function with array as [10]*100000 and \$N\$ as 9. It took 0.175681062359 seconds which is clearly not desirable. I do not understand where the time complexity increases. The for loop has \$O(M)\$ complexity because array has \$M\$ elements and even though max() has \$O(n)\$ complexity, that doesn't matter since I'm comparing just two elements. I looked at a solution by Andrei Simionescu and it looks awfully similar to mine -

def solution(n, arr): out = [0] * n m = 0 last = 0 for op in arr: op -= 1 if op == n: last = m continue out[op] = max(out[op], last) + 1 m = max(m, out[op]) for i in xrange(n): out[i] = max(out[i], last) return out 

I timed the above code and it took just 0.0276817503901 seconds on the same input. What is it that I'm doing wrong?

Answer 1

if a == N+1: counters = [maximum]*N #array initializer 

Looks \$O(N)\$ to me, making your total time complexity \$O(Nm)\$ worst case. The other guy has 2 separate loops, one \$O(m)\$ the next \$O(n)\$, for a total time complexity of \$O(m+n)\$ or \$O(max(n,m))\$.

Take home message:

Ditch the \$O(N)\$ array initializer, it's not worth it. Use 2 separate passes, like the other guy did.