MaxCounters solution
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.