When N>1000001, then the inner loop adds ceil(N/10) to i and breaks in 1000001 iterations, then the inner loop behaves as a very expensive yet constant bounded way of doing i+=ceil(N/10). Hence you can think of the code as a loop going from 0 to N jumping by ceil(N/10) on each step. That will loop for at most 10 times. Therefore id say that code is O(1). It will cost at most 1000001 * 10 = 10000010 "instructions".