Calculating the time complexity of this specific code

Question

Im trying to figure out the time complexity of the following code. lst_of_lsts contains m lists whose lengths are at most n.

This is what I think: all runs in O(m*n), minimum = O(m*n), remove = O(m*n).

Overall O(m*n)*O(m*n) = O(m^2*n^2)

Am I right?

 def multi_merge_v1(lst_of_lsts): 
      all = [e for lst in lst_of_lsts for e in lst] 
      merged = [] 
      while all != []: 
          minimum = min(all) 
          merged += [minimum] 
          all.remove(minimum) 
      return merged

Isn't this code review?

sshashank124
– sshashank124

2014-04-06 14:52:49 +00:00
Commented Apr 6, 2014 at 14:52 — sshashank124
– sshashank124, Commented Apr 6, 2014 at 14:52

Ilia Lebedev · Accepted Answer · 2014-04-06 15:08:28Z

1

def multi_merge_v1(lst_of_lsts): # M lists of N
  all = [e for lst in lst_of_lsts for e in lst]
  # this is messy. Enumerate all N*M list items --> O(nm)

  merged = [] # O(1)
  while all != []: # n*m items initially, decreases in size by 1 each iteration.
    minimum = min(all) # O(|all|).
    # |all| diminishes linearly from (nm) items, so this is O(nm).
    # can re-use work here. Should use a sorted data structure or heap

    merged += [minimum] # O(1) amortized due to table doubling
    all.remove(minimum) # O(|all|)
  return merged # O(1)

The overall complexity appears to be O(nm + nm*nm + 1) = O(n^2m^2+1). Yikes.

answered Apr 6, 2014 at 15:08

Ilia Lebedev

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