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Count the number of zeros..fill the array with count of zeros and remaining with 1's..

Above approach does involve two traversals of the array.
Alternative:
Use two pointers at the two ends of the array. Increment left pointer till you encounter a 1, Decrement right pointer till you encounter a 0 (just like Quick Sort).
If left pointer < right pointer, swap the elements.

here it is O(n/2) = O(n)

def segBinary():
  e = [0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0]
  
  j = len(e)-1
  for i in range(0, len(e)/2):
    if e[i] == 1:
      while e[j] != 0:
        j-=1
      one = e[i]
      e[i] = e[j]
      e[j] = one
  print e