My very simple Question
    How to fill all 1 between 1 and 1 in an array with only 0 and 1

    Given an array [1 1 0 1 0], fill as following

    0 1 0 1 0 
      ↑   ↑  
      1 1 1
    
 ⟹  0 1 1 1 0

    Given an array [0 1 0 0 1 1 0 1 0 1 0]

    0 1 0 0 1  1 0 1 0 1 0
      ↑     ↑  ↑   ↑   
      1 1 1 1  1 1 1
    0 1 1 1 1  1 1 1 0 1 0
                       ↑ 
                       + → Do nothing for this 1 


    Haskell partial solution (Use scanl1)
Warm up:
    scanl1 (+) [1, 2, 3]
    => 1 3 6
    scanl1 (-) [1, 2, 3]
    => 1 (1-2) (1-2-3)
    => 1  -1     -4

    s = [0, 1, 0, 1, 0]
    t = scanl1 (\a b -> a /= b) s 
    => 0 (0 /= 1)  (0 /= 1 /= 0)  (0 /= 1 /= 0 /= 1) (0 /= 1 /= 0 /= 1 /= 0)
    t = 0    1           1               0                  0

    Given an array s = [0 1 0 1 0]
    Assume there are even # of 1 in the s 
    s = 0  1  0  1  0
    t = 0  1  1  0  0  OR
    - - - - - - - - - - -
        0  1  1  1  0  

    It only works if there are EVEN # of 1 in the array

    r = zipWith(\a b -> a || b) s t
    r = [0 1  1  1  0]

    In APL
    (≠\∨⊣) 0 1 0 1 0
    0 1 1 1 0

    w ← 0 1 0 1 0
      (≠\) w 
    0 1 1 0 0

    w ∨ (≠\) w 
    0 1 1 1 0

    Or oneliner
    (≠\∨⊣) 0 1 0 1 0
    0 1 1 1 0