{id: 1,  point: [1,1], available: 1}, {id: 2,  point: [1,2], available: 1}, {id: 3,  point: [1,3], available: 1}, {id: 4,  point: [1,4], available: 0}, {id: 5,  point: [1,5], available: 1}, {id: 6,  point: [1,6], available: 1},
{id: 7,  point: [2,1], available: 1}, {id: 8,  point: [2,2], available: 1}, {id: 9,  point: [2,3], available: 0}, {id: 10, point: [2,4], available: 1}, {id: 11, point: [2,5], available: 1}, {id: 12, point: [2,6], available: 1},
{id: 13, point: [3,1], available: 1}, {id: 14, point: [3,2], available: 0}, {id: 15, point: [3,3], available: 1}, {id: 16, point: [3,4], available: 1}, {id: 17, point: [3,5], available: 1}, {id: 18, point: [3,6], available: 1},
{id: 19, point: [4,1], available: 1}, {id: 20, point: [4,2], available: 1}, {id: 21, point: [4,3], available: 0}, {id: 22, point: [4,4], available: 1}, {id: 23, point: [4,5], available: 1}, {id: 24, point: [4,6], available: 0},

I want 3 contigous block where available is 0. Contigous meaning that given entry a,b,c, a.point[0] = b.point[0] - 1 = c.point[0] - 2 (the analogous for point[1] may also hold if it's easier).  For this query above, I would expect to return at least:

(lists of ids)
[1,2,3]
[10,11,12]
[15,16,17]
[16,17,18]