python array challenge

Question

I have this array: [[0,10,20],[1,15,25],[2,20,40]]

0, 1, 2 would be ids and 10,20 15,25 20,40 would be their respective range.

So, I need to figure out the ranges and which ids has which range.

Each id would represent a flashlight, and each one has its own range, I need this to figure out which areas are being illuminated by which flashlights.

I don't know if was clear enough, so here is what I expect to be returned from the code:

[
 [0],[10,15],
 [0,1],[15,20],
 [1,2],[20,25],
 [2], [25,40]
]

Thanks in advance.

I'm confused about exactly what you need to do. What does the output represent? Also, do you need to have the 0, 1, 2 indexes inside the array? You can index the ranges with those numbers already without having them there. — ecline6
– ecline6, Commented Apr 12, 2013 at 5:05

Ilmo Euro · Accepted Answer · 2013-04-12 05:34:36Z

2

First, the real line must be segmented:

min_boundaries = set(min for id, min, max in orig)
max_boundaries = set(max for id, min, max in orig)
segment_boundaries = sorted(list(min_boundaries | max_boundaries))

Then, for each segment, check which lights illuminate that segment. A point in the middle of the segment is enough:

result = []
old_boundary = segment_boundaries[0]
for boundary in segment_boundaries[1:]:
    bounds = [old_boundary, boundary]
    middle = (old_boundary + boundary) / 2
    ids = [id for id, min, max in orig if min < middle < max]
    result.append(ids)
    result.append(bounds)
    old_boundary = boundary

answered Apr 12, 2013 at 5:34

Ilmo Euro

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vivek · Accepted Answer · 2013-04-12 06:43:55Z

0

With some changes in the data structure here is the simple solution:

adict = {0: [10, 20], 1: [15, 25], 2: [20, 40]}
data = [(10, 15), (15, 20), (20, 25), (25, 40)]

# This will check the membership 
# in_([3,4], [2,3]) returns True
def in_(val1 , val2):
    return val2[0] >= val1[0] and val2[1] <= val1[1]


res = {}
for each in keys:
    for key in adict.keys():
        if in_(adict[key], each):
            res.setdefault(each, []).append(key)

O/P

>>>res
{(10, 15): [0], (15, 20): [0, 1], (25, 40): [2], (20, 25): [1, 2]}

answered Apr 12, 2013 at 6:43

vivek

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