NetApp Interview Question for SDE-3s


Country: India




Comment hidden because of low score. Click to expand.
-1
of 1 vote

Go through the critical points - that is - end or start of range and check if the kth order statistic lies inside rectangle between this and previous critical point.

"""
0123456789ABC
XXXXXXXXXXXXX
   XXXX   XXX
          XX
"""
def kth(ranges, k):
    assert k >= 0
    events = []
    for start, end in ranges:
        events.extend([(start, 1), (end + 1, 0)])
    events.sort()
    depth = 0
    previous_pos = 0
    for pos, is_start in events:
        length = pos - previous_pos
        area = length * depth
        if depth and k < area:
            return previous_pos + k / depth
        k -= area
        if is_start:
            depth += 1
        else:
            depth -= 1
        previous_pos = pos
    raise ValueError("k is larger than length by", k)

def kth_stupid(ranges, k):
    lst = sum((range(s, e + 1) for (s, e) in ranges), [])
    return sorted(lst)[k]

import random
for i in xrange(500):
    ranges = []
    s = 0
    for n in xrange(50):
        left = random.randrange(99)
        right = random.randrange(left + 1, 100)
        s += right - left + 1
        ranges.append((left, right))
    for k in xrange(s):
        assert kth_stupid(ranges, k) == kth(ranges, k), (ranges, k)

- emb October 18, 2015 | Flag Reply


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