NetApp Interview Question for SDE-3s

Country: India

Comment hidden because of low score. Click to expand.
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.

def kth(ranges, k):
    assert k >= 0
    events = []
    for start, end in ranges:
        events.extend([(start, 1), (end + 1, 0)])
    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
            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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