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play.py
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play.py
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from bloom_filter import BloomFilter
exact = False
L = [[11, 20], [30, 40], [5, 10], [40, 30], [10, 5]]
sort_pair = sorted
S = set() if exact else BloomFilter(max_elements=10000, error_rate=0.1)
matches = []
for pair in L:
pair = tuple(sort_pair(pair))
if pair in S:
matches.append(pair)
S.add(pair)
print(matches)
import heapq
import numpy as np
L = list(range(20))
np.random.shuffle(L)
def iter_median(L_iter):
H_min = []
H_max = []
med = 0
L_dbg = []
for x in L_iter:
L_dbg.append(x)
#
if x <= med:
heapq.heappush(H_max, -x)
else:
heapq.heappush(H_min, x)
#
imbalance = len(H_max) - len(H_min)
if imbalance > 1:
x_mv = heapq.heappop(H_max)
heapq.heappush(H_min, -x_mv)
if imbalance < -1:
x_mv = heapq.heappop(H_min)
heapq.heappush(H_max, -x_mv)
#
imbalance = len(H_max) - len(H_min)
if imbalance == 0:
med = 0.5 * (H_min[0] - H_max[0])
elif imbalance == 1:
med = -H_max[0]
elif imbalance == -1:
med = H_min[0]
else:
assert False, ":("
med = float(med)
#
print("lower %s" % str(H_max))
print("upper %s" % str(H_min))
print("med %f %f" % (med, np.median(L_dbg)))
assert np.isclose(med, np.median(L_dbg))
return med
[0, 1, 1]
[0, 0, 1, 1, 2, 3, 3, 4, 4]
def find_index(L):
print("L %s" % L)
N = len(L)
# ii could be in [0, N-1], but will always be even index for solution
lower = 0
upper = N - 1
assert upper % 2 == 0, 'upper even'
while lower < upper:
mid = (lower + upper) // 2 # floordiv, but always even anyway
mid = mid - (mid % 2)
print(lower, mid, upper)
assert lower <= mid, 'lower range'
assert mid <= upper, 'upper range'
assert mid % 2 == 0, 'mid even'
#
if L[mid] != L[mid + 1]:
# then solution <= mid
upper = mid
else:
# then solution > mid
lower = mid + 2
assert lower == upper, 'closed'
return L[lower]
@given(lists(integers(-100, 100), unique=True, min_size=1))
def test_find_index(L):
solution = L[0]
L = sorted(L)
L = sum(([x, x] for x in L), [])
L.remove(solution)
#
y = find_index(L)
assert y == solution
def find_dominant_index(L):
if len(L) < 2:
return -1
# init
if L[0] <= L[1]:
gold = 1
silver = 0
else:
gold = 0
silver = 1
# Find top-2
for ii, xx in enumerate(L[2:], 2):
if xx >= L[silver]: # Could save in var if helps
if xx >= L[gold]:
silver = gold
gold = ii
else:
silver = ii
# check
if L[gold] >= 2 * L[silver]:
return gold
# or -1
return -1
@given(lists(integers(0, 99), min_size=2, max_size=50))
def test_find_dominant_index(L):
ii = find_dominant_index(L)
print(L, ii)
#
sorted_L = sorted(L)
if sorted_L[-1] >= 2 * sorted_L[-2]:
assert L[ii] == sorted_L[-1]
else:
assert ii == -1
def find_repeat(L):
N = len(L)
for ii in range(1, N): # Could stop before N
if N % ii == 0: # otherwise we know it can't be repeat
rr = N // ii
if L == L[:ii] * rr:
return True, (ii, rr)
return False, (0, 0)
@given(lists(integers(0, 99), min_size=1, max_size=50), integers(2, 20))
def test_rep(L, rep):
L2 = L * rep
assert find_repeat(L2)[0]
def find_sum(L, target):
D = {}
for ii, x in enumerate(L):
if x in D:
return D[x], ii
else:
D[target - x] = ii
assert False
def least_index_sum(L1, L2):
D1 = {ss: ii for ii, ss in enumerate(L1)}
best_el = None
best_score = len(L1) + len(L2) # upperbound can always beat
for ii, ss in enumerate(L2):
if (ss in D1) and (ii + D1[ss] < best_score):
best_score = ii + D1[ss]
best_el = ss
return best_el
def least_index_sum_2(L1, L2):
S = set(L1) & set(L2)
D1 = {ss: ii for ii, ss in enumerate(L1)}
D2 = {ss: ii for ii, ss in enumerate(L2)}
DS = {ss: D1[ss] + D2[ss] for ss in S}
key_min = None if len(DS) == 0 else min(DS.keys(), key=(lambda k: DS[k]))
return key_min
@given(lists(integers(0, 10)), lists(integers(0, 10)))
def test_min(L1, L2):
R1 = least_index_sum(L1, L2)
R2 = least_index_sum(L1, L2)
assert R1 == R2
from collections import deque
class RecentCounter(object):
def __init__(self):
self.q = deque()
#
def ping(self, t):
self.q.append(t)
#
expire_time = t - 3000
#
t_ = None
while (t_ is None) or (t_ < expire_time):
t_ = self.q.popleft()
assert t_ is not None
assert t_ >= expire_time
self.q.appendleft(t_) # put back last
print(list(self.q))
return len(self.q)
def segmenter(L, D):
if len(L) == 0: # base case
return []
#
max_try = len(L) + 1 # Could make this shorter based on lengths in D
best_sol = None
best_len = len(L) + 10 # some upper bound
for ii in range(1, max_try):
if L[:ii] in D: # could use continue to avoid nesting
sol = segmenter(L[ii:], D)
if (sol is not None) and len(sol) < best_len:
best_sol = [L[:ii]] + sol
best_len = len(sol)
print(best_sol, best_len)
return best_sol
def segmenter2(L, D, _len_list=None):
if len(L) == 0: # base case
return []
#
if _len_list is None:
_len_list = sorted(set(len(ss) for ss in D))
assert all(ss > 0 for ss in _len_list)
#
best_sol = None
best_len = len(L) + 10 # some upper bound
# or could do greedy approach of starting with longest and going smaller if doesn't fit
for ii in _len_list:
if L[:ii] in D: # could use continue to avoid nesting
sol = segmenter2(L[ii:], D, _len_list=_len_list)
if (sol is not None) and len(sol) < best_len:
best_sol = [L[:ii]] + sol
best_len = len(sol)
return best_sol
def segmenter_greedy(L, D, _len_list=None):
if len(L) == 0: # base case
return []
#
if _len_list is None:
_len_list = sorted(set(len(ss) for ss in D))[::-1]
assert all(ss > 0 for ss in _len_list)
#
for ii in _len_list:
if L[:ii] in D: # could use continue to avoid nesting
sol = segmenter_greedy(L[ii:], D, _len_list=_len_list)
if sol is not None:
return [L[:ii]] + sol
return None
@given(lists(lists(integers(0, 9), min_size=1).map(tuple), min_size=1, unique=True).map(set), lists(integers()))
def test_segmenter(D, idx):
DS = sorted(D)
L = sum([DS[ii % len(DS)] for ii in idx], ())
print('<', L, D)
#
sol = segmenter_greedy(L, D)
print('>', L, sol)
assert sum(sol, ()) == L
assert len(sol) <= len(L)