python – Page 3 – Code Says

Solution to omega2013 (Falling-Disks) by codility

4 Feb

Question: https://codility.com/demo/take-sample-test/falling_disks Question Name: omega2013 or FallingDisks

def solution(A, B):

# Find the smallest diameter to achieve each position

min_diameter_so_far = [A[0]] * len(A)

for index in xrange(1, len(A)):

min_diameter_so_far[index] = min(min_diameter_so_far[index-1], A[index])

rings_index = len(A) - 1 # Travel the well from bottom to top

fit_disks = 0 # The number of fitted disks

for disks_index in xrange(len(B)):

while min_diameter_so_far[rings_index] < B[disks_index]:

rings_index -= 1

if rings_index == -1:

# No way to fit into the well for current disk, and

# therefore the remaining disks.

return fit_disks

fit_disks += 1 # Current disk occupies this ring of well

rings_index -= 1

if rings_index == -1:

# The toppest ring of well is used. No way to fit any

# more disk.

break

return fit_disks

Solution to Equi by codility

2 Feb

Question: https://codility.com/demo/take-sample-test/equi Question Name: Equi The expected worst-case space complexity could be less than required.

def solution(A):

heading = 0 # The sum of A[0] + A[1] + ... + A[P-1]

tailing = sum(A) # The sum of A[P] + A[P+1] + ... + A[N-2] + A[N-1]

for index in xrange(len(A)):

tailing -= A[index] # The sum of A[P+1] + ... + A[N-2] + A[N-1]

if heading == tailing:

# A[0] + A[1] + ... + A[P-1] == A[P+1] + ... + A[N-2] + A[N-1]

return index

heading += A[index]

else:

# No equilibrium is found.

return -1

Solution to lithium2013 (Clocks) by codility

2 Feb

Question: https://codility.com/demo/take-sample-test/clocks Question Name: lithium2013 or Clocks At the very beginning, I found this task is very similar with the string rotation question in CTCI. And I did transfer this question into a string rotation one. The string rotation solution … Read More »

Solution to Tree-Height by codility

29 Jan

Question: https://codility.com/demo/take-sample-test/tree_height Question Name: TreeHeight Classic recursion problem.

def solution(T):

if T.l == None and T.r == None:

# Has no subtree

return 0

elif T.l == None:

# Only has right subtree

return 1 + solution(T.r)

elif T.r == None:

# Only has left subtree

return 1 + solution(T.l)

else:

# Have two subtrees

return 1 + max(solution(T.l), solution(T.r))

Solution to Abs-Distinct by codility

29 Jan

Question: https://codility.com/demo/take-sample-test/abs_distinct Question Name: AbsDistinct In the original requirement, expected worst-case space complexity is O(N). But actually, O(1) is enought.

def solution(A):

abs_distinct = 1

current = max(abs(A[0]), abs(A[-1]))

index_head = 0

index_tail = len(A)-1

while index_head <= index_tail:

# We travel the array from the greatest

# absolute value to the smallest.

former = abs(A[index_head])

if former == current:

# Skip the heading elements, whose

# absolute values are the same with

# current recording one.

index_head += 1

continue

latter = abs(A[index_tail])

if latter == current:

# Skip the tailing elements, whose

# absolute values are the same with

# current recording one.

index_tail -= 1

continue

# At this point, both the former and

# latter has different absolute value

# from current recorded one.

if former >= latter:

# The next greatest value is former

current = former

index_head += 1

else:

# The next greatest value is latter

current = latter

index_tail -= 1

# Meet with a new absolute value

abs_distinct += 1

return abs_distinct

Solution to Binary-Gap by codility

29 Jan

Question: https://codility.com/demo/take-sample-test/binary_gap Question Name: BinaryGap

def solution(N):

max_gap = 0

current_gap = 0

# Skip the tailing zero(s)

while N > 0 and N%2 == 0:

N //= 2

while N > 0:

remainder = N%2

if remainder == 0:

# Inside a gap

current_gap += 1

else:

# Gap ends

if current_gap != 0:

max_gap = max(current_gap, max_gap)

current_gap = 0

N //= 2

return max_gap

Solution to alpha2010 (Prefix-Set) by codility

29 Jan

Question: https://codility.com/demo/take-sample-test/prefix_set Question Name: alpha2010 or PrefixSet At the first glance, I think the set of Python is the best choice. Actually, the set solution passed all the test, and got 100/100 grade.

def solution(A):

alphabet = set()

for element in A:

alphabet.add(element)

for index in xrange(len(A)):

alphabet.discard(A[index])

if len(alphabet) == 0:

return index

BUT, acording to the time complexity … Read More »

Solution to Count-Semiprimes by codility

28 Jan

Question: https://codility.com/demo/take-sample-test/count_semiprimes Question Name: CountSemiprimes

def solution(N, P, Q):

from math import sqrt

# Find out all the primes with Sieve of Eratosthenes

prime_table = [False]*2+[True]*(N-1)

prime = []

prime_count = 0

for element in xrange(2, int(sqrt(N))+1):

if prime_table[element] == True:

prime.append(element)

prime_count += 1

multiple = element * element

while multiple <= N:

prime_table[multiple] = False

multiple += element

for element in xrange(int(sqrt(N))+1, N+1):

if prime_table[element] == True:

prime.append(element)

prime_count += 1

# Compute the semiprimes information

semiprime = [0] * (N+1)

# Find out all the semiprimes.

# semiprime[i] == 1 when i is semiprime, or

# 0 when i is not semiprime.

for index_former in xrange(prime_count-1):

for index_latter in xrange(index_former, prime_count):

if prime[index_former]*prime[index_latter] > N:

# So large that no need to record them

break

semiprime[prime[index_former]*prime[index_latter]] = 1

# Compute the number of semiprimes until each position.

# semiprime[i] == k means:

# in the range (0,i] there are k semiprimes.

for index in xrange(1, N+1):

semiprime[index] += semiprime[index-1]

# the number of semiprimes within the range [ P[K], Q[K] ]

# should be semiprime[Q[K]] - semiprime[P[K]-1]

question_len = len(P)

result = [0]*question_len

for index in xrange(question_len):

result[index] = semiprime[Q[index]] - semiprime[P[index]-1]

return result

Solution to Count-Non-Divisible by codility

28 Jan

Question: https://codility.com/demo/take-sample-test/count_non_divisible Question Name: CountNonDivisible

def solution(A):

from math import sqrt

A_max = max(A)

A_len = len(A)

# Compute the frequency of occurrence of each

# element in array A

count = {}

for element in A:

count[element] = count.get(element,0)+1

# Compute the divisors for each element in A

divisors = {}

for element in A:

# Every nature number has a divisor 1.

divisors[element] = [1]

# In this for loop, we only find out all the

# divisors less than sqrt(A_max), with brute

# force method.

for divisor in xrange(2, int(sqrt(A_max))+1):

multiple = divisor

while multiple <= A_max:

if multiple in divisors and not divisor in divisors[multiple]:

divisors[multiple].append(divisor)

multiple += divisor

# In this loop, we compute all the divisors

# greater than sqrt(A_max), filter out some

# duplicate ones, and combine them.

for element in divisors:

temp = [element/div for div in divisors[element]]

# Filter out the duplicate divisors

temp = [item for item in temp if item not in divisors[element]]

divisors[element].extend(temp)

# The result of each number should be, the array length minus

# the total number of occurances of its divisors.

result = []

for element in A:

result.append(A_len-sum([count.get(div,0) for div in divisors[element]]))

return result

Solution to boron2013 (Flags) by codility

27 Jan

Question: https://codility.com/demo/take-sample-test/flags Question Name: boron2013 or Flags The official solution is here. UPDATE 2014-10-02: thanks to Piotrek Martynowicz, a bug is found and fixed.

def solution(A):

from math import sqrt

A_len = len(A)

next_peak = [-1] * A_len

peaks_count = 0

first_peak = -1

# Generate the information, where the next peak is.

for index in xrange(A_len-2, 0, -1):

if A[index] > A[index+1] and A[index] > A[index-1]:

next_peak[index] = index

peaks_count += 1

first_peak = index

else:

next_peak[index] = next_peak[index+1]

if peaks_count < 2:

# There is no peak or only one.

return peaks_count

max_flags = 1

max_min_distance = int(sqrt(A_len))

for min_distance in xrange(max_min_distance + 1, 1, -1):

# Try for every possible distance.

flags_used = 1

flags_have = min_distance-1 # Use one flag at the first peak

pos = first_peak

while flags_have > 0:

if pos + min_distance >= A_len-1:

# Reach or beyond the end of the array

break

pos = next_peak[pos+min_distance]

if pos == -1:

# No peak available afterward

break

flags_used += 1

flags_have -= 1

max_flags = max(max_flags, flags_used)

return max_flags

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