Codility – Page 5

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

Solution to Peaks by codility

26 Jan

Question: http://codility.com/demo/take-sample-test/peaks Question Name: Peaks UPDATE on July 18th 2015: Many thanks to @Antonio Correia (https://34.145.67.234/2014/solution-to-peaks-by-codility/#comment-11725) !!! @Antonio Correia found a huge bug in my solution, even though it passed the codility.com. The bug has been fixed.

def solution(A):

from math import sqrt

A_len = len(A)

peaks_until_here = [0]*A_len

# Retrieve how many peaks exist from beginning to current

# position.

for index in range(1, A_len-1):

peaks_until_here[index] = peaks_until_here[index-1]

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

peaks_until_here[index] += 1

if A_len < 3 or peaks_until_here[-2] == 0:

# The array is too short to have a peak. OR

# There is no peak in this array.

return 0

peaks_until_here[-1] = peaks_until_here[-2]

max_blocks = 0

# Compute every possible partition plan, and find out the

# one with most blocks.

for candidate in range(1, int(sqrt(A_len))+1, 1):

if A_len % candidate == 0:

blocks, block_size = candidate, A_len//candidate

# Check the first block.

if peaks_until_here[0] < peaks_until_here[block_size-1]:

# Check the following blocks.

for each_block in range(block_size, A_len, block_size):

if peaks_until_here[each_block-1] == \

peaks_until_here[each_block+block_size-1]:

# No peak is found in the next block

# This partition plan is not accepted

break

else:

max_blocks = blocks

if candidate * candidate == A_len:

# If candidate is equal to sqrt(A_len) exactly,

# candidate would equal to A_len//candidate.

continue

block_size, blocks = candidate, A_len//candidate

# Check the first block.

if peaks_until_here[0] < peaks_until_here[block_size-1]:

# Check the following blocks.

for each_block in range(block_size, A_len, block_size):

if peaks_until_here[each_block-1] == \

peaks_until_here[each_block+block_size-1]:

# No peak is found in the next block

# This partition plan is not accepted

break

else:

return blocks

return max_blocks

Solution to Min-Perimeter-Rectangle by codility

26 Jan

Question: http://codility.com/demo/take-sample-test/min_perimeter_rectangle Question Name: MinPerimeterRectangle Let the length of one side be len_1, and the length of one adjacent side be len_2. For a rectangle with a constant area, the perimeter is minimized when the difference between len_1 and len_2, … Read More »

Solution to Max-Slice-Sum by codility

26 Jan

Question: http://codility.com/demo/take-sample-test/max_slice_sum Question Name: MaxSliceSum The classic maximum subarray problem. It should be the very first question in this lesson.

def solution(A):

max_slice_ending_here = A[0]

max_slice = A[0]

for element in A[1:]:

max_slice_ending_here = max(element, max_slice_ending_here+element)

max_slice = max(max_slice, max_slice_ending_here)

return max_slice

Solution to Max-Profit by codility

25 Jan

Question: http://codility.com/demo/take-sample-test/max_profit Question Name: MaxProfit For this question, we have to travel from the last element to the first one.

def solution(A):

days = len(A)

# If the number of days is zero or one, there

# is no time to get profit.

if days < 2:

return 0

max_price_from_here = A[days-1]

max_profit = 0

for index in xrange(days-2, -1, -1):

# max_price_from_here-A[index] means the maximum

# profit from current day to end.

max_profit = max(max_profit, max_price_from_here-A[index])

max_price_from_here = max(A[index], max_price_from_here)

return max_profit

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