Codility – Page 4

Solution to Chocolates-By-Numbers by codility

17 Feb

Question: https://codility.com/demo/take-sample-test/chocolates_by_numbers Question Name: Chocolates-By-Numbers or ChocolatesByNumbers When we met with an empty wrapper, we must have been this position for twice. We use i for the first time and j for the second time. Due to the modulo feature, … Read More »

Solution to upsilon2012 (Cartesian-Sequence) by codility

17 Feb

Question: https://codility.com/demo/take-sample-test/cartesian_sequence Question Name: upsilon2012 or CartesianSequence A straightforward solution would be construct the Cartesian tree, compute its height, and return the height plus one. But any left sub-tree would never be accessed or changed after its construction. And so … Read More »

Solution to chi2012 (Cannonballs) by codility

7 Feb

Question: https://codility.com/demo/take-sample-test/chi2012/ Question Name: chi2012 or Cannonballs

def solution(A, B):

to_fall = [-1] * (max(B) + 1) # Store where the ball should

# fall. to_fall[i] = j means,

# the cannonball, which is shot

# from heigth of i, will fall

# at position j.

# Scan and find the initial falling position for cannonballs

# from each different shooting height.

current_pos = 0 # Index for the landscape array A.

# For all shooting heigthes <= A[0], the cannonballs do not

# change the landscape. Keep the content being -1 and no need

# to try them.

for height in xrange( A[0]+1, max(B)+1 ):

while current_pos < len(A) and height > A[current_pos]:

current_pos += 1

if current_pos == len(A):

# No position could prevent the cannonball from flying

# beyond the bound. These cannonballs do not change

# the landscape. Keep the content being -1.

break

else:

# The cannonball meets with a high enought position,

# and falls at the previous position.

to_fall[height] = current_pos - 1

for cannonball in B:

if to_fall[cannonball] == -1:

# This cannonball would not change the landscape.

continue

else:

fall_pos = to_fall[cannonball]

A[ fall_pos ] += 1 # increases the ground by 1

# For shooting height > A[fall_pos], the increase

# is NOT enought to block/change their flying

# path.

# For shooting height < A[fall_pos] <= old A[fall_pos]

# they cannot fly beyond this position both

# before and after increase. NO change is

# made for them.

# ONLY for shooting height = A[fall_pos], its

# flying path might be changed due to the

# increase.

to_fall[ A[ fall_pos ] ] = min(

to_fall[ A[ fall_pos ] ], fall_pos-1)

return A

Solution to Min-Avg-Two-Slice by codility

6 Feb

Question: https://codility.com/demo/take-sample-test/min_avg_two_slice Question Name: MinAvgTwoSlice The key to solve this task is these two patterns: (1) There must be some slices, with length of two or three, having the minimal average value among all the slices. (2) And all … Read More »

Solution to gamma2011 (Count-Palindromic-Slices) by codility

5 Feb

Question: https://codility.com/demo/take-sample-test/count_palindromic_slices Question Name: CountPalindromicSlices This task is based on the longest palindromic substring question, which has a good solution naming Manacher’s algorithm. Some explanation about the algorithm could be found here (English) and here (Chinese).

def palindrome_substring(str):

''' Input: string

Attention: the input string will be extended. For example,

original string abc would be converted into #a#b#c#.

Output: array, to say palindrome_len.

palindrome_len[i] records the half width (include

the element in position i) of palindrome substring,

which centers in position i.

Method: Manacher's algorithm

Time complexity: O(n)

'''

str = "#" + "#".join(str) + "#" # Convert the original string so

# that, every palindrome substring

# in the new string has odd number

# of elements.

palindrome_len = [0] * len(str) # Store the half width (include the

# center point) of the longest

# palindrome substring with index

# being the substring's center.

# palindrome_len[i]-1 means the full

# length of palindrome substring in

# the original string.

bound = 0 # Record the first positin, that the previous computed

# palindrome substrings chould NOT achieve.

center = 0 # Record the center position of the substring, which

# is corresponding to "bound".

for index in xrange(len(str)):

if bound > index:

# Part of current substring has already been compared.

# For point "index", 2*center-index is the point

# symmetrical to the points "center".

palindrome_len[index] =

min( palindrome_len[2*center-index], bound-index )

else:

# None of current substring has been compared.

palindrome_len[index] = 1

# Compare the uncompared elements one by one.

while index-palindrome_len[index] >= 0 and

index+palindrome_len[index] < len(str) and

str[index-palindrome_len[index]] ==

str[index+palindrome_len[index]]:

palindrome_len[index] += 1

if bound < palindrome_len[index] + index:

# The bound has been extended.

center = index

bound = palindrome_len[index] + index

return palindrome_len

def solution(S):

# With center point i, if the longest palindrome substring

# has length of j, the number of palindrome substrings,

# with the same center, is j//2.

count = sum([

(length-1)/2 for length in palindrome_substring(S) if length>2

])

if count > 100000000:

return -1

else:

return count

Solution to Array-Inversion-Count by codility

4 Feb

Question: https://codility.com/demo/take-sample-test/array_inversion_count Question Name: ArrayInversionCount

def mergesort( aList, first, last ):

''' Modified merge sort algorithm.

Record the inversion count during sort.

'''

mid = ( first + last ) / 2

if first < last:

# Recursive calling

left_inver = mergesort( aList, first, mid )

right_inver = mergesort( aList, mid + 1, last )

else:

# Terminate condition

return 0

first_index = first # The index for the left part

second_index = mid + 1 # The index for the right part

temp = [0] * (last-first+1) # To hold the sorted content

temp_index = 0

merge_inver = 0 # Number of inversion in merging

while first_index <= mid and second_index <= last:

if aList[first_index] <= aList[second_index]:

# Less index indicates less value. No inversion.

temp[temp_index] = aList[first_index]

first_index += 1

else:

# Greater index has less value. Inversion exists.

# For exampe:

# [ 4, 5, 2, 3 ]

# | Left part | |Right Part|

# and first_index = 1, second_index = 3, mid = 2

# We need the item "2" to be the position 0. So it

# has to pass all the unwritten items in left part.

# Here these unwritten items are "4" and "5". So

# two more inversions are involved.

# In general, the left part is sorted. So all the

# elements, being and after first_index, are greater

# than element in position second_index. AND all of

# them have less indexes. As the result,

# there are mid-first_index+1 new reversions.

temp[temp_index] = aList[second_index]

second_index += 1

merge_inver += mid - first_index + 1

temp_index += 1

if first_index != mid+1:

# Some element in the left part left. They have less

# indexes, but greater values. Inversion involves.

# BUT these inversions have already been counted.

while first_index <= mid:

temp[temp_index] = aList[first_index]

first_index += 1

temp_index += 1

if second_index != last+1:

# Some element in the right part left. They have both

# greater indexes and values than all in the left part.

# No inversion is involved.

while second_index <= last:

temp[temp_index] = aList[second_index]

second_index += 1

temp_index += 1

# Rewrite the sorted content into the original array

aList[first:last+1] = temp[:]

return merge_inver + left_inver + right_inver

def solution(A):

return mergesort( A * 1, 0, len(A)-1)

UPDATE on July 18, 2014: Thanks to @Andrew, there is another solution with Binary Indexed Tree. UPDATE on October 8, 2014: Thanks to @Piotrek Martynowicz, there was a bug for the statement: “computes the … Read More »

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))

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