codility – Page 2

Solution to Ladder by codility

9 Mar

Question: https://codility.com/demo/take-sample-test/ladder Question Name: Ladder To solve this question, there are two key points. Firstly, for a given N rungs, the number of different ways of climbing is the (N+1)th element in the Fibonacci numbers. Naturally, we could get the … Read More »

Solution to nu2011 (Double-Median) by codility

5 Mar

Question: https://codility.com/demo/take-sample-test/double_median Question Name: nu2011 or DoubleMedian For the given queries, we need to find the median in two sorted arrays of different lengths. Then we need to find out the median of the query results. That is, we need … Read More »

Solution to oxygenium2014 (Count-Bounded-Slices) by codility

27 Feb

Question: https://codility.com/programmers/challenges/oxygenium2014 Question Name: oxygenium2014 or CountBoundedSlices For this question, I was unable to solve it with a golden answer. Instead, I tried to get two silver solutions. Both of them cheated the grading system to have the detected time … Read More »

Solution to Common-Prime-Divisors by codility

18 Feb

Question: https://codility.com/demo/take-sample-test/common_prime_divisors Question Name: CommonPrimeDivisors

def gcd(x, y):

''' Compute the greatest common divisor.

'''

if x%y == 0:

return y;

else:

return gcd(y, x%y)

def removeCommonPrimeDivisors(x, y):

''' Remove all prime divisors of x, which also exist in y. And

return the remaining part of x.

'''

while x != 1:

gcd_value = gcd(x, y)

if gcd_value == 1:

# x does not contain any more

# common prime divisors

break

x /= gcd_value

return x

def hasSamePrimeDivisors(x, y):

gcd_value = gcd(x, y) # The gcd contains all

# the common prime divisors

x = removeCommonPrimeDivisors(x, gcd_value)

if x != 1:

# If x and y have exactly the same common

# prime divisors, x must be composed by

# the prime divisors in gcd_value. So

# after previous loop, x must be one.

return False

y = removeCommonPrimeDivisors(y, gcd_value)

return y == 1

def solution(A, B):

count = 0

for x,y in zip(A,B):

if hasSamePrimeDivisors(x,y):

count += 1

return count

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 »

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