Codility – Page 6

Solution to Max-Double-Slice-Sum by codility

25 Jan

Question: http://codility.com/demo/take-sample-test/max_double_slice_sum Question Name: MaxDoubleSliceSum A variant of the classic maximum subarray problem. We should travel the array twice. For the first travel, we compute and record the maximum sub-array sum, which ends at each position. At the second reverse … Read More »

Solution to Equi-Leader by codility

24 Jan

Question: http://codility.com/demo/take-sample-test/equi_leader Question Name: EquiLeader A variant of the previous question.

def solution(A):

A_len = len(A)

candidate = -1

candidate_count = 0

candidate_index = -1

# Find out a leader candidate

for index in xrange(A_len):

if candidate_count == 0:

candidate = A[index]

candidate_index = index

candidate_count += 1

else:

if A[index] == candidate:

candidate_count += 1

else:

candidate_count -= 1

# Make sure the candidate is the leader

leader_count = len([number for number in A if number == candidate])

if leader_count <= A_len//2:

# The candidate is not the leader

return 0

else:

leader = candidate

equi_leaders = 0

leader_count_so_far = 0

for index in xrange(A_len):

if A[index] == leader:

leader_count_so_far += 1

if leader_count_so_far > (index+1)//2 and

leader_count-leader_count_so_far > (A_len-index-1)//2:

# Both the head and tail have leaders of the same value

# as "leader"

equi_leaders += 1

return equi_leaders

Solution to Dominator by codility

24 Jan

Question: http://codility.com/demo/take-sample-test/dominator Question Name: Dominator Same question in the introduction pdf.

def solution(A):

A_len = len(A)

candidate = -1

candidate_count = 0

candidate_index = -1

for index in xrange(A_len):

if candidate_count == 0:

candidate = A[index]

candidate_index = index

candidate_count += 1

else:

if A[index] == candidate:

candidate_count += 1

else:

candidate_count -= 1

if len([number for number in A if number == candidate]) <= A_len//2:

return -1

else:

return candidate_index

Solution to Fish by codility

23 Jan

Question: http://codility.com/demo/take-sample-test/fish Question Name: Fish UPDATE 03-14-2014: “downstream_count” is a redundant variable. It could be replaced with len(downstream) safely. (Thanks to the reminder from Max)

def solution(A, B):

alive_count = 0 # The number of fish that will stay alive

downstream = [] # To record the fishs flowing downstream

downstream_count = 0 # To record the number of elements in downstream

for index in xrange(len(A)):

# Compute for each fish

if B[index] == 1:

# This fish is flowing downstream. It would

# NEVER meet the previous fishs. But possibly

# it has to fight with the downstream fishs.

downstream.append(A[index])

downstream_count += 1

else:

# This fish is flowing upstream. It would either

# eat ALL the previous downstream-flow fishs,

# and stay alive.

# OR

# be eaten by ONE of the previous downstream-

# flow fishs, which is bigger, and died.

while downstream_count != 0:

# It has to fight with each previous living

# fish, with nearest first.

if downstream[-1] < A[index]:

# Win and to continue the next fight

downstream_count -= 1

downstream.pop()

else:

# Lose and die

break

else:

# This upstream-flow fish eat all the previous

# downstream-flow fishs. Win and stay alive.

alive_count += 1

# Currently, all the downstream-flow fishs in stack

# downstream will not meet with any fish. They will

# stay alive.

alive_count += len(downstream)

return alive_count

Solution to sigma2012 (Stone-Wall) by codility

23 Jan

Question: http://codility.com/demo/take-sample-test/stone_wall Question Name: sigma2012 or StoneWall

def solution(H):

stack = []

block_count = 0 # The number of needing blocks

for height in H:

while len(stack) != 0 and height < stack[-1]:

# If the height of current block is less than

# the previous ones, the previous ones have

# to end before current point. They have no

# chance to exist in the remaining part.

# So the previous blocks are completely finished.

stack.pop()

block_count += 1

if len(stack) == 0 or height > stack[-1]:

# If the height of current block is greater than

# the previous one, a new block is needed for

# current position.

stack.append(height)

# Else (the height of current block is same as that

# of previous one), they should be combined to

# one block.

# Some blocks with different heights are still in the stack.

block_count += len(stack)

return block_count

Solution to Nesting by codility

22 Jan

Question: http://codility.com/demo/take-sample-test/nesting Question Name: Nesting This question is nearly the same as the previous one, but does not need a stack.

def solution(S):

parentheses = 0

for element in S:

if element == "(":

parentheses += 1

else:

parentheses -= 1

if parentheses < 0:

return 0

if parentheses == 0:

return 1

else:

return 0

Solution to Brackets by codility

22 Jan

Question: http://codility.com/demo/take-sample-test/brackets Question Name: Brackets Classic application of stacks.

def solution(S):

if len(S) % 2 == 1: return 0

matched = {"]":"[", "}":"{", ")": "("}

to_push = ["[", "{", "("]

stack = []

for element in S:

if element in to_push:

stack.append(element)

else:

if len(stack) == 0:

return 0

elif matched[element] != stack.pop():

return 0

if len(stack) == 0:

return 1

else:

return 0

UPDATE (2016-07-19): Much appreciate Jeffrey’s comment and suggestion: I’m surprised so few people didn’t do a simple modulo at the start to check whether there is an even number of chars. … Read More »

Solution to beta2010 (Number-Of-Disc-Intersections) by codility

22 Jan

Question: http://codility.com/demo/take-sample-test/number_of_disc_intersections Question Name: beta2010 or NumberOfDiscIntersections Interesting question. Does geometry matter here? Not much.

def solution(A):

discs_count = len(A) # The total number of discs

range_upper = [0]*discs_count # The upper limit position of discs

range_lower = [0]*discs_count # The lower limit position of discs

# Fill the limit_upper and limit_lower

for index in xrange(0, discs_count):

range_upper[index] = index + A[index]

range_lower[index] = index - A[index]

range_upper.sort()

range_lower.sort()

range_lower_index = 0

intersect_count = 0

for range_upper_index in xrange(0, discs_count):

# Compute for each disc

while range_lower_index < discs_count and

range_upper[range_upper_index] >= range_lower[range_lower_index]:

# Find all the discs that:

# disc_center - disc_radius <= current_center + current_radius

range_lower_index += 1

# We must exclude some discs such that:

# disc_center - disc_radius <= current_center + current_radius

# AND

# disc_center + disc_radius <(=) current_center + current_radius

# These discs are not intersected with current disc, and below the

# current one completely.

# After removing, the left discs are intersected with current disc,

# and above the current one.

# Attention: the current disc is intersecting itself in this result

# set. But it should not be. So we need to minus one to fix it.

intersect_count += range_lower_index - range_upper_index -1

if intersect_count > 10000000:

return -1

return intersect_count

UPDATE on 2015/10/16: here is some pseudo-code to help you understand. Firstly, we convert the 2D problem into 1D. Because all the circles’ center … Read More »

Solution to Triangle by codility

21 Jan

Question: http://codility.com/demo/take-sample-test/triangle Question Name: Triangle Update: we need two parts to prove our solution. On one hand, there is no false triangular. Since the array is sorted, we already know A[index] < = A[index+1] <= A[index+2], and all values are … Read More »

Solution to Max-Product-Of-Three by codility

21 Jan

Question: http://codility.com/demo/take-sample-test/max_product_of_three Question Name: MaxProductOfThree An small trap: multiplying two negatives makes a positive number.

def solution(A):

A.sort()

return max(A[0]*A[1]*A[-1], A[-1]*A[-2]*A[-3])

UPDATE on 03-02-2015: Thanks to @Julian, there is an O(N) solution. We sort the input in the original solution, because we want the two … Read More »

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1. To post your code, please add the code inside a <pre> </pre> section (preferred), or <code> </code>. And inside the pre or code section, you do not need to escape < > and &, e.g. no need to use < instead of <.

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Ovi February 12, 2024 at 11:14 pm on Solution to Min-Perimeter-Rectangle by codilityExactly my question too. I tried something very similar to Sheng's solution, importing the math module and using sqrt(). Codility marked it only as 80%,...
Doron Ben Ari February 11, 2024 at 4:10 am on Solution to Min-Avg-Two-Slice by codilityInspired by Sheng's great solution, I wrote the following proof: Assume you found the slice with minimal average. Assume the first two consecutive elements of...
Ovi February 4, 2024 at 9:03 am on Solution to sigma2012 (Stone-Wall) by codility def solution(H): heights = H # geometric logic: the blocks need to be cuboid, all sides are rectagular. Nothing fancy, means that the angles...
Ovidiu Oprea February 2, 2024 at 11:06 am on Solution to Fish by codilityIt's the reason why it took me 2 weeks to finish this task (in the context of adult life with non-dev job). I was trying...
Ovi January 15, 2024 at 2:40 pm on Solution to Brackets by codilityThanks for this blog post! I was banging my head against the wall thinking this is a simple / short solution, where you just build...
parikshit verma July 5, 2023 at 8:26 am on Solution to Missing-Integer by codility def missingnum(a): st=min(a) en=max(a) b=set(range(1,en+2)) if (en<1 or st>1) : return 1 else: return min(b-set(a))
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Pavankumar Nandeshwar June 26, 2023 at 5:23 am on Solution to Dwarfs-Rafting by codilityCan someone explain what is wrong with my solution in C++ ? int solution(int N, string &S, string &T) { int q = N /...
Pavankumar Nandeshwar June 23, 2023 at 10:25 pm on Solution to Flood-Depth by codilityMy C++ Solution. O(N) #include <climits> int solution(vector<int> &A) { int max_depth {}; int max_lvl = {}; int min_lvl = INT_MAX; for (unsigned int i...