codility – Page 5

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 »

Solution to Genomic-Range-Query by codility

21 Jan

Question: http://codility.com/demo/take-sample-test/genomic_range_query Question Name: GenomicRangeQuery This is a typical case that uses more space for less time.

def solution(S, P, Q):

result = []

DNA_len = len(S)

mapping = {"A":1, "C":2, "G":3, "T":4}

# next_nucl is used to store the position information

# next_nucl[0] is about the "A" nucleotides, [1] about "C"

# [2] about "G", and [3] about "T"

# next_nucl[i][j] = k means: for the corresponding nucleotides i,

# at position j, the next corresponding nucleotides appears

# at position k (including j)

# k == -1 means: the next corresponding nucleotides does not exist

next_nucl = [[-1]*DNA_len, [-1]*DNA_len, [-1]*DNA_len, [-1]*DNA_len]

# Scan the whole DNA sequence, and retrieve the position information

next_nucl[mapping[S[-1]] - 1][-1] = DNA_len-1

for index in range(DNA_len-2,-1,-1):

next_nucl[0][index] = next_nucl[0][index+1]

next_nucl[1][index] = next_nucl[1][index+1]

next_nucl[2][index] = next_nucl[2][index+1]

next_nucl[3][index] = next_nucl[3][index+1]

next_nucl[mapping[S[index]] - 1][index] = index

for index in range(0,len(P)):

if next_nucl[0][P[index]] != -1 and next_nucl[0][P[index]] <= Q[index]:

result.append(1)

elif next_nucl[1][P[index]] != -1 and next_nucl[1][P[index]] <= Q[index]:

result.append(2)

elif next_nucl[2][P[index]] != -1 and next_nucl[2][P[index]] <= Q[index]:

result.append(3)

else:

result.append(4)

return result

Solution to Passing-Cars by codility

20 Jan

Question: http://codility.com/demo/take-sample-test/passing_cars Question Name: PassingCars

def solution(A):

west = 0 # The number of west-driving cars so far

passing = 0 # The number of passing

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

# Travel the list from the end to the beginning

if A[index] == 0: # A east-driving car

passing += west

if passing > 1000000000:

return -1

else: # A west-driving car

west += 1

return passing

Solution to Max-Counters by codility

19 Jan

Question: http://codility.com/demo/take-sample-test/max_counters Question Name: MaxCounters A straightforward solution is easy as following. But the expected worst-case time complexity cannot be guaranteed.

def solution(N, A):

result = [0]*N

for command in A:

if 1 <= command <= N:

result[command-1] += 1

else:

result[:] = [max(result)]*N

return result

We could use lazy-write to improve the performance. When receiving the max_counter command, we record the current-max value, … Read More »

Solution to Frog-River-One by codility

19 Jan

Question: https://codility.com/demo/take-sample-test/frog_river_one Question Name: FrogRiverOne If coding with C, bitmap is preferred to store which positions are covered.

def solution(X, A):

covered_time = [-1]*X # Record the time, each position is covered

uncovered = X # Record the number of uncovered position

for index in range(0,len(A)):

if covered_time[A[index]-1] != -1:

# This position is already covered

continue

else:

# This position is to be covered

covered_time[A[index]-1] = index

uncovered -= 1

if uncovered == 0:

# All positions are covered

return index

# Finally, some positions are not covered

return -1

Solution to Perm-Check by codility

19 Jan

Question: http://codility.com/demo/take-sample-test/perm_check Question Name: PermCheck This question is a simple variant of the counting question. The Python solution is:

def solution(A):

counter = [0]*len(A)

limit = len(A)

for element in A:

if not 1 <= element <= limit:

return 0

else:

if counter[element-1] != 0:

return 0

else:

counter[element-1] = 1

return 1

The Java solution is:

class Solution {

public static int solution(int[] A) {

int[] counter = new int [A.length];

for(int i= 0; i< A.length; i++){

if (A[i] < 1 || A[i] > A.length) {

// Out of range

return 0;

}

else if(counter[A[i]-1] == 1) {

// met before

return 0;

}

else {

// first time meet

counter[A[i]-1] = 1;

}

return 1;

}

Solution to Tape-Equilibrium by codility

16 Jan

Question: http://codility.com/demo/take-sample-test/tape_equilibrium Question Name: TapeEquilibrium The variable of head stores the sum of the heading part of the tape. And the variable of tail stores the sum of tailing part. Then, we move the index from 2nd position to the … Read More »

tl;dr: Please put your code into a <pre>YOUR CODE</pre> section.

Hello everyone!

If you want to ask a question about the solution. DO READ the post and comments firstly.

If you had some troubles in debugging your solution, please try to ask for help on StackOverflow, instead of here.

If you want to post some comments with code or symbol, here is the guidline.

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 <.

2. To use special symbols < and > outside the pre block, please use "<" and ">" instead.

3. If you have a comment with lots of < and >, you could add the major part of your comment into a <pre class="decode:true crayon-inline "> YOUR COMMENTS </pre> section.

Finally, if you are posting the first comment here, it usually needs moderation. Please be patient and stay tuned. Thanks!

CodeSays.com Admin

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