Codility linked list

Recently, during interview to some company, I got tested by Codility service. And here is my performance results. Generally this is copy&paste from their email. Hope this can be helpful for others who want to see what this tests are.

This is overal performance table:

Task name Correctness Performance Task score
1 PtrListLen 100 not assessed 100
2 BugfixingLeaderSorted 100 not assessed 100
3 DeepestPit 55 66 60
4 CountMultiplicativePairs 80 87 83
Total 84 N/A 343 / 400

PtrListLen – problem description

A pointer is called a linked list if:

· it is an empty pointer (it is then called a terminator or an empty list); or

· it points to a structure (called a node or the head) that contains a value and a linked list (called the tail).

The length of a list is defined as the total number of nodes it contains. In particular, an empty list has length 0.

For example, consider the following linked list:

A -> B -> C -> D ->

This list contains four nodes: A, B, C and D. Node D is the last node and its tail is the terminator. The length of this list is 4.

Assume that the following declarations are given:

classIntList(object):
value=0
next=None

Write a function:

def solution(L)

that, given a non-empty linked list L consisting of N nodes, returns its length.

For example, given list L shown in the example above, the function should return 4.

Assume that:

· N is an integer within the range [1..5,000];

· list L does not have a cycle (each non-empty pointer points to a different structure).

In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.

Copyright 20092015 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Tests

Score: 100 of 100

Estimated time complexity: None

Test name

Running time

Result

example
example, length=4

0.060s

OK

extreme_single_double
length=1

0.044s

OK

three_elems
length=3

0.044s

OK

twenty_elements
length=20

0.044s

OK

medium
length=93

0.044s

OK

medium2
length=999

0.044s

OK

1k_elements
length=1,000

0.044s

OK

quite_long
length=4,000

0.052s

OK

long
length=5,000

0.056s

OK

Solution (language: Python)

# you can use print for debugging purposes, e.g.
# print “this is a debug message”
def solution(L):
# write your code in Python 2.7
count = 0
while L:
count = count + 1
L = L.next
return count


BugfixingLeaderSorted – problem description

A non-empty zero-indexed array A consisting of N integers and sorted in a non-decreasing order is given. The leader of this array is the value that occurs in more than half of the elements of A.

You are given an implementation of a function:

def solution(A)

that, given a non-empty zero-indexed array A consisting of N integers, sorted in a non-decreasing order, returns the leader of array A. The function should return 1 if array A does not contain a leader.

For example, given array A consisting of ten elements such that:

A[0] = 2
A[1] = 2
A[2] = 2
A[3] = 2
A[4] = 2
A[5] = 3
A[6] = 4
A[7] = 4
A[8] = 4
A[9] = 6

the function should return 1, because the value that occurs most frequently in the array, 2, occurs five times, and 5 is not more than half of 10.

Given array A consisting of five elements such that:

A[0] = 1
A[1] = 1
A[2] = 1
A[3] = 1
A[4] = 50

the function should return 1.

Unfortunately, despite the fact that the function may return expected result for the example input, there is a bug in the implementation, which may produce incorrect results for other inputs. Find the bug and correct it. You should modify at most three lines of code.

Assume that:

· N is an integer within the range [1..100,000];

· each element of array A is an integer within the range [0..2,147,483,647];

· array A is sorted in non-decreasing order.

Complexity:

· expected worst-case time complexity is O(N);

· expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Copyright 20092015 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Tests

Score: 100 of 100

Estimated time complexity: None

Test name

Running time

Result

example1
first example test

0.044s

OK

example2
second example test

0.044s

OK

simple1
values from a continuous range

0.044s

OK

simple2
0s/1s only

0.044s

OK

single
one element

0.044s

OK

two_values
two different values

0.044s

OK

extreme_big_values
min/max values only

0.044s

OK

medium_1
small sequence repeated many times

0.044s

OK

medium_2
no leader and small sequence with values from a continuous range

0.044s

OK

cyclic_sequence
no leader and small sequence repeated many times

0.044s

OK

medium_random
random sequences

0.044s

OK

large
two different values, length = ~100,000

0.104s

OK

large_range
values from a continuous range, length = ~100,000

0.108s

OK

Solution (language: Python)

def solution(A):
n = len(A)
L = [-1] + A
count = 0
pos = (n + 1) // 2
candidate = L[pos]
for i in xrange(1, n + 1):
if (L[i] == candidate):
count = count + 1
if (2*count > n):
return candidate
return -1


DeepestPit – problem description

A non-empty zero-indexed array A consisting of N integers is given. A pit in this array is any triplet of integers (P, Q, R) such that:

· 0 P < Q < R < N;

· sequence [A[P], A[P+1], …, A[Q]] is strictly decreasing,
i.e. A[P] > A[P+1] > … > A[Q];

· sequence A[Q], A[Q+1], …, A[R] is strictly increasing,
i.e. A[Q] < A[Q+1] < … < A[R].

The depth of a pit (P, Q, R) is the number min{A[P] A[Q], A[R] A[Q]}.

For example, consider array A consisting of 10 elements such that:

A[0] = 0
A[1] = 1
A[2] = 3
A[3] = -2
A[4] = 0
A[5] = 1
A[6] = 0
A[7] = -3
A[8] = 2
A[9] = 3

Triplet (2, 3, 4) is one of pits in this array, because sequence [A[2], A[3]] is strictly decreasing (3 > 2) and sequence [A[3], A[4]] is strictly increasing (2 < 0). Its depth is min{A[2] A[3], A[4] A[3]} = 2. Triplet (2, 3, 5) is another pit with depth 3. Triplet (5, 7, 8) is yet another pit with depth 4. There is no pit in this array deeper (i.e. having depth greater) than 4.

Write a function:

def solution(A)

that, given a non-empty zero-indexed array A consisting of N integers, returns the depth of the deepest pit in array A. The function should return 1 if there are no pits in array A.

For example, consider array A consisting of 10 elements such that:

A[0] = 0
A[1] = 1
A[2] = 3
A[3] = -2
A[4] = 0
A[5] = 1
A[6] = 0
A[7] = -3
A[8] = 2
A[9] = 3

the function should return 4, as explained above.

Assume that:

· N is an integer within the range [1..1,000,000];

· each element of array A is an integer within the range [100,000,000..100,000,000].

Complexity:

· expected worst-case time complexity is O(N);

· expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Copyright 20092015 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Tests

Score: 60 of 100

Test name

Running time

Result

example
example test

0.044s

OK

extreme_no_pit
small test cases

0.044s

WRONG ANSWER
got -1 expected 1

extreme_depth

0.044s

WRONG ANSWER
got -1 expected 200000000

simple1
no pit

0.044s

WRONG ANSWER
got -1 expected 1

simple2
one pit

0.044s

WRONG ANSWER
got -1 expected 50

user
user-defined test case

0.044s

OK

simple3
`vulcano’ shape

0.044s

OK

retries
retries

0.044s

OK

medium1
medium correctness test

0.104s

OK

medium_pit
medium test one pit

0.044s

OK

large_pit_1
large test one pit 1

0.148s

WRONG ANSWER
got 43265 expected 43287

large_pit_2
large test one pit 2

0.152s

WRONG ANSWER
got 432950 expected 433170

big_pit_1
big test one pit 1

0.120s

OK

big_pit_2
big test one pit 1

0.132s

OK

big3_1
large random test

0.424s

OK

big3_2
big random test

1.336s

OK

Solution (language: Python)

# you can use print for debugging purposes, e.g.
# print “this is a debug message”
def solution(A):
deepest = 0
pit = lambda p, q, r: min(A[p] – A[q], A[r] – A[q])
p = 0
r = -1
q = -1
l = len(A)
for i in xrange(0, l):
if q<0 and A[i]>=A[i-1]:
q = i -1
if (q>=0 and r<0) and (A[i]<=A[i-1] or i+1==l):
r = i-1
deepest = max(deepest, pit(p, q, r))
p = i-1
q = -1
r = -1
return deepest if deepest else -1


CountMultiplicativePairs – problem description

Zero-indexed arrays A and B consisting of N non-negative integers are given. Together, they represent N real numbers, denoted as C[0], …, C[N1]. Elements of A represent the integer parts and the corresponding elements of B (divided by 1,000,000) represent the fractional parts of the elements of C. More formally, A[I] and B[I] represent C[I] = A[I] + B[I] / 1,000,000.

For example, consider the following arrays A and B:

A[0] = 0 B[0] = 500,000
A[1] = 1 B[1] = 500,000
A[2] = 2 B[2] = 0
A[3] = 2 B[3] = 0
A[4] = 3 B[4] = 0
A[5] = 5 B[5] = 20,000

They represent the following real numbers:

C[0] = 0.5
C[1] = 1.5
C[2] = 2.0
C[3] = 2.0
C[4] = 3.0
C[5] = 5.02

A pair of indices (P, Q) is multiplicative if 0 P < Q < N and C[P] * C[Q] C[P] + C[Q].

The above arrays yield the following multiplicative pairs:

· (1, 4), because 1.5 * 3.0 = 4.5 4.5 = 1.5 + 3.0,

· (1, 5), because 1.5 * 5.02 = 7.53 6.52 = 1.5 + 5.02,

· (2, 3), because 2.0 * 2.0 = 4.0 4.0 = 2.0 + 2.0,

· (2, 4) and (3, 4), because 2.0 * 3.0 = 6.0 5.0 = 2.0 + 3.0.

· (2, 5) and (3, 5), because 2.0 * 5.02 = 10.04 7.02 = 2.0 + 5.02.

· (4, 5), because 3.0 * 5.02 = 15.06 8.02 = 3.0 + 5.02.

Write a function:

def solution(A, B)

that, given zero-indexed arrays A and B, each containing N non-negative integers, returns the number of multiplicative pairs of indices.

If the number of multiplicative pairs is greater than 1,000,000,000, the function should return 1,000,000,000.

For example, given:

A[0] = 0 B[0] = 500,000
A[1] = 1 B[1] = 500,000
A[2] = 2 B[2] = 0
A[3] = 2 B[3] = 0
A[4] = 3 B[4] = 0
A[5] = 5 B[5] = 20,000

the function should return 8, as explained above.

Assume that:

· N is an integer within the range [0..100,000];

· each element of array A is an integer within the range [0..1,000];

· each element of array B is an integer within the range [0..999,999];

· real numbers created from arrays are sorted in non-decreasing order.

Complexity:

· expected worst-case time complexity is O(N);

· expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Copyright 20092015 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Tests

Score: 83 of 100

Estimated time complexity: O(N)

Test name

Running time

Result

example
example test

0.044s

OK

simple_diagonal
simple test with A = B

0.052s

OK

simple_equality
simple test with multiple equalities to be counted

0.052s

OK

mulitple_zeros
many zeros

0.052s

WRONG ANSWER
got 1 expected 2

extreme_empty
empty sequence + [1.5, 3.01]

0.044s

WRONG ANSWER
got -1 expected 0

extreme_single
singleton sequence + [1.4, 3.5]

0.044s

OK

doubles
2-element sequences, precise calculation

0.040s

OK

precision_small
precise calculation, N = 400

0.044s

OK

geometric_small
geometric sequence, N = 111

0.044s

OK

arithmetic_small
arithmetic sequence, N = ~300

0.044s

OK

random_small
random sequence, N = ~600

0.044s

OK

geometric_medium
geometric sequence, N = ~9,000

0.064s

OK

random_medium
random sequence, N = ~10,000

0.064s

OK

arithmetic_medium
arithmetic sequence, N = 20,000

0.088s

OK

geometric_large
geometric sequence, N = ~60,000

0.188s

OK

arithmetic_large
arithmetic sequence, N = 90,000

0.244s

OK

random_large
random sequence, N = 100,000

0.272s

OK

extreme_large
big numbers, N = 100,000

0.280s

OK

extreme_zeros
almost all zeros, N <= 100,000

0.224s

WRONG ANSWER
got 0 expected 1000000000

Solution (language: Python)

def solution(A, B):
# write your code in Python 2.7
if not len(A) or not len(B) or len(A) != len(B):
return -1
# make C and filter all values <= 1
C = [A[i]+float(B[i])/1000000 for i in range(len(A)) if A[i]+float(B[i])/1000000 > 1]
C.sort()
result = 0
p = 0 # position
l = len(C) – 1
while l > p:
res = C[l] / (C[l] – 1)
while (p < l and C[p] < res):
p = p + 1
if p == l:
break
result = result + (l-p)
if result > 1000000000:
return 1000000000
l = l-1
return result



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