Test results - Codility

Tasks Details

easy

1. Dominator

Find an index of an array such that its value occurs at more than half of indices in the array.

100%

An array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.

For example, consider array A such that

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

The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.

Write a function

def solution(A)

that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.

For example, given array A such that

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

the function may return 0, 2, 4, 6 or 7, as explained above.

Write an efficient algorithm for the following assumptions:

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

each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

Solution

Programming language used Python

Time spent on task 2 minutes

Notes

not defined yet

Task timeline

Code: 20:25:19 UTC, py, verify, result: Failed

1 2 3 4 5 6 7 8 9 10 11 12 13 14

import bisect;
import operator;

def solution(A):
    N = len(A)
    if N == 0: return -1
    ks, vs = zip(*sorted(zip([i for i in xrange(0, len(A))], A),  key=operator.itemgetter(1), reverse=False))
    half = N/2
    dominator = vs[half]
    lo = bisect.bisect_left(vs, dominator, 0, half)
    hi = bisect.bisect_right(vs, dominator, lo=half)  
    dominatorCount = hi - lo
    print dominatorCount
    return -1 if dominatorCount < half+1 else dominatorCount;

Analysis

▶

example
example test

✘

WRONG ANSWER
got 5, but element is not a dominator, value -1 has only 1 occurences (n=8)

Code: 20:26:02 UTC, py, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14

import bisect;
import operator;

def solution(A):
    N = len(A)
    if N == 0: return -1
    ks, vs = zip(*sorted(zip([i for i in xrange(0, len(A))], A),  key=operator.itemgetter(1), reverse=False))
    half = N/2
    dominator = vs[half]
    lo = bisect.bisect_left(vs, dominator, 0, half)
    hi = bisect.bisect_right(vs, dominator, lo=half)  
    dominatorCount = hi - lo
    print dominatorCount
    return -1 if dominatorCount < half+1 else ks[half];

Analysis

▶

example
example test

✔

Code: 20:26:14 UTC, py, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13

import bisect;
import operator;

def solution(A):
    N = len(A)
    if N == 0: return -1
    ks, vs = zip(*sorted(zip([i for i in xrange(0, len(A))], A),  key=operator.itemgetter(1), reverse=False))
    half = N/2
    dominator = vs[half]
    lo = bisect.bisect_left(vs, dominator, 0, half)
    hi = bisect.bisect_right(vs, dominator, lo=half)  
    dominatorCount = hi - lo
    return -1 if dominatorCount < half+1 else ks[half];

Analysis

▶

example
example test

✔

Code: 20:26:25 UTC, py, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13

import bisect;
import operator;

def solution(A):
    N = len(A)
    if N == 0: return -1
    ks, vs = zip(*sorted(zip([i for i in xrange(0, len(A))], A),  key=operator.itemgetter(1), reverse=False))
    half = N/2
    dominator = vs[half]
    lo = bisect.bisect_left(vs, dominator, 0, half)
    hi = bisect.bisect_right(vs, dominator, lo=half)  
    dominatorCount = hi - lo
    return -1 if dominatorCount < half+1 else ks[half];

Analysis

▶

example
example test

✔

Code: 20:26:30 UTC, py, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13

import bisect;
import operator;

def solution(A):
    N = len(A)
    if N == 0: return -1
    ks, vs = zip(*sorted(zip([i for i in xrange(0, len(A))], A),  key=operator.itemgetter(1), reverse=False))
    half = N/2
    dominator = vs[half]
    lo = bisect.bisect_left(vs, dominator, 0, half)
    hi = bisect.bisect_right(vs, dominator, lo=half)  
    dominatorCount = hi - lo
    return -1 if dominatorCount < half+1 else ks[half];

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example
example test

✔

▶

small_nondominator
all different and all the same elements

✔

▶

small_half_positions
half elements the same, and half + 1 elements the same

✔

▶

small
small test

✔

▶

small_pyramid
decreasing and plateau, small

✔

▶

extreme_empty_and_single_item
empty and single element arrays

✔

▶

extreme_half1
array with exactly N/2 values 1, N even + [0,0,1,1,1]

✔

▶

extreme_half2
array with exactly floor(N/2) values 1, N odd + [0,0,1,1,1]

✔

▶

extreme_half3
array with exactly ceil(N/2) values 1 + [0,0,1,1,1]

✔

▶

medium_pyramid
decreasing and plateau, medium

✔

▶

large_pyramid
decreasing and plateau, large

✔

▶

medium_random
random test with dominator, N = 10,000

✔

▶

large_random
random test with dominator, N = 100,000

✔