Tasks Details
easy
1.
AbsDistinct
Compute number of distinct absolute values of sorted array elements.
Task Score
100%
Correctness
100%
Performance
100%
A non-empty array A consisting of N numbers is given. The array is sorted in non-decreasing order. The absolute distinct count of this array is the number of distinct absolute values among the elements of the array.
For example, consider array A such that:
A[0] = -5 A[1] = -3 A[2] = -1 A[3] = 0 A[4] = 3 A[5] = 6The absolute distinct count of this array is 5, because there are 5 distinct absolute values among the elements of this array, namely 0, 1, 3, 5 and 6.
Write a function:
def solution(A)
that, given a non-empty array A consisting of N numbers, returns absolute distinct count of array A.
For example, given array A such that:
A[0] = -5 A[1] = -3 A[2] = -1 A[3] = 0 A[4] = 3 A[5] = 6the function should return 5, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647];
- array A is sorted in non-decreasing order.
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Solution
Programming language used Python
Time spent on task 1 minutes
Notes
not defined yet
Code: 20:28:16 UTC,
java,
autosave
Code: 20:28:30 UTC,
py,
verify,
result: Passed
def solution(A):
right=len(A)-1
left=0
count=len(A)
while (right!=left):
if abs(A[right])<abs(A[left]):
left +=1
if A[left]==A[left-1]: count -= 1
elif abs(A[right])>abs(A[left]):
right -=1
if A[right]==A[right+1]: count -= 1
elif abs(A[right])==abs(A[left]):
count -=1
left +=1
return(count)
Analysis
Code: 20:28:43 UTC,
py,
verify,
result: Passed
def solution(A):
right=len(A)-1
left=0
count=len(A)
while (right!=left):
if abs(A[right])<abs(A[left]):
left +=1
if A[left]==A[left-1]: count -= 1
elif abs(A[right])>abs(A[left]):
right -=1
if A[right]==A[right+1]: count -= 1
elif abs(A[right])==abs(A[left]):
count -=1
left +=1
return(count)
Analysis
Code: 20:28:45 UTC,
py,
final,
score: 
100
def solution(A):
right=len(A)-1
left=0
count=len(A)
while (right!=left):
if abs(A[right])<abs(A[left]):
left +=1
if A[left]==A[left-1]: count -= 1
elif abs(A[right])>abs(A[left]):
right -=1
if A[right]==A[right+1]: count -= 1
elif abs(A[right])==abs(A[left]):
count -=1
left +=1
return(count)
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N*log(N))
expand all
Correctness tests
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
4.
0.036 s
OK
5.
0.036 s
OK
6.
0.036 s
OK
7.
0.036 s
OK
8.
0.036 s
OK
9.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK