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:
int solution(vector<int> &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.
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used C++
Time spent on task 1 minutes
Notes
not defined yet
Code: 19:47:10 UTC,
java,
autosave
Code: 19:47:16 UTC,
cpp,
verify,
result: Failed
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
int result = 1;
const int N = A.size();
int index_head = 0;
int index_tail = N - 1;
// avoiding INT_MIN in the begininig
while (A[index_head] == INT_MIN) {
index_head++;
}
if (index_head > 0) {
result++;
}
int current = std::max(abs(A[index_head]), abs(A[index_tail]));
while(index_head <= index_tail) {
int former = abs(A[index_head]);
if (former == current) {
index_head++;
continue;
}
int latter = abs(A[index_tail]);
if (latter == current) {
index_tail--;
continue;
}
if (former >= latter) {
current = former;
index_head++;
} else {
current = latter;
index_tail--;
}
result++;
}
return result;
}
Analysis
Compile error
func.cpp: In function 'int solution(std::vector<int>&)':
func.cpp:17:29: error: 'INT_MIN' was not declared in this scope
while (A[index_head] == INT_MIN) {
^~~~~~~
Code: 19:47:40 UTC,
cpp,
verify,
result: Passed
#include <climits>
int solution(vector<int> &A) {
int result = 1;
const int N = A.size();
int index_head = 0;
int index_tail = N - 1;
// avoiding INT_MIN in the begininig
while (A[index_head] == INT_MIN) {
index_head++;
}
if (index_head > 0) {
result++;
}
int current = std::max(abs(A[index_head]), abs(A[index_tail]));
while(index_head <= index_tail) {
int former = abs(A[index_head]);
if (former == current) {
index_head++;
continue;
}
int latter = abs(A[index_tail]);
if (latter == current) {
index_tail--;
continue;
}
if (former >= latter) {
current = former;
index_head++;
} else {
current = latter;
index_tail--;
}
result++;
}
return result;
}
Analysis
Code: 19:47:46 UTC,
cpp,
verify,
result: Passed
#include <climits>
int solution(vector<int> &A) {
int result = 1;
const int N = A.size();
int index_head = 0;
int index_tail = N - 1;
// avoiding INT_MIN in the begininig
while (A[index_head] == INT_MIN) {
index_head++;
}
if (index_head > 0) {
result++;
}
int current = std::max(abs(A[index_head]), abs(A[index_tail]));
while(index_head <= index_tail) {
int former = abs(A[index_head]);
if (former == current) {
index_head++;
continue;
}
int latter = abs(A[index_tail]);
if (latter == current) {
index_tail--;
continue;
}
if (former >= latter) {
current = former;
index_head++;
} else {
current = latter;
index_tail--;
}
result++;
}
return result;
}
Analysis
Code: 19:47:50 UTC,
cpp,
final,
score:
100
#include <climits>
int solution(vector<int> &A) {
int result = 1;
const int N = A.size();
int index_head = 0;
int index_tail = N - 1;
// avoiding INT_MIN in the begininig
while (A[index_head] == INT_MIN) {
index_head++;
}
if (index_head > 0) {
result++;
}
int current = std::max(abs(A[index_head]), abs(A[index_tail]));
while(index_head <= index_tail) {
int former = abs(A[index_head]);
if (former == current) {
index_head++;
continue;
}
int latter = abs(A[index_tail]);
if (latter == current) {
index_tail--;
continue;
}
if (former >= latter) {
current = former;
index_head++;
} else {
current = latter;
index_tail--;
}
result++;
}
return result;
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N*log(N))
expand all
Correctness tests
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
4.
0.001 s
OK
5.
0.001 s
OK
6.
0.001 s
OK
7.
0.001 s
OK
8.
0.001 s
OK
9.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
1.
0.001 s
OK
1.
0.001 s
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1.
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OK
1.
0.001 s
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1.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
1.
0.001 s
OK
1.
0.001 s
OK