Tasks Details
easy
1.
Distinct
Compute number of distinct values in an array.
Task Score
100%
Correctness
100%
Performance
100%
Write a function
class Solution { public int solution(int[] A); }
that, given an array A consisting of N integers, returns the number of distinct values in array A.
For example, given array A consisting of six elements such that:
A[0] = 2 A[1] = 1 A[2] = 1 A[3] = 2 A[4] = 3 A[5] = 1the function should return 3, because there are 3 distinct values appearing in array A, namely 1, 2 and 3.
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 [−1,000,000..1,000,000].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 17 minutes
Notes
not defined yet
Task timeline
Code: 13:16:42 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = A.length;
// first part for negatives, second part for positives and adding
// counting zero as part of the positives section
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : element;
bitSet.set(index);
}
return bitSet.cardinality();
}
}
Analysis
Code: 13:17:24 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = A.length;
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : element;
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
expand all
User tests
1.
1.301 s
OK
function result: 1
function result: 1
1.
1.331 s
RUNTIME ERROR,
tested program terminated unexpectedly
stderr:
Exception in thread "main" java.lang.IndexOutOfBoundsException: bitIndex < 0: -1 at java.util.BitSet.set(BitSet.java:444) at Solution.solution(Solution.java:10) at wrapper.run(wrapper.java:40) at wrapper.main(wrapper.java:29)
Code: 13:17:48 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = A.length;
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1 );
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
Code: 13:19:06 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 100000;
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1 );
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
Code: 13:19:22 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 1000000;
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1 );
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
Code: 13:19:44 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 1000000;
BitSet bitSet = new BitSet( (offset * 2) + 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1 );
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
Code: 13:20:20 UTC,
java,
verify,
result: Failed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 1e6;
// first part for negatives, second part for positives and adding
// counting zero as part of the positives section
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1);
bitSet.set(index);
}
return bitSet.cardinality();
}
}
Analysis
Compile error
Solution.java:5: error: incompatible types: possible lossy conversion from double to int
int offset = 1e6;
^
1 error
Code: 13:22:23 UTC,
java,
verify,
result: Failed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 10e6;
// first part for negatives, second part for positives and adding
// counting zero as part of the positives section
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1);
bitSet.set(index);
}
return bitSet.cardinality();
}
}
Analysis
Compile error
Solution.java:5: error: incompatible types: possible lossy conversion from double to int
int offset = 10e6;
^
1 error
Code: 13:26:13 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 1_000_000;
// first part for negatives, second part for positives and adding
// counting zero as part of the positives section
BitSet bitSet = new BitSet( (offset * 2) - 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1);
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
Code: 13:26:54 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 1_000_000;
// first part for negatives, second part for positives and adding
// counting zero as part of the positives section
BitSet bitSet = new BitSet( (offset * 2) + 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1);
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
Code: 13:27:11 UTC,
java,
verify,
result: Passed
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 1_000_000;
// first part for negatives, second part for positives and adding
// counting zero as part of the positives section
BitSet bitSet = new BitSet( (offset * 2) + 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1);
bitSet.set(index);
}
return bitSet.cardinality();
}
}
User test case 1:
[0]
User test case 2:
[-1, 1, 0, 1]
Analysis
Code: 13:27:19 UTC,
java,
final,
score:
100
import java.util.BitSet;
class Solution {
public int solution(int[] A) {
int offset = 1_000_000;
// first part for negatives, second part for positives and adding
// counting zero as part of the positives section
BitSet bitSet = new BitSet( (offset * 2) + 1 );
for (int element : A ) {
int index = element >= 0 ? offset + element : (element * -1);
bitSet.set(index);
}
return bitSet.cardinality();
}
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N*log(N)) or O(N)
expand all
Correctness tests
1.
1.317 s
OK
1.
1.332 s
OK
2.
1.325 s
OK
1.
1.314 s
OK
1.
1.328 s
OK
1.
1.336 s
OK
1.
1.326 s
OK
1.
1.325 s
OK
1.
1.327 s
OK
1.
1.188 s
OK