Java solution to Codility PermCheck (Permuation Check) problem (Lesson 4 – Counting Elements) which scored 100%. The problem is to check if the given array is a permutation. The main strategy is to use two java.util.HashSets: 1) a perfect set and 2) the actual set and check for the missing element in the actual set that exists in the perfect set.
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import java.util.HashSet;
import java.util.Set;
public class PermutationCheck {
public int solution(int[] A) {
Set testedSet = new HashSet();
Set perfectSet = new HashSet();
for(int i=0; i<A.length; i++) {
testedSet.add(A[i]);
perfectSet.add(i+1);
}
for(int current : perfectSet) {
//as soon as find 1 element missing return false
if(!testedSet.contains(current)) return 0;
}
return 1;
}
}
TestNG test cases for this problem which all passed:
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import org.testng.Assert;
import org.testng.annotations.*;
import com.codility.lesson04.countingelements.PermutationCheck;
public class PermutationCheckTests {
private PermutationCheck solution;
@BeforeTest
public void setUp() {
solution = new PermutationCheck();
}
@DataProvider(name = "test1")
public Object [][] createData1() {
return new Object [][] {
new Object [] { new int [] { 1 }, 1 },
new Object [] { new int [] { 2 }, 0 },
new Object [] { new int [] { 1, 2 }, 1 },
new Object [] { new int [] { 2, 2 }, 0 },
new Object [] { new int [] { 1, 3, 2 }, 1 },
new Object [] { new int [] { 1, 3, 3 }, 0 },
new Object [] { new int [] { 4, 1, 3, 2 }, 1 },
new Object [] { new int [] { 4, 1, 3 }, 0 },
new Object [] { new int [] { 1, 3, 5, 4, 2 }, 1 },
new Object [] { new int [] { 1, 3, 6, 4, 1, 2 }, 0 },
new Object [] { new int [] { 1, 3, 5, 4, 6, 2 }, 1 },
new Object [] { new int [] { 1000000 }, 0 },
new Object [] { new int [] { 1, 1000000 }, 0 },
};
}
@Test(dataProvider = "test1")
public void verifySolution(int [] pA, int pExpected) {
Assert.assertEquals(solution.solution(pA), pExpected);
}
}