Tasks Details
easy
1.
PermCheck
Check whether array A is a permutation.
Task Score
100%
Correctness
100%
Performance
100%
A non-empty array A consisting of N integers is given.
A permutation is a sequence containing each element from 1 to N once, and only once.
For example, array A such that:
A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2is a permutation, but array A such that:
A[0] = 4 A[1] = 1 A[2] = 3is not a permutation, because value 2 is missing.
The goal is to check whether array A is a permutation.
Write a function:
class Solution { public int solution(int[] A); }
that, given an array A, returns 1 if array A is a permutation and 0 if it is not.
For example, given array A such that:
A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2the function should return 1.
Given array A such that:
A[0] = 4 A[1] = 1 A[2] = 3the function should return 0.
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 [1..1,000,000,000].
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 1 minutes
Notes
not defined yet
Task timeline
Code: 16:53:09 UTC,
java,
autosave
Code: 16:53:17 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
Set<Integer> nums = new HashSet<Integer>();
for (int i = 1; i <= A.length; i++) {
nums.add(i);
}
for (int a : A) {
if (!nums.contains(a)) {
return 0;
} else {
nums.remove(a);
}
}
return nums.isEmpty() ? 1 : 0;
}
}
Analysis
Code: 16:53:23 UTC,
java,
final,
score: 
100
// you can also use imports, for example:
import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
Set<Integer> nums = new HashSet<Integer>();
for (int i = 1; i <= A.length; i++) {
nums.add(i);
}
for (int a : A) {
if (!nums.contains(a)) {
return 0;
} else {
nums.remove(a);
}
}
return nums.isEmpty() ? 1 : 0;
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N * log(N))
expand all
Correctness tests
1.
0.008 s
OK
2.
0.008 s
OK
1.
0.004 s
OK
2.
0.008 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
4.
0.008 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
4.
0.008 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
permutations_of_ranges
permutations of sets like [2..100] for which the anwsers should be false
permutations of sets like [2..100] for which the anwsers should be false
✔
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
expand all
Performance tests
1.
0.060 s
OK
2.
0.064 s
OK
1.
0.372 s
OK
2.
0.696 s
OK
1.
0.612 s
OK
2.
0.604 s
OK
1.
0.692 s
OK
2.
0.692 s
OK
1.
0.008 s
OK
2.
0.280 s
OK
3.
0.008 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.064 s
OK
4.
0.692 s
OK
5.
0.612 s
OK