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 13 minutes
Notes
not defined yet
Task timeline
Code: 23:51:57 UTC,
java,
verify,
result: Failed
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
int[] counts = new int[A.length];
for(int i = 0; i < A.length; i++)
{
if(A[i] > A.length)
{
return 0;
}
else if(counts[A[i]] > 0)
{
return 0;
}
else
{
counts[A[i]] = 1;
}
}
return 0;
}
}
Analysis
expand all
Example tests
1.
0.680 s
RUNTIME ERROR,
tested program terminated unexpectedly
stdout:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 4 at Solution.solution(Solution.java:13) at wrapper.run(wrapper.java:27) at wrapper.main(wrapper.java:20)
1.
0.700 s
OK
Code: 23:52:39 UTC,
java,
verify,
result: Failed
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
int[] counts = new int[A.length];
for(int i = 0; i < A.length; i++)
{
if(A[i] > A.length)
{
return 0;
}
else if(counts[A[i] - 1] > 0)
{
return 0;
}
else
{
counts[A[i] - 1] = 1;
}
}
return 0;
}
}
Analysis
Code: 23:54:11 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
int[] counts = new int[A.length];
for(int i = 0; i < A.length; i++)
{
if(A[i] > A.length)
{
return 0;
}
else if(counts[A[i] - 1] > 0)
{
return 0;
}
else
{
counts[A[i] - 1] = 1;
}
}
return 1;
}
}
Analysis
Code: 23:54:28 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
int[] counts = new int[A.length];
for(int i = 0; i < A.length; i++)
{
if(A[i] > A.length)
{
return 0;
}
else if(counts[A[i] - 1] > 0)
{
return 0;
}
else
{
counts[A[i] - 1] = 1;
}
}
return 1;
}
}
Analysis
Code: 23:54:36 UTC,
java,
final,
score: 
100
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
int[] counts = new int[A.length];
for(int i = 0; i < A.length; i++)
{
if(A[i] > A.length)
{
return 0;
}
else if(counts[A[i] - 1] > 0)
{
return 0;
}
else
{
counts[A[i] - 1] = 1;
}
}
return 1;
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N * log(N))
expand all
Performance tests
1.
0.700 s
OK
antiSum2
total sum is corret (equals 1 + 2 + ... N), but it is not a permutation, N = ~100,000
total sum is corret (equals 1 + 2 + ... N), but it is not a permutation, N = ~100,000
✔
OK
1.
0.870 s
OK
1.
0.870 s
OK
1.
0.890 s
OK
1.
0.790 s
OK