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 2009–2025 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
Code: 20:17:32 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) {
int result=1;
int[] count = new int[A.length+1];
for(int i=0; i<A.length; i++) {
if(A[i]<=A.length && count[A[i]]==0) {
count[A[i]]++;
}
}
for(int i=1; i<count.length; i++) {
if(count[i]==0) {
result =0;
break;
}
}
return result;
}
}
Analysis
Code: 20:17:39 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) {
int result=1;
int[] count = new int[A.length+1];
for(int i=0; i<A.length; i++) {
if(A[i]<=A.length && count[A[i]]==0) {
count[A[i]]++;
}
}
for(int i=1; i<count.length; i++) {
if(count[i]==0) {
result =0;
break;
}
}
return result;
}
}
Analysis
Code: 20:17:47 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) {
int result=1;
int[] count = new int[A.length+1];
for(int i=0; i<A.length; i++) {
if(A[i]<=A.length && count[A[i]]==0) {
count[A[i]]++;
}
}
for(int i=1; i<count.length; i++) {
if(count[i]==0) {
result =0;
break;
}
}
return result;
}
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N * log(N)) or O(N)
expand all
Correctness tests
1.
1.501 s
OK
2.
1.506 s
OK
1.
1.490 s
OK
2.
1.467 s
OK
1.
1.515 s
OK
2.
1.462 s
OK
3.
1.478 s
OK
4.
1.490 s
OK
1.
1.488 s
OK
2.
1.468 s
OK
3.
1.480 s
OK
4.
1.486 s
OK
1.
1.507 s
OK
2.
1.483 s
OK
expand all
Performance tests
1.
1.512 s
OK
2.
1.499 s
OK
1.
1.727 s
OK
2.
1.723 s
OK
1.
1.750 s
OK
2.
1.724 s
OK
1.
1.733 s
OK
2.
1.738 s
OK
1.
1.381 s
OK
2.
1.620 s
OK
3.
1.507 s
OK