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:
function solution(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 JavaScript
Time spent on task 1 minutes
Notes
not defined yet
Code: 09:22:59 UTC,
js,
verify,
result: Passed
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var sum = 0;
var numbers = {};
for (var i = 0; i < A.length; i++) {
var a = A[i];
sum += a;
if (numbers[a] === 1) {
return 0;
}
else {
numbers[a] = 1;
}
}
var n = A.length;
var sum_n = (n * (n + 1)) / 2;
var difference = sum_n - sum;
if (difference !== 0) return 0;
return 1;
}
Analysis
Code: 09:23:03 UTC,
js,
verify,
result: Passed
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var sum = 0;
var numbers = {};
for (var i = 0; i < A.length; i++) {
var a = A[i];
sum += a;
if (numbers[a] === 1) {
return 0;
}
else {
numbers[a] = 1;
}
}
var n = A.length;
var sum_n = (n * (n + 1)) / 2;
var difference = sum_n - sum;
if (difference !== 0) return 0;
return 1;
}
Analysis
Code: 09:23:05 UTC,
js,
final,
score:
100
// you can use console.log for debugging purposes, i.e.
// console.log('this is a debug message');
function solution(A) {
// write your code in JavaScript (ECMA-262, 5th edition)
var sum = 0;
var numbers = {};
for (var i = 0; i < A.length; i++) {
var a = A[i];
sum += a;
if (numbers[a] === 1) {
return 0;
}
else {
numbers[a] = 1;
}
}
var n = A.length;
var sum_n = (n * (n + 1)) / 2;
var difference = sum_n - sum;
if (difference !== 0) return 0;
return 1;
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N * log(N))
expand all
Correctness tests
1.
0.072 s
OK
2.
0.072 s
OK
1.
0.072 s
OK
2.
0.072 s
OK
1.
0.073 s
OK
2.
0.072 s
OK
3.
0.072 s
OK
4.
0.072 s
OK
1.
0.085 s
OK
2.
0.072 s
OK
3.
0.072 s
OK
4.
0.072 s
OK
1.
0.072 s
OK
2.
0.073 s
OK
expand all
Performance tests
1.
0.087 s
OK
2.
0.087 s
OK
1.
0.131 s
OK
2.
0.145 s
OK
1.
0.153 s
OK
2.
0.150 s
OK
1.
0.144 s
OK
2.
0.159 s
OK
1.
0.072 s
OK
2.
0.115 s
OK
3.
0.072 s
OK