Tasks Details
medium
Find the smallest positive integer that does not occur in a given sequence.
Task Score
100%
Correctness
100%
Performance
100%
This is a demo task.
Write a function:
function solution(A);
that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.
For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.
Given A = [1, 2, 3], the function should return 4.
Given A = [−1, −3], the function should return 1.
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 [−1,000,000..1,000,000].
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: 10:26:16 UTC,
java,
autosave
Code: 10:26:28 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const sorted = A.sort((a, b) => a - b)
let current = 0
for (let i = 0; i < sorted.length; i++) {
if (sorted[i] - current > 1) {
return current + 1
}
if (sorted[i] > 0) {
current = sorted[i]
}
}
return current < 0 ? 1 : current + 1
}
Analysis
Code: 10:26:33 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const sorted = A.sort((a, b) => a - b)
let current = 0
for (let i = 0; i < sorted.length; i++) {
if (sorted[i] - current > 1) {
return current + 1
}
if (sorted[i] > 0) {
current = sorted[i]
}
}
return current < 0 ? 1 : current + 1
}
Analysis
Code: 10:26:35 UTC,
js,
final,
score: 
100
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const sorted = A.sort((a, b) => a - b)
let current = 0
for (let i = 0; i < sorted.length; i++) {
if (sorted[i] - current > 1) {
return current + 1
}
if (sorted[i] > 0) {
current = sorted[i]
}
}
return current < 0 ? 1 : current + 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
3.
0.072 s
OK
4.
0.068 s
OK
1.
0.068 s
OK
2.
0.068 s
OK
3.
0.068 s
OK
1.
0.068 s
OK
2.
0.068 s
OK
1.
0.068 s
OK
2.
0.072 s
OK
1.
0.072 s
OK
expand all
Performance tests
1.
0.112 s
OK
2.
0.108 s
OK
3.
0.116 s
OK
1.
0.192 s
OK
1.
0.204 s
OK
2.
0.204 s
OK
1.
0.156 s
OK