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:
class Solution { public int solution(int[] 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 Java 21
Time spent on task 24 minutes
Notes
not defined yet
Code: 00:14:52 UTC,
java,
autosave
Code: 00:33:38 UTC,
java,
verify,
result: Failed
// 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) {
boolean[] checker = new boolean[A.length + 1];
int num;
for (int i = 0; i < A.length; i++) {
num = A[i];
if (num > 0 && num < checker.length) checker[num] = true; //양수일때만 체크
}
for (int i = 1; i < checker.length; i++) {
if (false == checker[i]) return i;
}
return 1; //위에 해당하지 않는 경우 음수만 이루어진 경우 이므로 양의 최소수 1 리턴
}
}
Analysis
Code: 00:37:57 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) {
boolean[] checker = new boolean[A.length + 1];
int checkCount = 0;
int num;
for (int i = 0; i < A.length; i++) {
num = A[i];
if (num > 0 && num < checker.length) checker[num] = true; //양수 일 때만 체크
}
for (int i = 1; i < checker.length; i++) {
if(checker[i]) checkCount++;
else return i;
}
return checkCount == (checker.length - 1) ? checker.length : 1; //모두 만족하면 그 다음수 리턴 아니면 모두 음수이므로 양의 최소값 1 리턴
}
}
Analysis
Code: 00:38:08 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) {
boolean[] checker = new boolean[A.length + 1];
int checkCount = 0;
int num;
for (int i = 0; i < A.length; i++) {
num = A[i];
if (num > 0 && num < checker.length) checker[num] = true; //양수 일 때만 체크
}
for (int i = 1; i < checker.length; i++) {
if(checker[i]) checkCount++;
else return i;
}
return checkCount == (checker.length - 1) ? checker.length : 1; //모두 만족하면 그 다음수 리턴 아니면 모두 음수이므로 양의 최소값 1 리턴
}
}
Analysis
Code: 00:38:12 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) {
boolean[] checker = new boolean[A.length + 1];
int checkCount = 0;
int num;
for (int i = 0; i < A.length; i++) {
num = A[i];
if (num > 0 && num < checker.length) checker[num] = true; //양수 일 때만 체크
}
for (int i = 1; i < checker.length; i++) {
if(checker[i]) checkCount++;
else return i;
}
return checkCount == (checker.length - 1) ? checker.length : 1; //모두 만족하면 그 다음수 리턴 아니면 모두 음수이므로 양의 최소값 1 리턴
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N * log(N))
expand all
Correctness tests
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.008 s
OK
4.
0.004 s
OK
1.
0.008 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
1.
0.008 s
OK
2.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
1.
0.004 s
OK
expand all
Performance tests
1.
0.028 s
OK
2.
0.020 s
OK
3.
0.032 s
OK
1.
0.224 s
OK
1.
0.248 s
OK
2.
0.244 s
OK
1.
0.172 s
OK