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 8
Time spent on task 3 minutes
Notes
not defined yet
Code: 15:07:31 UTC,
java,
autosave
Code: 15:07:36 UTC,
java,
autosave
import java.util.*;
class Solution {
public int solution(int[] A) {
boolean[] array = new boolean[A.length + 1];
for (int a : A) {
if (a > 0 && a <= A.length) {
array[a] = true;
}
}
for (int i = 1; i<array.length; i++) {
if (!array[i]) {
return i;
}
}
return array.length;
}
}
Code: 15:09:28 UTC,
java,
autosave
import java.util.*;
class Solution {
public int solution(int[] A) {
boolean[] array = new boolean[A.length + 1];
for (int a : A) {
if (a > 0 && a <= A.length) {
array[a] = true;
}
}
for (int i = 1; i <array.length; i++) {
if (!array[i]) {
return i;
}
}
return array.length;
}
}
Code: 15:09:38 UTC,
java,
autosave
import java.util.*;
class Solution {
public int solution(int[] A) {
boolean[] array = new boolean[A.length + 1];
for (int a : A) {
if (a > 0 && a <= A.length) {
array[a] = true;
}
}
for (int i = 1; i < array.length; i++) {
if (!array[i]) {
return i;
}
}
return array.length;
}
}
Code: 15:09:46 UTC,
java,
verify,
result: Passed
import java.util.*;
class Solution {
public int solution(int[] A) {
boolean[] array = new boolean[A.length + 1];
for (int a : A) {
if (a > 0 && a <= A.length) {
array[a] = true;
}
}
for (int i = 1; i < array.length; i++) {
if (!array[i]) {
return i;
}
}
return array.length;
}
}
Analysis
Code: 15:09:56 UTC,
java,
verify,
result: Passed
import java.util.*;
class Solution {
public int solution(int[] A) {
boolean[] array = new boolean[A.length + 1];
for (int a : A) {
if (a > 0 && a <= A.length) {
array[a] = true;
}
}
for (int i = 1; i < array.length; i++) {
if (!array[i]) {
return i;
}
}
return array.length;
}
}
Analysis
Code: 15:10:02 UTC,
java,
final,
score:
100
import java.util.*;
class Solution {
public int solution(int[] A) {
boolean[] array = new boolean[A.length + 1];
for (int a : A) {
if (a > 0 && a <= A.length) {
array[a] = true;
}
}
for (int i = 1; i < array.length; i++) {
if (!array[i]) {
return i;
}
}
return array.length;
}
}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.
0.004 s
OK
4.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
1.
0.004 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.
0.020 s
OK
2.
0.020 s
OK
3.
0.032 s
OK
1.
0.220 s
OK
1.
0.244 s
OK
2.
0.244 s
OK
1.
0.168 s
OK