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 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 6 minutes
Notes
not defined yet
Task timeline
Code: 07:11:39 UTC,
java,
autosave
Code: 07:11:50 UTC,
java,
autosave
// 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 [] chkArr = new int[1000001];
int maxCnt=0;
int cnt=0;
for(int i=0; i<A.length; i++){
if(A[i] > 0){
chkArr[A[i]]++;
cnt++;
}
}
if(A.length == 1){
if(chkArr[1] == 1) return 2;
else return 1;
}
for(int j=1; j<cnt; j++){
if(chkArr[j] == 0){
return j;
}
maxCnt = j;
}
return maxCnt+1;
}
}
Code: 07:12:03 UTC,
java,
verify,
result: Failed
import java.util.Arrays;
class Solution {
public int solution(int[] A) {
int [] chkArr = new int[1000001];
int maxCnt=0;
int cnt=0;
for(int i=0; i<A.length; i++){
if(A[i] > 0){
chkArr[A[i]]++;
cnt++;
}
}
if(A.length == 1){
if(chkArr[1] == 1) return 2;
else return 1;
}
for(int j=1; j<cnt; j++){
if(chkArr[j] == 0){
return j;
}
maxCnt = j;
}
return maxCnt+1;
}
}
Analysis
Code: 07:16:07 UTC,
java,
autosave
import java.util.Arrays;
class Solution {
public int solution(int[] A) {
int [] chkArr = new int[1000001];
int maxCnt=0;
int cnt=0;
for(int i=0; i<A.length; i++){
if(A[i] > 0){
chkArr[A[i]]++;
cnt++;
}
}
if(A.length == 1){
if(chkArr[1] == 1) return 2;
else return 1;
}
for(int j=1; j<=cnt; j++){
if(chkArr[j] == 0){
return j;
}
maxCnt = j;
}
return maxCnt+1;
}
}
Code: 07:16:09 UTC,
java,
verify,
result: Passed
import java.util.Arrays;
class Solution {
public int solution(int[] A) {
int [] chkArr = new int[1000001];
int maxCnt=0;
int cnt=0;
for(int i=0; i<A.length; i++){
if(A[i] > 0){
chkArr[A[i]]++;
cnt++;
}
}
if(A.length == 1){
if(chkArr[1] == 1) return 2;
else return 1;
}
for(int j=1; j<=cnt; j++){
if(chkArr[j] == 0){
return j;
}
maxCnt = j;
}
return maxCnt+1;
}
}
Analysis
Code: 07:16:32 UTC,
java,
verify,
result: Passed
import java.util.Arrays;
class Solution {
public int solution(int[] A) {
int [] chkArr = new int[1000001];
int maxCnt=0;
int cnt=0;
for(int i=0; i<A.length; i++){
if(A[i] > 0){
chkArr[A[i]]++;
cnt++;
}
}
if(A.length == 1){
if(chkArr[1] == 1) return 2;
else return 1;
}
for(int j=1; j<=cnt; j++){
if(chkArr[j] == 0){
return j;
}
maxCnt = j;
}
return maxCnt+1;
}
}
Analysis
Code: 07:16:37 UTC,
java,
final,
score:
100
import java.util.Arrays;
class Solution {
public int solution(int[] A) {
int [] chkArr = new int[1000001];
int maxCnt=0;
int cnt=0;
for(int i=0; i<A.length; i++){
if(A[i] > 0){
chkArr[A[i]]++;
cnt++;
}
}
if(A.length == 1){
if(chkArr[1] == 1) return 2;
else return 1;
}
for(int j=1; j<=cnt; j++){
if(chkArr[j] == 0){
return j;
}
maxCnt = j;
}
return maxCnt+1;
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N * log(N))
expand all
Correctness tests
1.
0.008 s
OK
2.
0.012 s
OK
3.
0.012 s
OK
4.
0.008 s
OK
1.
0.012 s
OK
2.
0.008 s
OK
3.
0.012 s
OK
1.
0.008 s
OK
2.
0.004 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
1.
0.008 s
OK
expand all
Performance tests
1.
0.032 s
OK
2.
0.024 s
OK
3.
0.036 s
OK
1.
0.228 s
OK
1.
0.252 s
OK
2.
0.248 s
OK
1.
0.180 s
OK