Tasks Details
easy
Find the missing element in a given permutation.
Task Score
100%
Correctness
100%
Performance
100%
An array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.
Your goal is to find that missing element.
Write a function:
int solution(vector<int> &A);
that, given an array A, returns the value of the missing element.
For example, given array A such that:
A[0] = 2 A[1] = 3 A[2] = 1 A[3] = 5the function should return 4, as it is the missing element.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..100,000];
- the elements of A are all distinct;
- each element of array A is an integer within the range [1..(N + 1)].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used C++
Time spent on task 6 minutes
Notes
not defined yet
Code: 23:01:02 UTC,
java,
autosave
Code: 23:01:12 UTC,
cpp,
autosave
Code: 23:02:04 UTC,
cpp,
autosave
#include <iostream>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
//10만개를 sort하면 최소 time이 약 O(10만*log(10만))
bool number[100001] = {false,};
for(int i =0; i<A.size(); i++){
number[A[i]] = true;
}
for(int i = 1; i<A.size()+2; i++){
if(!number[i]) return i;
}
return 0;
}
Code: 23:02:14 UTC,
cpp,
verify,
result: Passed
#include <iostream>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
//10만개를 sort하면 최소 time이 약 O(10만*log(10만))
bool number[100002] = {false,};
for(int i =0; i<A.size(); i++){
number[A[i]] = true;
}
for(int i = 1; i<A.size()+2; i++){
if(!number[i]) return i;
}
return 0;
}
Analysis
Code: 23:02:49 UTC,
cpp,
autosave
#include <iostream>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
//10만개를 sort하면 최소 time이 약 O(10만*log(10만))
bool number[100002] = {false,};
for(int i =0; i<A.size(); i++){
number[A[i]] = true;
}
for(int i = 1; i<A.size()+2; i++){
if(!number[i]) return i;
}
return 0;
}
Code: 23:03:00 UTC,
cpp,
autosave
#include <iostream>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
//10만개를 sort하면 최소 time이 약 O(10만*log(10만))
bool number[100002] = {false,};
for(int i =0; i<A.size(); i++){
number[A[i]] = true;
}
for(int i = 1; i<A.size()+2; i++){
if(!number[i]) return i;
}
return 0;
}
Code: 23:03:10 UTC,
cpp,
verify,
result: Passed
#include <iostream>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
//10만개를 sort하면 최소 time이 약 O(10만*log(10만))
bool number[100002] = {false,};
for(int i =0; i<A.size(); i++){
number[A[i]] = true;
}
for(int i = 1; i<A.size()+2; i++){
if(!number[i]) return i;
}
return 0;
}
Analysis
User test case 1:
[100001]
Code: 23:06:02 UTC,
cpp,
verify,
result: Passed
#include <iostream>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
//10만개를 sort하면 최소 time이 약 O(10만*log(10만))
bool number[100002] = {false,};
for(int i =0; i<A.size(); i++){
number[A[i]] = true;
}
for(int i = 1; i<A.size()+2; i++){
if(!number[i]) return i;
}
return 0;
}
Analysis
User test case 1:
[100001]
Code: 23:06:06 UTC,
cpp,
final,
score:
100
#include <iostream>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
//10만개를 sort하면 최소 time이 약 O(10만*log(10만))
bool number[100002] = {false,};
for(int i =0; i<A.size(); i++){
number[A[i]] = true;
}
for(int i = 1; i<A.size()+2; i++){
if(!number[i]) return i;
}
return 0;
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N) or O(N * log(N))
expand all
Correctness tests
1.
0.001 s
OK
2.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
1.
0.001 s
OK
expand all
Performance tests
1.
0.001 s
OK
1.
0.001 s
OK
1.
0.008 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
1.
0.008 s
OK
1.
0.004 s
OK