Test results - Codility

Tasks Details

easy

1. PermCheck

Check whether array A is a permutation.

Task Score

100%

Correctness

100%

Performance

100%

A non-empty array A consisting of N integers is given.

A permutation is a sequence containing each element from 1 to N once, and only once.

For example, array A such that:

A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2

is a permutation, but array A such that:

A[0] = 4 A[1] = 1 A[2] = 3

is not a permutation, because value 2 is missing.

The goal is to check whether array A is a permutation.

Write a function:

int solution(vector<int> &A);

that, given an array A, returns 1 if array A is a permutation and 0 if it is not.

For example, given array A such that:

A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2

the function should return 1.

Given array A such that:

A[0] = 4 A[1] = 1 A[2] = 3

the function should return 0.

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..1,000,000,000].

Solution

Programming language used C++

Time spent on task 5 minutes

Notes

not defined yet

Code: 06:59:03 UTC, java, autosave

show code in pop-up

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// 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) {
        // write your code in Java SE 8
    }
}

Code: 06:59:14 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(A.size()+1, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Code: 07:00:27 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(A.size()+1, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Code: 07:00:48 UTC, cpp, verify, result: Passed

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(A.size()+1, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.001 s

▶

example2
the second example test

✔

0.001 s

expand all User tests

▶

test_data1

✘

RUNTIME ERROR
tested program terminated with exit code 1

0.001 s

RUNTIME ERROR, tested program terminated with exit code 1

stderr:

Segmentation Fault

▶

test_data2

✔

0.001 s

OK
function result: 1

User test case 1:

[1000000001]

User test case 2:

[1]

Code: 07:01:02 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(A.size()+1, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Code: 07:01:03 UTC, cpp, verify, result: Passed

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(A.size()+1, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.001 s

▶

example2
the second example test

✔

0.001 s

expand all User tests

▶

test_data1

✘

RUNTIME ERROR
tested program terminated with exit code 1

0.001 s

RUNTIME ERROR, tested program terminated with exit code 1

stderr:

Segmentation Fault

▶

test_data2

✔

0.001 s

OK
function result: 1

User test case 1:

[1000000000]

User test case 2:

[1]

Code: 07:02:35 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(100, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Code: 07:02:46 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(1000000001, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Code: 07:02:54 UTC, cpp, verify, result: Passed

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(1000000001, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.068 s

▶

example2
the second example test

✔

0.068 s

expand all User tests

▶

test_data1

✔

0.068 s

OK
function result: 0

▶

test_data2

✔

0.068 s

OK
function result: 1

User test case 1:

[1000000000]

User test case 2:

[1]

Code: 07:03:17 UTC, cpp, verify, result: Passed

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(1000000001, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.076 s

▶

example2
the second example test

✔

0.076 s

expand all User tests

▶

test_data1

✔

0.076 s

OK
function result: 0

▶

test_data2

✔

0.076 s

OK
function result: 1

User test case 1:

[1000000000]

User test case 2:

[1]

Code: 07:03:20 UTC, cpp, final, score: 100

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    vector<bool> number(1000000001, false);

    for(int i = 0; i<A.size(); i++){
        if(!number[A[i]]) number[A[i]] = true;
        else return 0;
    }

    for(int i =1; i<A.size()+1; i++){
        if(!number[i]) return 0;
    }

    return 1;
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N) or O(N * log(N))

expand all Example tests

▶

example1
the first example test

✔

0.068 s

▶

example2
the second example test

✔

0.068 s

expand all Correctness tests

▶

extreme_min_max
single element with minimal/maximal value

✔

0.068 s

▶

single
single element

✔

0.068 s

▶

double
two elements

✔

0.068 s

▶

antiSum1
total sum is correct, but it is not a permutation, N <= 10

✔

0.068 s

▶

small_permutation
permutation + one element occurs twice, N = ~100

✔

0.068 s

▶

permutations_of_ranges
permutations of sets like [2..100] for which the anwsers should be false

✔

0.068 s

0.072 s

0.068 s

expand all Performance tests

▶

medium_permutation
permutation + few elements occur twice, N = ~10,000

✔

0.072 s

0.068 s

▶

antiSum2
total sum is correct, but it is not a permutation, N = ~100,000

✔

0.076 s

▶

large_not_permutation
permutation + one element occurs three times, N = ~100,000

✔

0.076 s

▶

large_range
sequence 1, 2, ..., N, N = ~100,000

✔

0.076 s

▶

extreme_values
all the same values, N = ~100,000

✔

0.068 s

0.072 s

0.068 s

▶

various_permutations
all sequences are permutations

✔

0.068 s

0.076 s