Test results - Codility

Tasks Details

easy

1. PermCheck

Check whether array A is a permutation.

Task Score

88%

Correctness

100%

Performance

80%

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 40 minutes

Notes

not defined yet

Code: 06:09:25 UTC, cpp, verify, result: Failed

show code in pop-up

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

// you can also use includes, for example:
// #include <algorithm>
#include<string.h>
int solution(vector<int> &A) {
    // write your code in C++11
    int n = A.size();
    int* hash  = new int[n + 1];
    memset(hash, 0, 4*(n + 1));
    for (int i = 0; i < n; ++i)
    {
        *(hash + i) = A[i] % (n + 1);
    }
    for (int i = 1; i < n + 1; ++i)
    {
        if (*(hash + i) != 1)
            return 0;
    }
    return 1;   
}

Analysis

expand all Example tests

▶

example1
the first example test

✘

WRONG ANSWER
got 0 expected 1

0.004 s

WRONG ANSWER, got 0 expected 1

▶

example2
the second example test

✔

0.008 s

Code: 06:19:34 UTC, cpp, verify, result: Failed

show code in pop-up

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

// you can also use includes, for example:
// #include <algorithm>
#include<string.h>
int solution(vector<int> &A) {
    // write your code in C++11
    int n = A.size();
    int* hash  = new int[n + 1];
    memset(hash, 0, 4*(n + 1));
    for (int i = 0; i < n; ++i)
    {
        *(hash + i) = A[i] % (n + 1);
    }
    for (int i = 1; i < n + 1; ++i)
    {
        if (*(hash + i) != 1)
            return 0;
    }
    return 1;   
}

Analysis

expand all Example tests

▶

example1
the first example test

✘

WRONG ANSWER
got 0 expected 1

0.008 s

WRONG ANSWER, got 0 expected 1

▶

example2
the second example test

✔

0.008 s

Code: 06:19:39 UTC, cpp, verify, result: Failed

show code in pop-up

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

// you can also use includes, for example:
// #include <algorithm>
#include<string.h>
int solution(vector<int> &A) {
    // write your code in C++11
    int n = A.size();
    int* hash  = new int[n + 1];
    memset(hash, 0, 4*(n + 1));
    for (int i = 0; i < n; ++i)
    {
        *(hash + i) = A[i] % (n + 1);
    }
    for (int i = 1; i < n + 1; ++i)
    {
        if (*(hash + i) != 1)
            return 0;
    }
    return 1;   
}

Analysis

expand all Example tests

▶

example1
the first example test

✘

WRONG ANSWER
got 0 expected 1

0.004 s

WRONG ANSWER, got 0 expected 1

▶

example2
the second example test

✔

0.004 s

Code: 06:32:48 UTC, cpp, verify, result: Failed

show code in pop-up

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

// you can also use includes, for example:
// #include <algorithm>
#include<string.h>
int solution(vector<int> &A) {
    // write your code in C++11
    int n = A.size();
    int* hash  = new int[n];
    memset(hash, 0, 4*(n));
    for (int i = 0; i < n; ++i)
    {
        *(hash + i) = A[i] % (n) + 1;
    }
    for (int i = 0; i < n; ++i)
    {
        if (*(hash + i) == 0)
            return 0;
    }
    return 1;    
}

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.004 s

▶

example2
the second example test

✘

WRONG ANSWER
got 1 expected 0

0.008 s

WRONG ANSWER, got 1 expected 0

Code: 06:35:06 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 20 21 22 23 24 25

// you can also use includes, for example:
// #include <algorithm>
#include<string.h>
int solution(vector<int> &A) {
    // write your code in C++11
    int n = A.size();
    int* hash  = new int[n];
    memset(hash, 0, 4*(n));
    int max = 0;
    for (int i = 0; i < n; ++i)
        if (A[i] > max)
            max = A[i];
    if (max != n)
        return 0;
    for (int i = 0; i < n; ++i)
    {
        *(hash + i) = A[i] % (n) + 1;
    }
    for (int i = 0; i < n; ++i)
    {
        if (*(hash + i) == 0)
            return 0;
    }
    return 1;    
}

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.004 s

▶

example2
the second example test

✔

0.004 s

Code: 06:35:54 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 20 21 22 23 24 25

// you can also use includes, for example:
// #include <algorithm>
#include<string.h>
int solution(vector<int> &A) {
    // write your code in C++11
    int n = A.size();
    int* hash  = new int[n];
    memset(hash, 0, 4*(n));
    int max = 0;
    for (int i = 0; i < n; ++i)
        if (A[i] > max)
            max = A[i];
    if (max != n)
        return 0;
    for (int i = 0; i < n; ++i)
    {
        *(hash + i) = A[i] % (n) + 1;
    }
    for (int i = 0; i < n; ++i)
    {
        if (*(hash + i) == 0)
            return 0;
    }
    return 1;    
}

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.004 s

▶

example2
the second example test

✔

0.004 s

Code: 06:35:59 UTC, cpp, final, score: 88

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

// you can also use includes, for example:
// #include <algorithm>
#include<string.h>
int solution(vector<int> &A) {
    // write your code in C++11
    int n = A.size();
    int* hash  = new int[n];
    memset(hash, 0, 4*(n));
    int max = 0;
    for (int i = 0; i < n; ++i)
        if (A[i] > max)
            max = A[i];
    if (max != n)
        return 0;
    for (int i = 0; i < n; ++i)
    {
        *(hash + i) = A[i] % (n) + 1;
    }
    for (int i = 0; i < n; ++i)
    {
        if (*(hash + i) == 0)
            return 0;
    }
    return 1;    
}

Analysis summary

The following issues have been detected: wrong answers.

Analysis

Detected time complexity:

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

expand all Example tests

▶

example1
the first example test

✔

0.008 s

▶

example2
the second example test

✔

0.008 s

expand all Correctness tests

▶

extreme_max
single element with maximal value

✔

0.004 s

▶

single
single element

✔

0.008 s

▶

double
two elements

✔

0.004 s

▶

antiSum1
total sum is corret (equals 1 + 2 + ... N), but it is not a permutation, N = 3

✔

0.004 s

expand all Performance tests

▶

medium_permutation
permutation, N = ~10,000

✔

0.008 s

▶

antiSum2
total sum is corret (equals 1 + 2 + ... N), but it is not a permutation, N = ~100,000

✘

WRONG ANSWER
got 1 expected 0

0.044 s

WRONG ANSWER, got 1 expected 0

▶

large_permutation
large permutation, N = ~100,000

✔

0.044 s

▶

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

✔

0.044 s

▶

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

✔

0.032 s