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Tasks Details

easy

1. PermCheck

Check whether array A is a permutation.

Task Score

100%

Correctness

100%

Performance

100%

Task description

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:

def solution(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 Python

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

‌

Code: 07:05:48 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: 07:05:53 UTC, py, autosave

show code in pop-up

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# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(A):
    # write your code in Python 3.6
    pass

Code: 07:05:56 UTC, py, 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

def solution(ar):
    """
    idea is use xor - xor of two same number cancel the same bits and make that to zero
    do xor of all elements of array
    do xor of 1 to length plus 1
    at last xor of all should be zero

    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.
    :param ar:
    :return:
    """
    xor_sum = 0
    # xor of 1 to len ar
    # xor of all elements of existing array
    for index in range(1, len(ar) + 1):
        xor_sum = xor_sum ^ index ^ ar[index - 1]
    # at the end xor sum should be zero
    if xor_sum == 0:
        return 1
    else:
        return 0

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.036 s

▶

example2
the second example test

✔

0.036 s

Code: 07:06:00 UTC, py, 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

def solution(ar):
    """
    idea is use xor - xor of two same number cancel the same bits and make that to zero
    do xor of all elements of array
    do xor of 1 to length plus 1
    at last xor of all should be zero

    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.
    :param ar:
    :return:
    """
    xor_sum = 0
    # xor of 1 to len ar
    # xor of all elements of existing array
    for index in range(1, len(ar) + 1):
        xor_sum = xor_sum ^ index ^ ar[index - 1]
    # at the end xor sum should be zero
    if xor_sum == 0:
        return 1
    else:
        return 0

Analysis

expand all Example tests

▶

example1
the first example test

✔

0.036 s

▶

example2
the second example test

✔

0.036 s

Code: 07:06:02 UTC, py, final, score: 100

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

def solution(ar):
    """
    idea is use xor - xor of two same number cancel the same bits and make that to zero
    do xor of all elements of array
    do xor of 1 to length plus 1
    at last xor of all should be zero

    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.
    :param ar:
    :return:
    """
    xor_sum = 0
    # xor of 1 to len ar
    # xor of all elements of existing array
    for index in range(1, len(ar) + 1):
        xor_sum = xor_sum ^ index ^ ar[index - 1]
    # at the end xor sum should be zero
    if xor_sum == 0:
        return 1
    else:
        return 0

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

✔

0.036 s

▶

example2
the second example test

✔

0.036 s

expand all Correctness tests

▶

extreme_min_max
single element with minimal/maximal value

✔

0.036 s

▶

single
single element

✔

0.036 s

▶

double
two elements

✔

0.036 s

▶

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

✔

0.036 s

▶

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

✔

0.036 s

▶

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

✔

0.036 s

expand all Performance tests

▶

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

✔

0.048 s

0.044 s

▶

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

✔

0.148 s

▶

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

✔

0.144 s

0.140 s

▶

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

✔

0.140 s

0.152 s

▶

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

✔

0.036 s

0.148 s

0.036 s

▶

various_permutations
all sequences are permutations

✔

0.036 s

0.048 s

0.152 s

0.140 s