Test results - Codility

Tasks Details

medium

1. ArrayInversionCount

Compute number of inversion in an array.

Task Score

100%

Correctness

100%

Performance

100%

An array A consisting of N integers is given. An inversion is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

def solution(A)

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

For example, in the following array:

A[0] = -1 A[1] = 6 A[2] = 3 A[3] = 4 A[4] = 7 A[5] = 4

there are four inversions:

(1,2) (1,3) (1,5) (4,5)

so the function should return 4.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

Solution

Programming language used Python

Time spent on task 1 minutes

Notes

not defined yet

Code: 01:17:06 UTC, java, autosave

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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: 01:17:09 UTC, py, autosave

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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: 01:17:16 UTC, py, autosave

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

 -*- coding：utf-8 -*-
# &Author  AnFany
# Lesson 99：Future training
# P 99.4 ArrayInversionCount


def solution_direct(A):
    """
    返回数组A中逆序索引对的个数
    时间复杂度O(N**2)
    :param A: 数组
    :return: 逆序索引对的个数
    """
    return len([0 for i in range(len(A)) for j in range(i, len(A)) if A[i] > A[j]])


def merge(a):
    """
    实现归并排序
    :param a:数组
    :return:逆序索引对的个数
    """
    # 总的逆序索引对的个数
    length = len(a)
    if length <= 1:
        return a, 0
    # 分割点
    middle = length // 2
    # 左边
    left_list, left_count = merge(a[: middle])
    # 右边
    right_list, right_count = merge(a[middle:])
    # 左右之和
    count = left_count + right_count
    # 合并在一起的序列，也就是排序后的系列
    sorted_list = []
    #  合并时候的逆序数
    while len(left_list) and len(right_list):
        if left_list[0] > right_list[0]:
            len_left = len(left_list)
            count += len_left
            small = right_list.pop(0)
            sorted_list.append(small)
        else:
            small = left_list.pop(0)
            sorted_list.append(small)

    if left_list:
        sorted_list += left_list
    else:
        sorted_list += right_list
    return sorted_list, count


def solution(A):
    """
    返回数组A中逆序索引对的个数
    利用归并排序的思想，最终的逆序数对的个数=左序列排完序得到的逆序数+右序列排完序得到的逆序数+合并左右得到的逆序数
    其实所有的逆序数归根结底都是合并左右得来的逆序数
    时间复杂度：O(N*log(N))
    :param A: 数组
    :return: 逆序索引对的个数
    """
    length = len(A)
    if length <= 1:
        return 0
    count = merge(A)[1]
    if count > 1e9:
        return -1
    return count

Code: 01:17:18 UTC, py, verify, result: Failed

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

 -*- coding：utf-8 -*-
# &Author  AnFany
# Lesson 99：Future training
# P 99.4 ArrayInversionCount



def merge(a):
    """
    实现归并排序
    :param a:数组
    :return:逆序索引对的个数
    """
    # 总的逆序索引对的个数
    length = len(a)
    if length <= 1:
        return a, 0
    # 分割点
    middle = length // 2
    # 左边
    left_list, left_count = merge(a[: middle])
    # 右边
    right_list, right_count = merge(a[middle:])
    # 左右之和
    count = left_count + right_count
    # 合并在一起的序列，也就是排序后的系列
    sorted_list = []
    #  合并时候的逆序数
    while len(left_list) and len(right_list):
        if left_list[0] > right_list[0]:
            len_left = len(left_list)
            count += len_left
            small = right_list.pop(0)
            sorted_list.append(small)
        else:
            small = left_list.pop(0)
            sorted_list.append(small)

    if left_list:
        sorted_list += left_list
    else:
        sorted_list += right_list
    return sorted_list, count


def solution(A):
    """
    返回数组A中逆序索引对的个数
    利用归并排序的思想，最终的逆序数对的个数=左序列排完序得到的逆序数+右序列排完序得到的逆序数+合并左右得到的逆序数
    其实所有的逆序数归根结底都是合并左右得来的逆序数
    时间复杂度：O(N*log(N))
    :param A: 数组
    :return: 逆序索引对的个数
    """
    length = len(A)
    if length <= 1:
        return 0
    count = merge(A)[1]
    if count > 1e9:
        return -1
    return count

Analysis

expand all Example tests

▶

example1
example test

✘

RUNTIME ERROR
tested program terminated with exit code 1

0.036 s

RUNTIME ERROR, tested program terminated with exit code 1

stderr:

Traceback (most recent call last):
  File "exec.py", line 129, in <module>
    main()
  File "exec.py", line 70, in main
    sol = __import__('solution')
  File "/tmp/solution.py", line 4
    -*- coding\uff1autf-8 -*-
    ^
IndentationError: unexpected indent

Code: 01:17:26 UTC, py, verify, result: Passed

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

# -*- coding：utf-8 -*-
# &Author  AnFany
# Lesson 99：Future training
# P 99.4 ArrayInversionCount



def merge(a):
    """
    实现归并排序
    :param a:数组
    :return:逆序索引对的个数
    """
    # 总的逆序索引对的个数
    length = len(a)
    if length <= 1:
        return a, 0
    # 分割点
    middle = length // 2
    # 左边
    left_list, left_count = merge(a[: middle])
    # 右边
    right_list, right_count = merge(a[middle:])
    # 左右之和
    count = left_count + right_count
    # 合并在一起的序列，也就是排序后的系列
    sorted_list = []
    #  合并时候的逆序数
    while len(left_list) and len(right_list):
        if left_list[0] > right_list[0]:
            len_left = len(left_list)
            count += len_left
            small = right_list.pop(0)
            sorted_list.append(small)
        else:
            small = left_list.pop(0)
            sorted_list.append(small)

    if left_list:
        sorted_list += left_list
    else:
        sorted_list += right_list
    return sorted_list, count


def solution(A):
    """
    返回数组A中逆序索引对的个数
    利用归并排序的思想，最终的逆序数对的个数=左序列排完序得到的逆序数+右序列排完序得到的逆序数+合并左右得到的逆序数
    其实所有的逆序数归根结底都是合并左右得来的逆序数
    时间复杂度：O(N*log(N))
    :param A: 数组
    :return: 逆序索引对的个数
    """
    length = len(A)
    if length <= 1:
        return 0
    count = merge(A)[1]
    if count > 1e9:
        return -1
    return count

Analysis

expand all Example tests

▶

example1
example test

✔

0.036 s

Code: 01:17:29 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 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64

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

# -*- coding：utf-8 -*-
# &Author  AnFany
# Lesson 99：Future training
# P 99.4 ArrayInversionCount



def merge(a):
    """
    实现归并排序
    :param a:数组
    :return:逆序索引对的个数
    """
    # 总的逆序索引对的个数
    length = len(a)
    if length <= 1:
        return a, 0
    # 分割点
    middle = length // 2
    # 左边
    left_list, left_count = merge(a[: middle])
    # 右边
    right_list, right_count = merge(a[middle:])
    # 左右之和
    count = left_count + right_count
    # 合并在一起的序列，也就是排序后的系列
    sorted_list = []
    #  合并时候的逆序数
    while len(left_list) and len(right_list):
        if left_list[0] > right_list[0]:
            len_left = len(left_list)
            count += len_left
            small = right_list.pop(0)
            sorted_list.append(small)
        else:
            small = left_list.pop(0)
            sorted_list.append(small)

    if left_list:
        sorted_list += left_list
    else:
        sorted_list += right_list
    return sorted_list, count


def solution(A):
    """
    返回数组A中逆序索引对的个数
    利用归并排序的思想，最终的逆序数对的个数=左序列排完序得到的逆序数+右序列排完序得到的逆序数+合并左右得到的逆序数
    其实所有的逆序数归根结底都是合并左右得来的逆序数
    时间复杂度：O(N*log(N))
    :param A: 数组
    :return: 逆序索引对的个数
    """
    length = len(A)
    if length <= 1:
        return 0
    count = merge(A)[1]
    if count > 1e9:
        return -1
    return count

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N*log(N))

expand all Example tests

▶

example1
example test

✔

0.036 s

expand all Correctness tests

▶

simple1

✔

0.036 s

▶

simple2

✔

0.036 s

▶

simple3

✔

0.036 s

▶

extreme_0_inv
[0], [], [1,2,3], [1,1,1]

✔

0.036 s

0.040 s

0.036 s

▶

medium1
n=100

✔

0.036 s

▶

medium2
n=200

✔

0.040 s

expand all Performance tests

▶

medium3
n=1000

✔

0.044 s

▶

big1
n=10000

✔

0.148 s

▶

big2
n=20000

✔

0.300 s

▶

big3
n=30000

✔

0.496 s

▶

big_monotonic
long descending and non-ascending sequence

✔

0.512 s

0.500 s

0.508 s