Test results - Codility

Tasks Details

easy

1. Brackets

Determine whether a given string of parentheses (multiple types) is properly nested.

100%

A string S consisting of N characters is considered to be properly nested if any of the following conditions is true:

S is empty;

S has the form "(U)" or "[U]" or "{U}" where U is a properly nested string;

S has the form "VW" where V and W are properly nested strings.

For example, the string "{[()()]}" is properly nested but "([)()]" is not.

Write a function:

def solution(S)

that, given a string S consisting of N characters, returns 1 if S is properly nested and 0 otherwise.

For example, given S = "{[()()]}", the function should return 1 and given S = "([)()]", the function should return 0, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..200,000];

string S is made only of the following characters: '(', '{', '[', ']', '}' and/or ')'.

Solution

Programming language used Python

Time spent on task 1 minutes

Notes

not defined yet

Code: 14:06:04 UTC, py, verify, result: Passed

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

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

# -*- coding：utf-8 -*-
# &Author  AnFany
# Lesson 7：Stacks and Queues
# P 7.1 Brackets


def solution(S):
    """
    判断字符串S中的符号是否为正确的嵌套
    :param S: 字符串
    :return: 判断是否为正确的嵌套
    """
    left = '([{'
    brackers_dict = {'(': ')', '[': ']', '{': '}'}
    brackers = []
    if len(S) == 0:
        return 1
    else:
        for i in S:
            if i in left:
                brackers.append(i)
            else:
                try:
                    if brackers_dict[brackers[-1]] == i:
                        brackers.pop(-1)
                    else:
                        return 0
                except IndexError:
                    return 0
        if len(brackers) == 0:
            return 1
        else:
            return 0

Analysis

▶

example1
example test 1

✔

▶

example2
example test 2

✔

Code: 14:06:07 UTC, py, final, score: 100

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

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

# -*- coding：utf-8 -*-
# &Author  AnFany
# Lesson 7：Stacks and Queues
# P 7.1 Brackets


def solution(S):
    """
    判断字符串S中的符号是否为正确的嵌套
    :param S: 字符串
    :return: 判断是否为正确的嵌套
    """
    left = '([{'
    brackers_dict = {'(': ')', '[': ']', '{': '}'}
    brackers = []
    if len(S) == 0:
        return 1
    else:
        for i in S:
            if i in left:
                brackers.append(i)
            else:
                try:
                    if brackers_dict[brackers[-1]] == i:
                        brackers.pop(-1)
                    else:
                        return 0
                except IndexError:
                    return 0
        if len(brackers) == 0:
            return 1
        else:
            return 0

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example1
example test 1

✔

▶

example2
example test 2

✔

▶

negative_match
invalid structures

✔

▶

empty
empty string

✔

▶

simple_grouped
simple grouped positive and negative test, length=22

✔

▶

large1
simple large positive test, 100K ('s followed by 100K )'s + )(

✔

▶

large2
simple large negative test, 10K+1 ('s followed by 10K )'s + )( + ()

✔

▶

large_full_ternary_tree
tree of the form T=(TTT) and depth 11, length=177K+

✔

▶

multiple_full_binary_trees
sequence of full trees of the form T=(TT), depths [1..10..1], with/without some brackets at the end, length=49K+

✔

▶

broad_tree_with_deep_paths
string of the form [TTT...T] of 300 T's, each T being '{{{...}}}' nested 200-fold, length=120K+

✔