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:

class Solution { public int solution(String 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 Java 8

Time spent on task 1 minutes

Notes

not defined yet

Code: 15:44:19 UTC, java, verify, result: Passed

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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(String S) {
        boolean isProperlyNested = true;
        Stack<Character> characterStack = new Stack<>();
        for (Character character : S.toCharArray()) {
            if (isLeftCharacter(character)) {
                characterStack.push(character);
            } else {
                if (!characterStack.isEmpty()) {
                    Character lastCharacter = characterStack.peek();
                    if (properlyNested(lastCharacter, character)) {
                        characterStack.pop();
                    } else {
                        isProperlyNested = false;
                        break;
                    }
                } else {
                    isProperlyNested = false;
                    break;
                }
            }
        }
        if (!characterStack.isEmpty()) {
            isProperlyNested = false;
        }
        return isProperlyNested ? 1 : 0;
    }

    private boolean isLeftCharacter(Character character) {
        return character == '(' || character == '{' || character == '[';
    }

    private boolean properlyNested(Character lastCharacter, Character character) {
        if (lastCharacter == '(') {
            return character == ')';
        }
        if (lastCharacter == '{') {
            return character == '}';
        }
        return character == ']';
    }
}

Analysis

▶

example1
example test 1

✔

▶

example2
example test 2

✔

Code: 15:44:23 UTC, java, final, score: 100

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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(String S) {
        boolean isProperlyNested = true;
        Stack<Character> characterStack = new Stack<>();
        for (Character character : S.toCharArray()) {
            if (isLeftCharacter(character)) {
                characterStack.push(character);
            } else {
                if (!characterStack.isEmpty()) {
                    Character lastCharacter = characterStack.peek();
                    if (properlyNested(lastCharacter, character)) {
                        characterStack.pop();
                    } else {
                        isProperlyNested = false;
                        break;
                    }
                } else {
                    isProperlyNested = false;
                    break;
                }
            }
        }
        if (!characterStack.isEmpty()) {
            isProperlyNested = false;
        }
        return isProperlyNested ? 1 : 0;
    }

    private boolean isLeftCharacter(Character character) {
        return character == '(' || character == '{' || character == '[';
    }

    private boolean properlyNested(Character lastCharacter, Character character) {
        if (lastCharacter == '(') {
            return character == ')';
        }
        if (lastCharacter == '{') {
            return character == '}';
        }
        return character == ']';
    }
}

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+

✔