Solution to Brackets by codility | Code Says

Solution to Brackets by codility | Code Says

Question: http://codility.com/demo/take-sample-test/brackets

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:

int solution(char *S);

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

// use hashset

I’m surprised so few people didn’t do a simple modulo at the start to check whether there is an even number of chars. In production, you wouldn’t want this algorithm to run against a big string and only terminate at the end when one closing token is missing. You’d want it to terminate as soon as it’s possible

def solution(S):

    matched = {"]":"[", "}":"{", ")": "("}

    to_push = ["[", "{", "("]

    stack = []

    for element in S:

        if element in to_push:

            stack.append(element)

        else:

            if len(stack) == 0:

                return 0

            elif matched[element] != stack.pop():

                return 0

    if len(stack) == 0:

        return 1

    else:

        return 0

https://github.com/acprimer/Codility/blob/master/src/Lesson5/Brackets.java
Use linkedlist as stack, stack in java is deprecated
   public int solution(String S) {
        Stack<Character> stack = new Stack<Character>();
        for (int i = 0; i < S.length(); i++) {
            switch (S.charAt(i)) {
                case '(':
                case '[':
                case '{':
                    stack.push(S.charAt(i));
                    break;
                case ')':
                    if (stack.isEmpty() || stack.peek() != '(') return 0;
                    else stack.pop();
                    break;
                case ']':
                    if (stack.isEmpty() || stack.peek() != '[') return 0;
                    else stack.pop();
                    break;
                case '}':
                    if (stack.isEmpty() || stack.peek() != '{') return 0;
                    else stack.pop();
                    break;
            }
        }
        return stack.isEmpty() ? 1 : 0;
    }

http://www.martinkysel.com/codility-brackets-solution/
http://codility-lessons.blogspot.com/2015/02/lesson-5-brackets.html
https://github.com/yiqin/Codility-Practice/blob/master/Brackets.java
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