[100%] Brackets (codility lessons)

Last updated at 2021-09-29Posted at 2021-09-28

Lesson7

Stacks and Queues

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Easy

Brackets

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

Task description

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 consists only of the following characters: "(", "{", "[", "]", "}" and/or ")".

Source code

class Solution {
    public int solution(String S) {
        java.util.Map<String, String> charPairs = new java.util.HashMap<>();
        charPairs.put(")", "(");
        charPairs.put("}", "{");
        charPairs.put("]", "[");

        java.util.Stack<String> stack = new java.util.Stack();
        for (char c : S.toCharArray()) {
            String currentFlag = String.valueOf(c);
            if (charPairs.containsKey(currentFlag)) {
                // this is an ending flag
                if (stack.isEmpty()) return 0;

                String s = stack.pop();
                if (!charPairs.get(currentFlag).equals(s)) {
                    return 0;
                }
            } else {
                stack.push(currentFlag);
            }
        }

        if (stack.isEmpty()) return 1;

        return 0;
    }
}

Detected time complexity:

O(N)

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