Codility---Brackets

Posted 2020-09-15 Hugh

Task description

A string S consisting of N characters is considered to be properly nestedif 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.

Assume that:

• N is an integer within the range [0..200,000];
• string S consists only of the following characters: "(", "{", "[", "]", "}" and/or ")".

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N) (not counting the storage required for input arguments).

 

 

Solution

 

Programming language used: Java

Total time used: 14 minutes

Code: 16:26:24 UTC, java, final, score:  100

show code in pop-up

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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");
import java.util.Stack;
class Solution {
    public int solution(String S) {
        // write your code in Java SE 8
        if (S.length() % 2 != 0) {
            return 0;
        }

        Character openingBrace = new Character(‘{‘);
        Character openingBracket = new Character(‘[‘);
        Character openingParen = new Character(‘(‘);
        Stack<Character> openingStack = new Stack<Character>();

        for (int i = 0; i < S.length(); i++) {
            char c = S.charAt(i);
            if (c == openingBrace || c == openingBracket || c == openingParen) {
                openingStack.push(c);
            } else  {
                if (i == S.length()-1 && openingStack.size() != 1) {
                    return 0;
                }
                if (openingStack.isEmpty()) {
                    return 0;
                }
                Character openingCharacter = openingStack.pop();
                switch (c) {
                case ‘}‘:
                    if (!openingCharacter.equals(openingBrace)) {
                        return 0;
                    }
                    break;
                case ‘]‘:
                    if (!openingCharacter.equals(openingBracket)) {
                        return 0;
                    }
                    break;
                case ‘)‘:
                    if (!openingCharacter.equals(openingParen)) {
                        return 0;
                    }
                    break;

                default:
                    break;
                }
            } 
        }
        if (! openingStack.isEmpty()) {
            return 0;
        }

        return 1;
    }
}


https://codility.com/demo/results/training87ME5J-MVG/