Codility---Nesting

Task description

A string S consisting of N characters is called properly nested if:

• S is empty;
• S has the form "(U)" where U is a properly nested string;
• S has the form "VW" where V and W are properly nested strings.

For example, string "(()(())())" is properly nested but string "())" isn't.

Write a function:

class Solution { public int solution(String S); }

that, given a string S consisting of N characters, returns 1 if string 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..1,000,000];
• string S consists only of the characters "(" and/or ")".

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(1) (not counting the storage required for input arguments).

 

Solution

 

Programming language used: Java

Total time used: 23 minutes

Effective time used: 23 minutes

Code: 15:25:23 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");
import java.util.Stack;
class Solution {
    public int solution(String S) {
        // write your code in Java SE 8
        Stack<Character> st = new Stack<Character>();
        if(S == null)
            return 1;
        for(int i=0; i<S.length();i++){
            if(S.charAt(i) == '(')
                st.push('(');
            else if(S.charAt(i) == ')' && !st.empty())
                st.pop();
            else if(S.charAt(i) == ')' && st.empty())
                return 0;
        }
        if(st.empty())
            return 1;
        else
            return 0;
    }
}


https://codility.com/demo/results/trainingUYAFS5-NWU/