Codility : Brackets Determine whether a given string of parentheses is properly nested

Your first condition in the closing brackets block checks whether your stack has the size != 1. I assume this is meant to check that you don't have any leftover opening brackets, which is a good idea. However, you'll miss this entire check if your last char isn't a closing bracket/paren/..

This for example would fail for an input like (((.

A simple fix would be replacing this condition with a check after the loop ends that the stack is indeed empty.

This is my C# simple solution which got 100% for Correctness and 100% Performance. Time Complexity is O(N). https://codility.com/demo/results/trainingRVS3SF-DC6/

using System;
using System.Collections.Generic;

class Solution {

    public int solution(string S) {

        // Return 1 if string size is 0 
        if(S.Length==0) return 1;

        Stack<char> brackets = new Stack<char>();

        foreach(char c in S){
            if(c=='['||c=='{'||c=='('){
                brackets.Push(c);
            }
            else{
                // return 0 if no opening brackets found and 
                // first bracket is a closing bracket
                if(brackets.Count==0) return 0;

                if(c==')'){
                    if(brackets.Peek()=='(') brackets.Pop();
                    else return 0;
                }

                if(c=='}'){
                    if(brackets.Peek()=='{') brackets.Pop();
                    else return 0;
                }

                if(c==']'){
                    if(brackets.Peek()=='[') brackets.Pop();
                    else return 0;
                }
            }
        }

        if(brackets.Count==0) return 1;

        return 0;
    }
}

Simple java solution, 100/100

public int solution(String S) {
        Deque<Character> stack = new ArrayDeque<Character>();

        for(int i = 0; i < S.length(); i++) {
            char c = S.charAt(i);

            switch (c) {
                case ')':
                    if (stack.isEmpty() || stack.pop() != '(')
                        return 0;
                    break;
                case ']':
                    if (stack.isEmpty() || stack.pop() != '[')
                        return 0;
                    break;
                case '}':
                    if(stack.isEmpty() || stack.pop() != '{')
                        return 0;
                    break;
                default:
                    stack.push(c);
                    break;
            }
        }

        return stack.isEmpty() ? 1 : 0;
    }