Test results - Codility

Tasks Details

easy

1. Nesting

Determine whether a given string of parentheses (single type) is properly nested.

100%

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.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..1,000,000];

string S is made only of the characters '(' and/or ')'.

Solution

Programming language used Java 8

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 07:59:04 UTC, java, verify, result: Passed

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import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> sstack = new Stack<Character>();
        
        if(S.equals("")){
            return 1;
        }

        for(int i=0; i<S.length();i++){
            if(S.charAt(i)==')'){
                if(sstack.empty()){
                    return 0;
                }
                if(sstack.pop() != '('){
                    return 0;
                }
            }else{
                sstack.push(S.charAt(i));
            }
        }
        if(!sstack.empty()){
            return 0;
        }
        return 1;
    }
}

Analysis

▶

example1
example test

✔

▶

example2
example test2

✔

Code: 07:59:10 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> sstack = new Stack<Character>();
        
        if(S.equals("")){
            return 1;
        }

        for(int i=0; i<S.length();i++){
            if(S.charAt(i)==')'){
                if(sstack.empty()){
                    return 0;
                }
                if(sstack.pop() != '('){
                    return 0;
                }
            }else{
                sstack.push(S.charAt(i));
            }
        }
        if(!sstack.empty()){
            return 0;
        }
        return 1;
    }
}

Analysis

▶

example1
example test

✔

▶

example2
example test2

✔

Code: 07:59:14 UTC, java, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> sstack = new Stack<Character>();
        
        if(S.equals("")){
            return 1;
        }

        for(int i=0; i<S.length();i++){
            if(S.charAt(i)==')'){
                if(sstack.empty()){
                    return 0;
                }
                if(sstack.pop() != '('){
                    return 0;
                }
            }else{
                sstack.push(S.charAt(i));
            }
        }
        if(!sstack.empty()){
            return 0;
        }
        return 1;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example1
example test

✔

▶

example2
example test2

✔

▶

negative_match
invalid structure, but the number of parentheses matches

✔

▶

empty
empty string

✔

▶

simple_grouped
simple grouped positive and negative test, length=22

✔

▶

small_random

✔

▶

large1
simple large positive and negative test, 10K or 10K+1 ('s followed by 10K )'s

✔

▶

large_full_ternary_tree
tree of the form T=(TTT) and depth 11, length=177K+

✔

▶

multiple_full_binary_trees
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+

✔

▶

broad_tree_with_deep_paths
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million

✔