Test results - Codility

Tasks Details

easy

1. Brackets

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

Task Score

100%

Correctness

100%

Performance

100%

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 is made only of the following characters: '(', '{', '[', ']', '}' and/or ')'.

Solution

Programming language used Java 8

Time spent on task 1 minutes

Notes

not defined yet

Code: 02:06:14 UTC, java, autosave

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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");

class Solution {
    public int solution(String S) {
        // write your code in Java SE 8
    }
}

Code: 02:06:18 UTC, java, autosave

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import java.util.*;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        
        for (char s : S.toCharArray()) {
            if ('(' == s || '[' == s || '{' == s) {
                stack.push(s);   
            } else if (!stack.empty()) {
                char pop = stack.pop();
                if (('(' == pop && ')' != s) ||
                    ('[' == pop && ']' != s) ||
                    ('{' == pop && '}' != s)
                ) {
                    return 0;
                }
            }
        }
        
        return stack.empty() ? 1 : 0;
    }
}

Code: 02:06:46 UTC, java, autosave

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import java.util.*;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        
        for (char s : S.toCharArray()) {
            if ('(' == s || '[' == s || '{' == s) {
                stack.push(s);   
            } else if (stack.empty()) {
                return 0;    
            } else {
                char pop = stack.pop();
                if (('(' == pop && ')' != s) ||
                    ('[' == pop && ']' != s) ||
                    ('{' == pop && '}' != s)
                ) {
                    return 0;
                }
            }
        }
        
        return stack.empty() ? 1 : 0;
    }
}

Code: 02:07:00 UTC, java, verify, result: Passed

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import java.util.*;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        
        for (char s : S.toCharArray()) {
            if ('(' == s || '[' == s || '{' == s) {
                stack.push(s);   
            } else if (stack.empty()) {
                return 0;    
            } else {
                char pop = stack.pop();
                if (('(' == pop && ')' != s) ||
                    ('[' == pop && ']' != s) ||
                    ('{' == pop && '}' != s)
                ) {
                    return 0;
                }
            }
        }
        
        return stack.empty() ? 1 : 0;
    }
}

User test case 1:

['{{{']

User test case 2:

['}{']

Analysis

expand all Example tests

▶

example1
example test 1

✔

0.004 s

▶

example2
example test 2

✔

0.004 s

expand all User tests

▶

test_data1

✔

0.008 s

OK
function result: 0

▶

test_data2

✔

0.004 s

OK
function result: 0

Code: 02:07:07 UTC, java, verify, result: Passed

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import java.util.*;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        
        for (char s : S.toCharArray()) {
            if ('(' == s || '[' == s || '{' == s) {
                stack.push(s);   
            } else if (stack.empty()) {
                return 0;    
            } else {
                char pop = stack.pop();
                if (('(' == pop && ')' != s) ||
                    ('[' == pop && ']' != s) ||
                    ('{' == pop && '}' != s)
                ) {
                    return 0;
                }
            }
        }
        
        return stack.empty() ? 1 : 0;
    }
}

User test case 1:

['{{{']

User test case 2:

['}{']

Analysis

expand all Example tests

▶

example1
example test 1

✔

0.008 s

▶

example2
example test 2

✔

0.008 s

expand all User tests

▶

test_data1

✔

0.008 s

OK
function result: 0

▶

test_data2

✔

0.004 s

OK
function result: 0

Code: 02:07:13 UTC, java, final, score: 100

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import java.util.*;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        
        for (char s : S.toCharArray()) {
            if ('(' == s || '[' == s || '{' == s) {
                stack.push(s);   
            } else if (stack.empty()) {
                return 0;    
            } else {
                char pop = stack.pop();
                if (('(' == pop && ')' != s) ||
                    ('[' == pop && ']' != s) ||
                    ('{' == pop && '}' != s)
                ) {
                    return 0;
                }
            }
        }
        
        return stack.empty() ? 1 : 0;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

expand all Example tests

▶

example1
example test 1

✔

0.004 s

▶

example2
example test 2

✔

0.004 s

expand all Correctness tests

▶

negative_match
invalid structures

✔

0.004 s

0.008 s

0.004 s

▶

empty
empty string

✔

0.004 s

▶

simple_grouped
simple grouped positive and negative test, length=22

✔

0.004 s

expand all Performance tests

▶

large1
simple large positive test, 100K ('s followed by 100K )'s + )(

✔

0.228 s

0.008 s

0.020 s

▶

large2
simple large negative test, 10K+1 ('s followed by 10K )'s + )( + ()

✔

0.028 s

0.004 s

▶

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

✔

0.204 s

▶

multiple_full_binary_trees
sequence of full trees of the form T=(TT), depths [1..10..1], with/without some brackets at the end, length=49K+

✔

0.060 s

0.028 s

▶

broad_tree_with_deep_paths
string of the form [TTT...T] of 300 T's, each T being '{{{...}}}' nested 200-fold, length=120K+

✔

0.136 s