Test results - Codility

Tasks Details

easy

1. Nesting

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

Task Score

100%

Correctness

100%

Performance

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 21

Time spent on task 3 minutes

Notes

not defined yet

Code: 05:17:06 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: 05:17:36 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");

import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        for (cha)
    }
}

Code: 05:18:07 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");

import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        for (char c : S.toCharArray()) {
            switch (c) {
                case '(': stack.push(c); break;
                case ')':
                    if (stack.is)
            }
        }
    }
}

Code: 05:18:30 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");

import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        for (char c : S.toCharArray()) {
            switch (c) {
                case '(': stack.push(c); break;
                case ')':
                    if (stack.isEmpty() || statck.pop() != '(') {
                        return 0;
                    }
            }
        }
    }
}

Code: 05:18:46 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");

import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();
        for (char c : S.toCharArray()) {
            switch (c) {
                case '(':
                    stack.push(c);
                    break;
                case ')':
                    if (stack.isEmpty() || statck.pop() != '(') {
                        return 0;
                    }
                default:
                    break;
            }
        }
    }
}

Code: 05:18:56 UTC, java, verify, result: Failed

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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) {
        Stack<Character> stack = new Stack<>();
        for (char c : S.toCharArray()) {
            switch (c) {
                case '(':
                    stack.push(c);
                    break;
                case ')':
                    if (stack.isEmpty() || statck.pop() != '(') {
                        return 0;
                    }
                default:
                    break;
            }
        }
        return stack.isEmpty() ? 1 : 0;
    }
}

Analysis

Compile error

Solution.java:18: error: cannot find symbol
                    if (stack.isEmpty() || statck.pop() != '(') {
                                           ^
  symbol:   variable statck
  location: class Solution
1 error

Code: 05:19:05 UTC, java, verify, result: Passed

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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) {
        Stack<Character> stack = new Stack<>();
        for (char c : S.toCharArray()) {
            switch (c) {
                case '(':
                    stack.push(c);
                    break;
                case ')':
                    if (stack.isEmpty() || stack.pop() != '(') {
                        return 0;
                    }
                default:
                    break;
            }
        }
        return stack.isEmpty() ? 1 : 0;
    }
}

Analysis

expand all Example tests

▶

example1
example test

✔

0.012 s

▶

example2
example test2

✔

0.012 s

Code: 05:19:13 UTC, java, verify, result: Passed

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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) {
        Stack<Character> stack = new Stack<>();
        for (char c : S.toCharArray()) {
            switch (c) {
                case '(':
                    stack.push(c);
                    break;
                case ')':
                    if (stack.isEmpty() || stack.pop() != '(') {
                        return 0;
                    }
                default:
                    break;
            }
        }
        return stack.isEmpty() ? 1 : 0;
    }
}

Analysis

expand all Example tests

▶

example1
example test

✔

0.004 s

▶

example2
example test2

✔

0.004 s

Code: 05:19:17 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) {
        Stack<Character> stack = new Stack<>();
        for (char c : S.toCharArray()) {
            switch (c) {
                case '(':
                    stack.push(c);
                    break;
                case ')':
                    if (stack.isEmpty() || stack.pop() != '(') {
                        return 0;
                    }
                default:
                    break;
            }
        }
        return stack.isEmpty() ? 1 : 0;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

expand all Example tests

▶

example1
example test

✔

0.004 s

▶

example2
example test2

✔

0.008 s

expand all Correctness tests

▶

negative_match
invalid structure, but the number of parentheses matches

✔

0.008 s

0.012 s

▶

empty
empty string

✔

0.004 s

▶

simple_grouped
simple grouped positive and negative test, length=22

✔

0.008 s

▶

small_random

✔

0.008 s

0.012 s

0.008 s

expand all Performance tests

▶

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

✔

0.032 s

0.008 s

▶

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

✔

0.204 s

0.008 s

▶

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+

✔

0.060 s

0.064 s

0.008 s

▶

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

✔

1.064 s

0.012 s