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 ')'.
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 3 minutes
Notes
not defined yet
Task timeline
Code: 05:17:06 UTC,
java,
autosave
Code: 05:17:36 UTC,
java,
autosave
// 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
// 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
// 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
// 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
// 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
// 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
Code: 05:19:13 UTC,
java,
verify,
result: Passed
// 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
Code: 05:19:17 UTC,
java,
final,
score:
100
// 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
Correctness tests
1.
0.008 s
OK
2.
0.012 s
OK
1.
0.004 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
1.
0.008 s
OK
2.
0.012 s
OK
3.
0.008 s
OK
expand all
Performance tests
1.
0.032 s
OK
2.
0.032 s
OK
3.
0.008 s
OK
1.
0.204 s
OK
2.
0.008 s
OK
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+
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+
✔
OK
1.
0.060 s
OK
2.
0.064 s
OK
3.
0.008 s
OK
broad_tree_with_deep_paths
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
✔
OK
1.
1.064 s
OK
2.
0.012 s
OK