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 ')'.
Copyright 2009–2025 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
Code: 19:38:16 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) {
if(S.isEmpty()) return 1;
if(S.length()==1) return 0;
Stack<Character> stack = new Stack<>();
for(int i=0; i<S.length(); i++) {
char current = S.charAt(i);
char peek = stack.peek();
switch (current) {
case '}':
if (!stack.isEmpty() && peek == '{') {
stack.pop();
}
else return 0;
break;
case ']':
if (!stack.isEmpty() && peek == '[') {
stack.pop();
}
else return 0;
break;
case ')':
if (!stack.isEmpty() && peek == '(') {
stack.pop();
}
else return 0;
break;
default:
stack.push(current);
break;
}
}
return stack.size()==0 ? 1 : 0;
}
}
Analysis
expand all
Example tests
1.
0.772 s
RUNTIME ERROR,
tested program terminated unexpectedly
stderr:
Exception in thread "main" java.util.EmptyStackException at java.util.Stack.peek(Stack.java:102) at Solution.solution(Solution.java:18) at wrapper.run(wrapper.java:40) at wrapper.main(wrapper.java:29)
1.
0.458 s
RUNTIME ERROR,
tested program terminated unexpectedly
stderr:
Exception in thread "main" java.util.EmptyStackException at java.util.Stack.peek(Stack.java:102) at Solution.solution(Solution.java:18) at wrapper.run(wrapper.java:40) at wrapper.main(wrapper.java:29)
Code: 19:40:06 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) {
if(S.isEmpty()) return 1;
if(S.length()==1) return 0;
Stack<Character> stack = new Stack<>();
for(int i=0; i<S.length(); i++) {
char current = S.charAt(i);
switch (current) {
case '}':
if (!stack.isEmpty() && stack.peek() == '{') {
stack.pop();
}
else return 0;
break;
case ']':
if (!stack.isEmpty() && stack.peek() == '[') {
stack.pop();
}
else return 0;
break;
case ')':
if (!stack.isEmpty() && stack.peek() == '(') {
stack.pop();
}
else return 0;
break;
default:
stack.push(current);
break;
}
}
return stack.size()==0 ? 1 : 0;
}
}
Analysis
Code: 19:40:16 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) {
if(S.isEmpty()) return 1;
if(S.length()==1) return 0;
Stack<Character> stack = new Stack<>();
for(int i=0; i<S.length(); i++) {
char current = S.charAt(i);
switch (current) {
case '}':
if (!stack.isEmpty() && stack.peek() == '{') {
stack.pop();
}
else return 0;
break;
case ']':
if (!stack.isEmpty() && stack.peek() == '[') {
stack.pop();
}
else return 0;
break;
case ')':
if (!stack.isEmpty() && stack.peek() == '(') {
stack.pop();
}
else return 0;
break;
default:
stack.push(current);
break;
}
}
return stack.size()==0 ? 1 : 0;
}
}
Analysis
Code: 19:40:20 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) {
if(S.isEmpty()) return 1;
if(S.length()==1) return 0;
Stack<Character> stack = new Stack<>();
for(int i=0; i<S.length(); i++) {
char current = S.charAt(i);
switch (current) {
case '}':
if (!stack.isEmpty() && stack.peek() == '{') {
stack.pop();
}
else return 0;
break;
case ']':
if (!stack.isEmpty() && stack.peek() == '[') {
stack.pop();
}
else return 0;
break;
case ')':
if (!stack.isEmpty() && stack.peek() == '(') {
stack.pop();
}
else return 0;
break;
default:
stack.push(current);
break;
}
}
return stack.size()==0 ? 1 : 0;
}
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
1.490 s
OK
2.
1.484 s
OK
3.
1.495 s
OK
4.
1.474 s
OK
5.
1.505 s
OK
1.
1.413 s
OK
1.
1.503 s
OK
2.
1.477 s
OK
3.
1.487 s
OK
4.
1.475 s
OK
5.
1.488 s
OK
expand all
Performance tests
1.
1.699 s
OK
2.
1.479 s
OK
3.
1.504 s
OK
1.
1.505 s
OK
2.
1.476 s
OK
3.
1.485 s
OK
1.
1.745 s
OK
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+
sequence of full trees of the form T=(TT), depths [1..10..1], with/without some brackets at the end, length=49K+
✔
OK
1.
1.563 s
OK
2.
1.539 s
OK
3.
1.549 s
OK
4.
1.562 s
OK
5.
1.501 s
OK
broad_tree_with_deep_paths
string of the form [TTT...T] of 300 T's, each T being '{{{...}}}' nested 200-fold, length=120K+
string of the form [TTT...T] of 300 T's, each T being '{{{...}}}' nested 200-fold, length=120K+
✔
OK
1.
1.645 s
OK
2.
1.670 s
OK