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 24 minutes
Notes
not defined yet
Code: 10:44:54 UTC,
java,
autosave
Code: 11:07:55 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");
class Solution {
public int solution(String S) {
Stack<Character> stack = new Stack<>();
if (S.length() == 0) {
return 1;
}
for (int i=0; i<S.length(); i++) {
if (S.charAt(i) == '(' || S.charAt(i) == '{' || S.charAt(i) == '[') {
stack.push(S.charAt(i));
} else {
if (stack.isEmpty()) {
return 0;
}
char preChar = stack.pop();
if (S.charAt(i) == ')' && preChar != '(') {
return 0;
}
if (S.charAt(i) == '}' && preChar != '{') {
return 0;
}
if (S.charAt(i) == ']' && preChar != '[') {
return 0;
}
}
}
if (!stack.isEmpty()) {
return 0;
}
return 1;
}
}
Code: 11:08:01 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
import java.util.Stack;
// 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) {
Stack<Character> stack = new Stack<>();
if (S.length() == 0) {
return 1;
}
for (int i=0; i<S.length(); i++) {
if (S.charAt(i) == '(' || S.charAt(i) == '{' || S.charAt(i) == '[') {
stack.push(S.charAt(i));
} else {
if (stack.isEmpty()) {
return 0;
}
char preChar = stack.pop();
if (S.charAt(i) == ')' && preChar != '(') {
return 0;
}
if (S.charAt(i) == '}' && preChar != '{') {
return 0;
}
if (S.charAt(i) == ']' && preChar != '[') {
return 0;
}
}
}
if (!stack.isEmpty()) {
return 0;
}
return 1;
}
}
Analysis
Code: 11:08:06 UTC,
java,
final,
score: 
100
// you can also use imports, for example:
// import java.util.*;
import java.util.Stack;
// 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) {
Stack<Character> stack = new Stack<>();
if (S.length() == 0) {
return 1;
}
for (int i=0; i<S.length(); i++) {
if (S.charAt(i) == '(' || S.charAt(i) == '{' || S.charAt(i) == '[') {
stack.push(S.charAt(i));
} else {
if (stack.isEmpty()) {
return 0;
}
char preChar = stack.pop();
if (S.charAt(i) == ')' && preChar != '(') {
return 0;
}
if (S.charAt(i) == '}' && preChar != '{') {
return 0;
}
if (S.charAt(i) == ']' && preChar != '[') {
return 0;
}
}
}
if (!stack.isEmpty()) {
return 0;
}
return 1;
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.004 s
OK
2.
0.008 s
OK
3.
0.004 s
OK
4.
0.008 s
OK
5.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
4.
0.004 s
OK
5.
0.008 s
OK
expand all
Performance tests
1.
0.296 s
OK
2.
0.008 s
OK
3.
0.024 s
OK
1.
0.036 s
OK
2.
0.004 s
OK
3.
0.008 s
OK
1.
0.264 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.
0.076 s
OK
2.
0.076 s
OK
3.
0.076 s
OK
4.
0.084 s
OK
5.
0.028 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.
0.184 s
OK
2.
0.184 s
OK