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 1 minutes
Notes
not defined yet
Code: 12:02:03 UTC,
java,
autosave
Code: 12:02:09 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");
class Solution {
public int solution(String S) {
java.util.Map<String, String> charPairs = new java.util.HashMap<>();
charPairs.put(")", "(");
charPairs.put("}", "{");
charPairs.put("]", "[");
java.util.Stack<String> stack = new java.util.Stack();
for (char c : S.toCharArray()) {
String currentFlag = String.valueOf(c);
if (charPairs.containsKey(currentFlag)) {
// this is an ending flag
if (stack.isEmpty()) return 0;
String s = stack.pop();
if (!charPairs.get(currentFlag).equals(s)) {
return 0;
}
} else {
stack.push(currentFlag);
}
}
if (stack.isEmpty()) return 1;
return 0;
}
}
Analysis
Code: 12:02:31 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");
class Solution {
public int solution(String S) {
java.util.Map<String, String> charPairs = new java.util.HashMap<>();
charPairs.put(")", "(");
charPairs.put("}", "{");
charPairs.put("]", "[");
java.util.Stack<String> stack = new java.util.Stack();
for (char c : S.toCharArray()) {
String currentFlag = String.valueOf(c);
if (charPairs.containsKey(currentFlag)) {
// this is an ending flag
if (stack.isEmpty()) return 0;
String s = stack.pop();
if (!charPairs.get(currentFlag).equals(s)) {
return 0;
}
} else {
stack.push(currentFlag);
}
}
if (stack.isEmpty()) return 1;
return 0;
}
}
Analysis
Code: 12:02:36 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");
class Solution {
public int solution(String S) {
java.util.Map<String, String> charPairs = new java.util.HashMap<>();
charPairs.put(")", "(");
charPairs.put("}", "{");
charPairs.put("]", "[");
java.util.Stack<String> stack = new java.util.Stack();
for (char c : S.toCharArray()) {
String currentFlag = String.valueOf(c);
if (charPairs.containsKey(currentFlag)) {
// this is an ending flag
if (stack.isEmpty()) return 0;
String s = stack.pop();
if (!charPairs.get(currentFlag).equals(s)) {
return 0;
}
} else {
stack.push(currentFlag);
}
}
if (stack.isEmpty()) return 1;
return 0;
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.012 s
OK
2.
0.012 s
OK
3.
0.008 s
OK
4.
0.012 s
OK
5.
0.012 s
OK
1.
0.012 s
OK
1.
0.008 s
OK
2.
0.012 s
OK
3.
0.004 s
OK
4.
0.008 s
OK
5.
0.008 s
OK
expand all
Performance tests
1.
0.520 s
OK
2.
0.012 s
OK
3.
0.044 s
OK
1.
0.060 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
1.
0.460 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.
0.136 s
OK
2.
0.136 s
OK
3.
0.124 s
OK
4.
0.136 s
OK
5.
0.036 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.
0.308 s
OK
2.
0.304 s
OK