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 21
Time spent on task 1 minutes
Notes
not defined yet
Code: 14:33:12 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<Integer> st = new Stack<>();
//ASCII code () 40/41, [] 91/93, {} 123/125
for (int i = 0; i < S.length(); i++) {
int tmp = S.charAt(i);
if (tmp == 40 || tmp == 91 || tmp == 123) {
st.push(tmp);
} else if (tmp == 41 || tmp == 93 || tmp == 125) {
if (st.empty())
return 0;
switch (tmp) {
case 41:
if (st.pop() != 40) {
return 0;
}
break;
case 93:
if (st.pop() != 91) {
return 0;
}
break;
case 125:
if (st.pop() != 123) {
return 0;
}
break;
}
}
}
if (st.size() != 0) {
return 0;
} else {
return 1;
}
}
}
}
Code: 14:33:24 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) {
Stack<Integer> st = new Stack<>();
//ASCII code () 40/41, [] 91/93, {} 123/125
for (int i = 0; i < S.length(); i++) {
int tmp = S.charAt(i);
if (tmp == 40 || tmp == 91 || tmp == 123) {
st.push(tmp);
} else if (tmp == 41 || tmp == 93 || tmp == 125) {
if (st.empty())
return 0;
switch (tmp) {
case 41:
if (st.pop() != 40) {
return 0;
}
break;
case 93:
if (st.pop() != 91) {
return 0;
}
break;
case 125:
if (st.pop() != 123) {
return 0;
}
break;
}
}
}
if (st.size() != 0) {
return 0;
} else {
return 1;
}
}
}
Analysis
Code: 14:33:31 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) {
Stack<Integer> st = new Stack<>();
//ASCII code () 40/41, [] 91/93, {} 123/125
for (int i = 0; i < S.length(); i++) {
int tmp = S.charAt(i);
if (tmp == 40 || tmp == 91 || tmp == 123) {
st.push(tmp);
} else if (tmp == 41 || tmp == 93 || tmp == 125) {
if (st.empty())
return 0;
switch (tmp) {
case 41:
if (st.pop() != 40) {
return 0;
}
break;
case 93:
if (st.pop() != 91) {
return 0;
}
break;
case 125:
if (st.pop() != 123) {
return 0;
}
break;
}
}
}
if (st.size() != 0) {
return 0;
} else {
return 1;
}
}
}
Analysis
Code: 14:33:36 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) {
Stack<Integer> st = new Stack<>();
//ASCII code () 40/41, [] 91/93, {} 123/125
for (int i = 0; i < S.length(); i++) {
int tmp = S.charAt(i);
if (tmp == 40 || tmp == 91 || tmp == 123) {
st.push(tmp);
} else if (tmp == 41 || tmp == 93 || tmp == 125) {
if (st.empty())
return 0;
switch (tmp) {
case 41:
if (st.pop() != 40) {
return 0;
}
break;
case 93:
if (st.pop() != 91) {
return 0;
}
break;
case 125:
if (st.pop() != 123) {
return 0;
}
break;
}
}
}
if (st.size() != 0) {
return 0;
} else {
return 1;
}
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
4.
0.008 s
OK
5.
0.008 s
OK
1.
0.008 s
OK
1.
0.012 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
4.
0.008 s
OK
5.
0.008 s
OK
expand all
Performance tests
1.
0.264 s
OK
2.
0.008 s
OK
3.
0.028 s
OK
1.
0.036 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
1.
0.260 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.
0.072 s
OK
2.
0.076 s
OK
3.
0.072 s
OK
4.
0.072 s
OK
5.
0.032 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.
0.164 s
OK
2.
0.164 s
OK