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 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 4 minutes
Notes
not defined yet
Task timeline
Code: 16:25:25 UTC,
java,
verify,
result: Passed
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859
// 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) {
// write your code in Java SE 8
if (S.length() % 2 != 0) {
return 0;
}
Character openingBrace = new Character('{');
Character openingBracket = new Character('[');
Character openingParen = new Character('(');
Stack<Character> openingStack = new Stack<Character>();
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
if (c == openingBrace || c == openingBracket || c == openingParen) {
openingStack.push(c);
} else {
if (i == S.length()-1 && openingStack.size() != 1) {
return 0;
}
if (openingStack.isEmpty()) {
return 0;
}
Character openingCharacter = openingStack.pop();
switch (c) {
case '}':
if (!openingCharacter.equals(openingBrace)) {
return 0;
}
break;
case ']':
if (!openingCharacter.equals(openingBracket)) {
return 0;
}
break;
case ')':
if (!openingCharacter.equals(openingParen)) {
return 0;
}
break;
default:
break;
}
}
}
if (! openingStack.isEmpty()) {
return 0;
}
return 1;
}
}
Analysis
Code: 16:25:51 UTC,
java,
verify,
result: Passed
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859
// 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) {
// write your code in Java SE 8
if (S.length() % 2 != 0) {
return 0;
}
Character openingBrace = new Character('{');
Character openingBracket = new Character('[');
Character openingParen = new Character('(');
Stack<Character> openingStack = new Stack<Character>();
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
if (c == openingBrace || c == openingBracket || c == openingParen) {
openingStack.push(c);
} else {
if (i == S.length()-1 && openingStack.size() != 1) {
return 0;
}
if (openingStack.isEmpty()) {
return 0;
}
Character openingCharacter = openingStack.pop();
switch (c) {
case '}':
if (!openingCharacter.equals(openingBrace)) {
return 0;
}
break;
case ']':
if (!openingCharacter.equals(openingBracket)) {
return 0;
}
break;
case ')':
if (!openingCharacter.equals(openingParen)) {
return 0;
}
break;
default:
break;
}
}
}
if (! openingStack.isEmpty()) {
return 0;
}
return 1;
}
}
Analysis
Code: 16:26:24 UTC,
java,
final,
score:
100
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859
// 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) {
// write your code in Java SE 8
if (S.length() % 2 != 0) {
return 0;
}
Character openingBrace = new Character('{');
Character openingBracket = new Character('[');
Character openingParen = new Character('(');
Stack<Character> openingStack = new Stack<Character>();
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
if (c == openingBrace || c == openingBracket || c == openingParen) {
openingStack.push(c);
} else {
if (i == S.length()-1 && openingStack.size() != 1) {
return 0;
}
if (openingStack.isEmpty()) {
return 0;
}
Character openingCharacter = openingStack.pop();
switch (c) {
case '}':
if (!openingCharacter.equals(openingBrace)) {
return 0;
}
break;
case ']':
if (!openingCharacter.equals(openingBracket)) {
return 0;
}
break;
case ')':
if (!openingCharacter.equals(openingParen)) {
return 0;
}
break;
default:
break;
}
}
}
if (! openingStack.isEmpty()) {
return 0;
}
return 1;
}
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.034 s
OK
2.
0.120 s
OK
3.
0.045 s
OK
4.
0.082 s
OK
5.
0.076 s
OK
1.
0.034 s
OK
1.
0.101 s
OK
2.
0.051 s
OK
3.
0.119 s
OK
4.
0.035 s
OK
5.
0.107 s
OK
expand all
Performance tests
1.
0.691 s
OK
2.
0.098 s
OK
3.
0.201 s
OK
1.
0.091 s
OK
2.
0.032 s
OK
3.
0.046 s
OK
1.
0.614 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.
0.338 s
OK
2.
0.057 s
OK
3.
0.361 s
OK
4.
0.445 s
OK
5.
0.131 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.
0.457 s
OK
2.
0.528 s
OK