Tasks Details
easy
1.
Nesting
Determine whether a given string of parentheses (single type) is properly nested.
Task Score
100%
Correctness
100%
Performance
100%
A string S consisting of N characters is called properly nested if:
- S is empty;
- S has the form "(U)" where U is a properly nested string;
- S has the form "VW" where V and W are properly nested strings.
For example, string "(()(())())" is properly nested but string "())" isn't.
Write a function:
class Solution { public int solution(String S); }
that, given a string S consisting of N characters, returns 1 if string 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..1,000,000];
- string S is made only of the 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 1 minutes
Notes
not defined yet
Task timeline
Code: 14:11:06 UTC,
java,
autosave
Code: 14:11:33 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");
class Solution {
public int solution(String S) {
// write your code in Java SE 8
int res=0;
Stack s = new Stack();
int n= S.length();
if (n%2!=0)
return 0;
for (char c: S.toCharArray()) {
if (isOpenBracket(c))
s.push(c);
else{
if (s.isEmpty())
return 0;
char l = (char) s.pop();
if (!isPair(l,c))
return 0;
}
}
if (s.isEmpty())
return 1;
else
return 0;
}
static Boolean isOpenBracket( char c){
if ( c=='(')
return true;
else
return false;
}
public static Boolean isPair(char l, char r){
return l == '(' && r == ')';
}
}
Analysis
Compile error
Solution.java:11: error: cannot find symbol
Stack s = new Stack();
^
symbol: class Stack
location: class Solution
Solution.java:11: error: cannot find symbol
Stack s = new Stack();
^
symbol: class Stack
location: class Solution
2 errors
Code: 14:11:43 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) {
// write your code in Java SE 8
int res=0;
Stack s = new Stack();
int n= S.length();
if (n%2!=0)
return 0;
for (char c: S.toCharArray()) {
if (isOpenBracket(c))
s.push(c);
else{
if (s.isEmpty())
return 0;
char l = (char) s.pop();
if (!isPair(l,c))
return 0;
}
}
if (s.isEmpty())
return 1;
else
return 0;
}
static Boolean isOpenBracket( char c){
if ( c=='(')
return true;
else
return false;
}
public static Boolean isPair(char l, char r){
return l == '(' && r == ')';
}
}
Code: 14:11:44 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) {
// write your code in Java SE 8
int res=0;
Stack s = new Stack();
int n= S.length();
if (n%2!=0)
return 0;
for (char c: S.toCharArray()) {
if (isOpenBracket(c))
s.push(c);
else{
if (s.isEmpty())
return 0;
char l = (char) s.pop();
if (!isPair(l,c))
return 0;
}
}
if (s.isEmpty())
return 1;
else
return 0;
}
static Boolean isOpenBracket( char c){
if ( c=='(')
return true;
else
return false;
}
public static Boolean isPair(char l, char r){
return l == '(' && r == ')';
}
}
Analysis
Code: 14:11:52 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) {
// write your code in Java SE 8
int res=0;
Stack s = new Stack();
int n= S.length();
if (n%2!=0)
return 0;
for (char c: S.toCharArray()) {
if (isOpenBracket(c))
s.push(c);
else{
if (s.isEmpty())
return 0;
char l = (char) s.pop();
if (!isPair(l,c))
return 0;
}
}
if (s.isEmpty())
return 1;
else
return 0;
}
static Boolean isOpenBracket( char c){
if ( c=='(')
return true;
else
return false;
}
public static Boolean isPair(char l, char r){
return l == '(' && r == ')';
}
}
Analysis
Code: 14:11:57 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) {
// write your code in Java SE 8
int res=0;
Stack s = new Stack();
int n= S.length();
if (n%2!=0)
return 0;
for (char c: S.toCharArray()) {
if (isOpenBracket(c))
s.push(c);
else{
if (s.isEmpty())
return 0;
char l = (char) s.pop();
if (!isPair(l,c))
return 0;
}
}
if (s.isEmpty())
return 1;
else
return 0;
}
static Boolean isOpenBracket( char c){
if ( c=='(')
return true;
else
return false;
}
public static Boolean isPair(char l, char r){
return l == '(' && r == ')';
}
}
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
1.
0.008 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.008 s
OK
expand all
Performance tests
1.
0.036 s
OK
2.
0.016 s
OK
3.
0.004 s
OK
1.
0.248 s
OK
2.
0.008 s
OK
multiple_full_binary_trees
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+
✔
OK
1.
0.076 s
OK
2.
0.032 s
OK
3.
0.008 s
OK
broad_tree_with_deep_paths
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
✔
OK
1.
1.312 s
OK
2.
0.008 s
OK