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:
function solution(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 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used JavaScript
Time spent on task 4 minutes
Notes
not defined yet
Code: 09:13:05 UTC,
java,
autosave
Code: 09:14:34 UTC,
js,
autosave
Code: 09:14:44 UTC,
js,
autosave
Code: 09:15:03 UTC,
js,
autosave
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length > 0 && stack.pop()) {}
}
}
}
Code: 09:15:20 UTC,
js,
verify,
result: Failed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0) {
return 0;
}
}
}
return 1;
}
Analysis
Code: 09:15:54 UTC,
js,
autosave
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0 && sta) {
return 0;
}
}
}
return 1;
}
Code: 09:15:56 UTC,
js,
verify,
result: Failed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0 && stack.pop()) {
return 0;
}
}
}
return 1;
}
Analysis
Code: 09:16:12 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0) {
return 0;
} else {
stack.pop();
}
}
}
return 1;
}
Analysis
Code: 09:16:24 UTC,
js,
autosave
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0) {
return 0;
} else {
stack.pop();
}
}
}
if (stack.le)
return 1;
}
Code: 09:16:35 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0) {
return 0;
} else {
stack.pop();
}
}
}
if (stack.length !== 0) {
return 0;
}
return 1;
}
Analysis
Code: 09:16:37 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0) {
return 0;
} else {
stack.pop();
}
}
}
if (stack.length !== 0) {
return 0;
}
return 1;
}
Analysis
Code: 09:16:40 UTC,
js,
final,
score: 
100
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const N = S.length;
const stack = [];
for (let i = 0; i < N; i++) {
if (S[i] === '(') {
stack.push(')');
} else {
if (stack.length === 0) {
return 0;
} else {
stack.pop();
}
}
}
if (stack.length !== 0) {
return 0;
}
return 1;
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.068 s
OK
2.
0.068 s
OK
1.
0.068 s
OK
1.
0.068 s
OK
2.
0.068 s
OK
3.
0.068 s
OK
1.
0.092 s
OK
2.
0.068 s
OK
3.
0.068 s
OK
expand all
Performance tests
1.
0.072 s
OK
2.
0.072 s
OK
3.
0.068 s
OK
1.
0.072 s
OK
2.
0.068 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.
0.072 s
OK
2.
0.072 s
OK
3.
0.068 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.
0.084 s
OK
2.
0.068 s
OK