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:
function solution(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 JavaScript
Time spent on task 1 minutes
Notes
not defined yet
Task timeline
Code: 16:34:14 UTC,
java,
autosave
Code: 16:34:14 UTC,
java,
autosave
Code: 16:34:26 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const splited = S.split('')
const stack = []
for (let i of splited) {
// 여는 거
if (i === '{' || i === '[' || i ==='(') {
stack.push(i)
} else {
if (stack.size === 0) return 0
// 닫는 것이라면 가장 최근에 열었던 것이랑 비교 한다.
const pop = stack.pop()
if (i === ')') {
if (pop !== '(') {
return 0
}
}
if (i === '}') {
if (pop !== '{') {
return 0
}
}
if (i === ']') {
if (pop !== '[') {
return 0
}
}
}
}
return 1
}
User test case 1:
['{{{{{{{{{{']
Analysis
Code: 16:34:42 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const splited = S.split('')
const stack = []
for (let i of splited) {
// 여는 거
if (i === '{' || i === '[' || i ==='(') {
stack.push(i)
} else {
if (stack.size === 0) return 0
// 닫는 것이라면 가장 최근에 열었던 것이랑 비교 한다.
const pop = stack.pop()
if (i === ')') {
if (pop !== '(') {
return 0
}
}
if (i === '}') {
if (pop !== '{') {
return 0
}
}
if (i === ']') {
if (pop !== '[') {
return 0
}
}
}
}
return stack.length === 0 ? 1 : 0
}
User test case 1:
['{{{{{{{{{{']
Analysis
Code: 16:34:49 UTC,
js,
autosave
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const splited = S.split('')
const stack = []
for (let i of splited) {
// 여는 거
if (i === '{' || i === '[' || i ==='(') {
stack.push(i)
} else {
if (stack.size === 0) return 0
// 닫는 것이라면 가장 최근에 열었던 것이랑 비교 한다.
const pop = stack.pop()
if (i === ')') {
if (pop !== '(') {
return 0
}
}
if (i === '}') {
if (pop !== '{') {
return 0
}
}
if (i === ']') {
if (pop !== '[') {
return 0
}
}
}
}
return stack.length === 0 ? 1 : 0
}
Code: 16:34:59 UTC,
js,
autosave
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const splited = S.split('')
const stack = []
for (let i of splited) {
// 여는 거
if (i === '{' || i === '[' || i ==='(') {
stack.push(i)
} else {
if (stack.size === 0) return 0
// 닫는 것이라면 가장 최근에 열었던 것이랑 비교 한다.
const pop = stack.pop()
if (i === ')') {
if (pop !== '(') {
return 0
}
}
if (i === '}') {
if (pop !== '{') {
return 0
}
}
if (i === ']') {
if (pop !== '[') {
return 0
}
}
}
}
return stack.length === 0 ? 1 : 0
}
Code: 16:35:00 UTC,
js,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const splited = S.split('')
const stack = []
for (let i of splited) {
// 여는 거
if (i === '{' || i === '[' || i ==='(') {
stack.push(i)
} else {
if (stack.size === 0) return 0
// 닫는 것이라면 가장 최근에 열었던 것이랑 비교 한다.
const pop = stack.pop()
if (i === ')') {
if (pop !== '(') {
return 0
}
}
if (i === '}') {
if (pop !== '{') {
return 0
}
}
if (i === ']') {
if (pop !== '[') {
return 0
}
}
}
}
return stack.length === 0 ? 1 : 0
}
User test case 1:
['{{{{{{{{{{']
Analysis
Code: 16:35:02 UTC,
js,
final,
score:
100
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(S) {
const splited = S.split('')
const stack = []
for (let i of splited) {
// 여는 거
if (i === '{' || i === '[' || i ==='(') {
stack.push(i)
} else {
if (stack.size === 0) return 0
// 닫는 것이라면 가장 최근에 열었던 것이랑 비교 한다.
const pop = stack.pop()
if (i === ')') {
if (pop !== '(') {
return 0
}
}
if (i === '}') {
if (pop !== '{') {
return 0
}
}
if (i === ']') {
if (pop !== '[') {
return 0
}
}
}
}
return stack.length === 0 ? 1 : 0
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.072 s
OK
2.
0.072 s
OK
3.
0.072 s
OK
4.
0.072 s
OK
5.
0.072 s
OK
1.
0.072 s
OK
1.
0.072 s
OK
2.
0.072 s
OK
3.
0.072 s
OK
4.
0.072 s
OK
5.
0.072 s
OK
expand all
Performance tests
1.
0.088 s
OK
2.
0.068 s
OK
3.
0.072 s
OK
1.
0.080 s
OK
2.
0.076 s
OK
3.
0.072 s
OK
1.
0.084 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.
0.080 s
OK
2.
0.080 s
OK
3.
0.080 s
OK
4.
0.080 s
OK
5.
0.072 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.
0.084 s
OK
2.
0.084 s
OK