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:
func Solution(S string) int
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 Go
Time spent on task 1 minutes
Notes
not defined yet
Task timeline
Code: 07:04:36 UTC,
go,
verify,
result: Passed
package solution
// you can also use imports, for example:
// import "fmt"
// import "os"
// you can write to stdout for debugging purposes, e.g.
// fmt.Println("this is a debug message")
func Solution(S string) int {
b := []byte(S)
stack := make([]byte, 0)
open := byte('(')
close := byte(')')
for i := range b {
s := b[i]
ln := len(stack)
if ln > 0 && stack[ln-1] == open && s == close {
stack = stack[:ln-1]
continue
}
stack = append(stack, s)
}
if len(stack) == 0 {
return 1
} else {
return 0
}
}
Analysis
Code: 07:04:45 UTC,
go,
verify,
result: Passed
package solution
// you can also use imports, for example:
// import "fmt"
// import "os"
// you can write to stdout for debugging purposes, e.g.
// fmt.Println("this is a debug message")
func Solution(S string) int {
b := []byte(S)
stack := make([]byte, 0)
open := byte('(')
close := byte(')')
for i := range b {
s := b[i]
ln := len(stack)
if ln > 0 && stack[ln-1] == open && s == close {
stack = stack[:ln-1]
continue
}
stack = append(stack, s)
}
if len(stack) == 0 {
return 1
} else {
return 0
}
}
Analysis
Code: 07:04:48 UTC,
go,
final,
score:
100
package solution
// you can also use imports, for example:
// import "fmt"
// import "os"
// you can write to stdout for debugging purposes, e.g.
// fmt.Println("this is a debug message")
func Solution(S string) int {
b := []byte(S)
stack := make([]byte, 0)
open := byte('(')
close := byte(')')
for i := range b {
s := b[i]
ln := len(stack)
if ln > 0 && stack[ln-1] == open && s == close {
stack = stack[:ln-1]
continue
}
stack = append(stack, s)
}
if len(stack) == 0 {
return 1
} else {
return 0
}
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.009 s
OK
2.
0.009 s
OK
1.
0.009 s
OK
1.
0.009 s
OK
2.
0.010 s
OK
3.
0.009 s
OK
1.
0.010 s
OK
2.
0.010 s
OK
3.
0.010 s
OK
expand all
Performance tests
1.
0.011 s
OK
2.
0.011 s
OK
3.
0.009 s
OK
1.
0.016 s
OK
2.
0.009 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.
0.011 s
OK
2.
0.012 s
OK
3.
0.009 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.
0.041 s
OK
2.
0.009 s
OK