forked from sachuverma/DataStructures-Algorithms
-
Notifications
You must be signed in to change notification settings - Fork 0
/
Parenthesis Checker.cpp
73 lines (65 loc) · 1.66 KB
/
Parenthesis Checker.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
/*
Parenthesis Checker
===================
Given an expression string x. Examine whether the pairs and the orders of “{“,”}”,”(“,”)”,”[“,”]” are correct in exp.
For example, the function should return 'true' for exp = “[()]{}{[()()]()}” and 'false' for exp = “[(])”.
Example 1:
Input:
{([])}
Output:
true
Explanation:
{ ( [ ] ) }. Same colored brackets can form
balaced pairs, with 0 number of
unbalanced bracket.
Example 2:
Input:
()
Output:
true
Explanation:
(). Same bracket can form balanced pairs,
and here only 1 type of bracket is
present and in balanced way.
Example 3:
Input:
([]
Output:
false
Explanation:
([]. Here square bracket is balanced but
the small bracket is not balanced and
Hence , the output will be unbalanced.
Your Task:
This is a function problem. You only need to complete the function ispar() that takes a string as a parameter and returns a boolean value true if brackets are balanced else returns false. The printing is done automatically by the driver code.
Expected Time Complexity: O(|x|)
Expected Auixilliary Space: O(|x|)
Constraints:
1 ≤ |x| ≤ 32000
Note: The drive code prints "balanced" if function return true, otherwise it prints "not balanced".
*/
bool ispar(string x)
{
stack<char> s;
for (auto &i : x)
{
if (i == '(' || i == '{' || i == '[')
s.push(i);
else
{
if (!s.size())
return false;
if (i == ')' && s.top() == '(')
s.pop();
else if (i == ']' && s.top() == '[')
s.pop();
else if (i == '}' && s.top() == '{')
s.pop();
else
return false;
}
}
if (s.size())
return false;
return true;
}