# Brackets Problem(Determine whether a given string of parentheses is properly nested)

Problem:

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:
int solution(char *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.
Assume that:
N is an integer within the range [0..200,000];
string S consists only of the following characters: “(“, “{“, “[“, “]”, “}” and/or “)”.
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N) (not counting the storage required for input arguments).

Solution of the Problem:

private int characterSymmatry(String str){
Stack<Character> stack = new Stack<Character>();

for (int i = 0; i < str.length(); i++) {
char characters = str.charAt(i);
switch (characters) {
case '{':
stack.push(characters);
break;
case '[':
stack.push(characters);
break;

case '(':
stack.push(characters);
break;

case ')':
if(!stack.isEmpty()){
char c= stack.pop();
if(c!='('){
return 0;
}
}

break;
case ']':
if(!stack.isEmpty()){
char c2 = stack.pop();
if(c2!='['){
return 0;
}
}
break;
case '}':
if(!stack.isEmpty()){
char c3 = stack.pop();
if(c3!='{'){
return 0;
}
}
break;

default:
break;
}
}

if(stack.isEmpty()){
return 1;
}

return 0;
}

1. Alene