Parenthesis Checker gfgs [closed] - c++

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The following code is giving segmentation fault can anyone tell why. this is a geeks for geeks practice problem (Parenthesis Checker) .
Q- 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 = “[(])”.
bool check(string s)
{
stack<char> save;
int x = s.size();
for(int i=0;i<x;i++)
{
if(s[i]=='{' || s[i]=='['|| s[i]=='(')
{
save.push(s[i]);
}
else if(s[i]=='}'){
if(save.top()=='{')
{
save.pop();
}
else{
return false;
}
}
else if(s[i]==']')
{
if(save.top()=='[')
{
save.pop();
}
else{
return false;
}
}
else if(s[i]==')')
{
if(save.top()=='(')
{
save.pop();
}
else{
return false;
}
}
}
if(!save.empty())
{
return false;
}
else{
return true;
}
}

Answer 1

You have a chance to seg-fault when the stack is empty and the next character in the string is a closing paren.
E.g. if the input string is ")" or "())" your program will not behave as intended.
You will need to change the if statements from
if (save.top() == '('){...} to if (!save.empty() && save.top() == '('){...}
And similarly for every other case where you check save.top (The preceeding code does not guarantee that the stack is not empty at that given point).
Note: You would also need to do this for each occurrence of pop as well, but the guards before top() will guarantee that the stack is not empty when you then pop.