Write a piece of code which returns true if the string contains balanced parenthesis, false otherwise.

[answerhappygod]

a)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() == null)
{
return false;
}
stk.pop();
}
}
return true;
}
b)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() != null)
{
return true;
}
stk.pop();
}
}
return false;
}
c)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == ')')
stk.push(i);
else if (ch == '(')
{
if(stk.peek() == null)
{
return false;
}
stk.pop();
}
}
return true;
}
d)
public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() != null)
{
return false;
}
stk.pop();
}
}
return true;
}
View Answer
Answer: a
Explanation: Whenever a ‘(‘ is encountered, push it into the stack, and when a ‘)’ is encountered check the top of the stack to see if there is a matching ‘(‘, if not return false, continue this till the entire string is processed and then return true.



public boolean isBalanced(String exp)
{
int len = exp.length();
Stack<Integer> stk = new Stack<Integer>();
for(int i = 0; i < len; i++)
{
char ch = exp.charAt(i);
if (ch == '(')
stk.push(i);
else if (ch == ')')
{
if(stk.peek() == null)
{
return false;
}
stk.pop();
}
}
return true;
}
a) O(logn)
b) O(n)
c) O(1)
d) O(nlogn)