System Engineer

Declare a character stack S.
Now traverse the expression string exp. 
If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack.
If the current character is a closing bracket (‘)’ or ‘}’ or ‘]’) then pop from stack and if the popped character is the matching starting bracket then fine else brackets are not balanced.
After complete traversal, if there is some starting bracket left in stack then “not balanced”

The strategy here uses a temporary dummy node as the start of the result list. The pointer Tail always points to the last node in the result list, so appending new nodes is easy. 

The dummy node gives the tail something to point to initially when the result list is empty. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’, and adding it to the tail. When 

We are done, the result is in dummy.next.

The idea is to traverse every array element and find the highest bars on the left and right sides. Take the smaller of two heights. The difference between the smaller height and height of the current element is the amount of water that can be stored in this array element.