i cant figure out how to do this question using stacks or queues
Redundant parentheses
Hi @sktg99
The idea is to use stack. Iterate through the given expression and for each character in the expression, if the character is a open parenthesis ‘(‘ or any of the operators or operands, push it to the top of the stack. If the character is close parenthesis ‘)’, then pop characters from the stack till matching open parenthesis ‘(‘ is found and a counter is used, whose value is incremented for every character encountered till the opening parenthesis ‘(‘ is found. If the number of characters encountered between the opening and closing parenthesis pair, which is equal to the value of the counter, is less than or equal to 1, then a pair of duplicate parenthesis is found else there is no occurrence of redundant parenthesis pairs. For example, (((a+b))+c) has duplicate brackets around “a+b”. When the second “)” after a+b is encountered, the stack contains “((“. Since the top of stack is a opening bracket, it can be concluded that there are duplicate brackets.
Hope it helps
Mark resolved if satisfied 
why counter == 1 is also included in duplication case? i think it will always be 0 when there is duplication