Given a balanced expression, find if it contains duplicate parenthesis or not. A set of parenthesis are duplicate if the same subexpression is surrounded by multiple parenthesis.
Below expressions have duplicate parenthesis - ((a+b)+((c+d))) The subexpression "c+d" is surrounded by two pairs of brackets. (((a+(b)))+(c+d)) The subexpression "a+(b)" is surrounded by two pairs of brackets. (((a+(b))+c+d)) The whole expression is surrounded by two pairs of brackets. Below expressions don't have any duplicate parenthesis - ((a+b)+(c+d)) No subsexpression is surrounded by duplicate brackets. ((a+(b))+(c+d)) No subsexpression is surrounded by duplicate brackets.
It may be assumed that the given expression is valid and there are not any white spaces present.
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.
Below is the implementation of above idea :
Time complexity of above solution is O(n).
Auxiliary space used by the program is O(n).
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