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Given a balanced expression that can contain opening and closing parenthesis, check if it contains any duplicate parenthesis or not. If the expression contains duplicate parenthesis, the program outputs TRUE. Else, FALSE. Notes: o assume that the input expression is valid o the expected average and worst complexity is O (n) Examples: Input: ((x+y))+z Output: true 1. Explanation: Duplicate () found in subexpression ((x+y)) Input: (x+y) 2. Output: false Explanation: No duplicate () is found Input: (x+y)+((z))) 3. Output: true Explanation: Duplicate () found in subexpression ((z)) Hint: We can use a stack to solve this problem. The idea is to traverse the given expression and • If the current character in the expression is not a closing parenthesis )', push the character into the stack. • If the current character in the expression is a closing parenthesis '), check if the topmost element in the stack is an opening parenthesis or not. If it is an opening parenthesis, then the subexpression ending at the current character is of the form ((exp)): otherwi continue popping characters from the stack till matching (" is found for current )".

Given a balanced expression that can contain opening and closing parenthesis, check if it contains any duplicate parenthesis or not. If the expression contains duplicate parenthesis, the program outputs TRUE. Else, FALSE. Notes: o assume that the input expression is valid o the expected average and worst complexity is O (n) Examples: Input: ((x+y))+z Output: true 1. Explanation: Duplicate () found in subexpression ((x+y)) Input: (x+y) 2. Output: false Explanation: No duplicate () is found Input: (x+y)+((z))) 3. Output: true Explanation: Duplicate () found in subexpression ((z)) Hint: We can use a stack to solve this problem. The idea is to traverse the given expression and • If the current character in the expression is not a closing parenthesis )', push the character into the stack. • If the current character in the expression is a closing parenthesis '), check if the topmost element in the stack is an opening parenthesis or not. If it is an opening parenthesis, then the subexpression ending at the current character is of the form ((exp)): otherwi continue popping characters from the stack till matching (" is found for current )".

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Given a balanced expression that can contain opening and closing parenthesis, check if it contains any duplicate parenthesis or not. If the expression contains duplicate parenthesis, the program outputs TRUE. Else, FALSE. Notes: o assume that the input expression is valid o the expected average and worst complexity is O ( n) Examples: Input: ((x+y))+z Output: true 1. Explanation: Duplicate () found in subexpression ((x+y)) Input: (x+y) 2. Output: false Explanation: No duplicate () is found Input: ((x+y)+((z))) 3. Output: true Explanation: Duplicate () found in subexpression ((z)) Hint: We can use a stack to solve this problem. The idea is to traverse the given expression and • If the current character in the expression is not a closing parenthesis ')', push the character into the stack. • If the current character in the expression is a closing parenthesis ')', check if the topmost element in the stack is an opening parenthesis or not. If it is an opening parenthesis, then the subexpression ending at the current character is of the form ((exp)); otherwise, continue popping characters from the stack till matching (' is found for current ')'.
Transcribed Image Text:Given a balanced expression that can contain opening and closing parenthesis, check if it contains any duplicate parenthesis or not. If the expression contains duplicate parenthesis, the program outputs TRUE. Else, FALSE. Notes: o assume that the input expression is valid o the expected average and worst complexity is O ( n) Examples: Input: ((x+y))+z Output: true 1. Explanation: Duplicate () found in subexpression ((x+y)) Input: (x+y) 2. Output: false Explanation: No duplicate () is found Input: ((x+y)+((z))) 3. Output: true Explanation: Duplicate () found in subexpression ((z)) Hint: We can use a stack to solve this problem. The idea is to traverse the given expression and • If the current character in the expression is not a closing parenthesis ')', push the character into the stack. • If the current character in the expression is a closing parenthesis ')', check if the topmost element in the stack is an opening parenthesis or not. If it is an opening parenthesis, then the subexpression ending at the current character is of the form ((exp)); otherwise, continue popping characters from the stack till matching (' is found for current ')'.
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