Remove Minimum Parantheses

Problem Statement

Given a string s of ‘(‘ , ‘)’ and lowercase English characters.

Your task is to remove the minimum number of parentheses ( ‘(‘ or ‘)’, in any positions ) so that the resulting parentheses string is valid and return any valid string.

Formally, a parentheses string is valid if and only if:

It is the empty string, contains only lowercase characters, or It can be written as AB (A concatenated with B), where A and B are valid strings, or It can be written as (A), where A is a valid string.

Example 1:
Input: path = s = “lee(t(c)o)de)”
Output: “lee(t(c)o)de”
Explanation: “lee(t(co)de)” , “lee(t(c)ode)” would also be accepted.

Example 2: Input: s = “a)b(c)d”
Output: “ab(c)d”

Example 3: Input: s = “))((“
Output: ““

Constraints:

• 1 <= s.length <= 105
• s[i] is either’(‘ , ‘)’, or lowercase English letter.

Approach

A classic Stack problem.

• Convert input string to list, it’ll be easier to access the elements using index.
• Lopping through the string list.
• If the element is ( we append it in the stack.
• Else if it’s ) then we check if top indexed element from stack in the string is ( and if it is we pop the index from stack.
• Else we append it to the stack.
• Finally we delete the elements from string which are at indices stored in stack.
• Return String.

Solution

def simplifyPath(string):
    stack = []
    string = list(string)

    for i in range(len(string)):
        if string[i] == "(":
            stack.append(i)
        elif string[i] == ")":
            if stack and string[stack[-1]]=="(":
                stack.pop()
            else:
                stack.append(i)
        while stack:
            del string[stack[-1]]
            stack.pop()
        
        return "".join(string)

Conclusion

Time and Space Complexity both are O(N) where N is length of the string.
Reason for that is there is one pass over the string and stack of maximum length N.

Refrence

• Code with Alisha