Removing Enclosed Parentheses- Google Top Interview Questions


Problem Statement :


You are given a string s containing valid number of brackets "(", ")" and lowercase alphabet characters. Return the string after removing all outermost brackets that enclose the whole string.

Constraints

n ≤ 100,000 where n is the length of s

Example 1

Input

s = "(((abc)))"

Output

"abc"

Explanation

The three "((()))" brackets are outermost so we remove them.


Example 2

Input

s = "(((abc)))(d)"

Output

"(((abc)))(d)"

Explanation

There's no outermost brackets enclosing the string, so we return the same string.



Example 3

Input

s = "()"

Output

""

Explanation

We remove the outermost brackets.



Solution :



title-img




                        Solution in C++ :

string solve(string s) {
    int pref = 0;
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == '(') {
            pref++;
        } else {
            break;
        }
    }
    int suff = 0;
    for (int i = 0; i < pref; i++) {
        if (s[s.size() - 1 - i] == ')') {
            suff++;
        } else {
            break;
        }
    }
    int currDepth = pref;
    int minDepth = min(pref, suff);
    for (int i = pref; i < s.size() - pref; i++) {
        if (s[i] == '(') {
            currDepth++;
        } else if (s[i] == ')') {
            currDepth--;
            minDepth = min(minDepth, currDepth);
        }
    }
    return s.substr(minDepth, s.size() - 2 * minDepth);
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public String solve(String s) {
        int A[] = new int[s.length()];
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                stack.push(i);
            } else if (c == ')') {
                int match = stack.pop();
                A[i] = match;
            }
        }
        int i = 0, j = s.length() - 1;
        while (s.charAt(i) == '(' && s.charAt(j) == ')' && A[j] == i) {
            i++;
            j--;
        }
        return s.substring(i, j + 1);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, s):
        """
        (((abc)))
        123   321

        (((abc)))(d)
        123   3211 1
        """
        N = len(s)

        open_map = defaultdict(list)

        closed_map = defaultdict(list)

        curr_count = 0

        for idx, ch in enumerate(s):
            if ch == "(":
                curr_count += 1

                open_map[curr_count].append(idx)
            elif ch == ")":
                closed_map[curr_count].append(idx)

                curr_count -= 1

        if not open_map:
            return s

        to_remove = set()
        """
            The below code starts verifying the saved indices in the hash_maps
        """
        i = 0
        j = N - 1

        for open_ct in open_map:

            curr_open_arr = open_map[open_ct]
            curr_closed_arr = closed_map[open_ct]
            """
                The below code checks if There's only one open/closed brace and whether those braces have appeared at the ends (0, N -1), (1, N - 2), (2, N - 3) ---
            """
            if len(curr_open_arr) == 1 and curr_open_arr[0] == i and curr_closed_arr[0] == j:
                to_remove.add(curr_open_arr[0])
                to_remove.add(curr_closed_arr[0])
                i += 1
                j -= 1
            else:
                """
                If any of those are not valid, then the string is not enclosed
                """
                break

        if len(to_remove) == 0:
            return s

        new_s = []

        for idx, ch in enumerate(s):
            if idx in to_remove:
                continue
            new_s.append(ch)

        return "".join(new_s)
                    


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