**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 :

` ````
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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