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# Remove Outermost Parentheses.

**Problem of the day** - Remove Outermost Parentheses

**Tag** - Easy

A valid parentheses string is either empty `""`

, `"(" + A + ")"`

, or `A + B`

, where `A`

and `B`

are valid parentheses strings, and `+`

represents string concatenation.

- For example,
`""`

,`"()"`

,`"(())()"`

, and`"(()(()))"`

are all valid parentheses strings.

A valid parentheses string `s`

is primitive if it is nonempty, and there does not exist a way to split it into `s = A + B`

, with `A`

and `B`

nonempty valid parentheses strings.

Given a valid parentheses string `s`

, consider its primitive decomposition: `s = P1 + P2 + ... + Pk`

, where `Pi`

are primitive valid parentheses strings.

Return `s`

*after removing the outermost parentheses of every primitive string in the primitive decomposition of* `s`

.

**Example 1:**

**Input:** s = "(()())(())"
**Output:** "()()()"
**Explanation:**
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Description is pretty much clear. So, we just need to keep on tracking of the end of each valid sub-parentheses and skip the first and last element of it.

Algorithm is a bit tricky. Skipping first and last element of the valid sub-parentheses requires some extra attention.

Well, let's boogie!

`class Solution {public: string removeOuterParentheses(string s) { int count = 0; string res = ""; for(char c: s) { if(count == 0) { count += (c == '(' ? 1 : -1); } else { count += (c == '(' ? 1 : -1); if(count != 0) { res += c; } } } return res; }};`

I don't think, code requires some extra explanation. Still, if you feel, I am missing on something, feel free to reach out to me

I welcome your suggestions to improve it further!

Thanks for being part of my daily-code-workout journey. As always, if you have any thoughts about anything shared above, don't hesitate to reach out.

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