Remove Outermost Parentheses
Problem Statement - link #
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.
Examples #
Example 1:
Input: s = "(()())(())"
Output: "()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
Example 2:
Input: s = "(()())(())(()(()))"
Output: "()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
Example 3:
Input: s = "()()"
Output: ""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".
Constraints #
1 <= s.length <= 10^5
s[i]
is either'('
or')'
s
is a valid parentheses string
Solutions #
class Solution {
public:
string removeOuterParentheses(string s) {
string result = "";
int index1 = 0;
stack<char> st;
for(int i=0; i<s.length(); i++ ) {
if( s[i]=='(' ) {
st.push('(');
}
else {
st.pop();
if(st.empty()) {
result += s.substr( index1+1, (i-1-index1) );
index1 = i+1;
}
}
}
return result;
}
};