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 = P_1 + P_2 + ... + P_k, where P_i 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:
"(()())(())"
Output:
"()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
Example 2:
Input:
"(()())(())(()(()))"
Output:
"()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
Example 3:
Input:
"()()"
Output:
""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".
Note:
S.length <= 10000S[i] is "(" or ")"S is a valid parentheses string
class Solution {
public String removeOuterParentheses(String S) {
int lv = 0;
String result = "";
for(char c : S.toCharArray()){
if(c == '('){
if(lv > 0){
result += c;
}
lv ++;
} else if(c == ')'){
lv--;
if(lv > 0){
result += c;
}
}
}
return result;
}
}
* Using a StringBuilder will somehow improve the performance.
Thoughts:
Since we can assume the input is valid, just use a counter to mark the depth. Send char to result only when depth >= 1
If I’m a Interviewer
- What if the input string is not always valid?
- What if we upgrade parentheses to brackets ()[]{} ?
- Use Stack and Stack.size() instead of level counter
- What if we remove inner most (max depth) parentheses instead?
- Go through twice, 1) find max depth, 2) copy to result and skip max depth.
Well we can always construct an array of ints (levels) as a form of representation.
For example, “(()())(())(()(()))” -> [1,2,2,1,2,1,2,2,3]
Then we can easily remove any levels (outer/inner/somewhere in the middle).
- Remove Outer: remove 1, everything else —
- Remove Inner: remove max
- Remove Level2: remove 2, everything > 2 decrease by 1.