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Minimum Remove to Make Valid Parentheses in C++
Suppose we have a string s of '(' , ')' and lowercase English characters. We have to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parenthese string is valid and return any valid string. A parentheses string is valid when all of these criteria are fulfilled −
It is the empty string, contains lowercase characters only, or
It can be written as the form of AB (A concatenated with B), where A and B are valid strings, or
It can be written as the form of (A), where A is a valid string.
So if the input is like “a)b(c)d”, then the output will be “ab(c)d”
To solve this, we will follow these steps −
Define a stack st
for i in range 0 to size of s
if s[i] = ‘(’, then insert i into st
otherwise when s[i] is ‘)’, then
if stack is not empty, then pop from stack, otherwise s[i] = ‘*’
while st is not empty,
s[top element of stack] = ‘*’
pop from stack
ans := empty string
for i in range 0 to size of s – 1
if s[i] is not ‘*’, then ans := ans + s[i]
return ans
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
string minRemoveToMakeValid(string s) {
stack <int> st;
for(int i = 0; i < s.size(); i++){
if(s[i] == '(')st.push(i);
else if(s[i] == ')'){
if(!st.empty())st.pop();
else s[i] = '*';
}
}
while(!st.empty()){
s[st.top()] = '*';
st.pop();
}
string ans = "";
for(int i = 0; i < s.size(); i++){
if(s[i] != '*')ans += s[i];
}
return ans;
}
};
main(){
Solution ob;
cout << (ob.minRemoveToMakeValid("a)b(c)d"));
}Input
"a)b(c)d"
Output
ab(c)d
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