Minimum Remove to Make Valid Parentheses

class Solution {
    /**
    * A valid pair is considered, when I have seen a left paren first and then I saw a right paren. 
    * 1) find the total count (n) of properly paired parenthesis, so there are n number of '(' and n number of ')'
    * 2) iterate over the string, if you see an '(', add it and decrement counter
            * If you see an ')', add it as well and decrement the counter
            * once the counter reaches 0, do not add any of the respective
    **/
    public String minRemoveToMakeValid(String s) {
        if (s == null || s.length() == 0)
            return s;
        
        int pairCount = getValidParenPairCount(s) ;
        
        return getStringWithValidParen(s, pairCount);
    }
    
    private String getStringWithValidParen(String s, int pairCount) {
        int leftCount = pairCount;
        int rightCount = pairCount;
        
        StringBuilder sb = new StringBuilder();
        int i = 0;
        
        while(i < s.length()) {
            char c = s.charAt(i);
            
            switch(c) {
                case '(':
                    if (leftCount > 0) {
                        sb.append(c);
                        leftCount--;
                    }
                    break;
                case ')':
                    // I have to use a left paren first and then I can take a right paren
                    if (rightCount > 0 && rightCount > leftCount) {
                        sb.append(c);
                        rightCount--;
                    }
                    break;
                default:
                    sb.append(c);
                    break;
            }
            i++;
        }
        return sb.toString();
    }
    
    private int getValidParenPairCount(String s) {
        Stack<Character> st = new Stack<Character>();
        
        int count = 0;
        int i = 0;
        
        while(i < s.length()) {
            char c = s.charAt(i);
            
            switch(c) {
                case '(':
                    st.push(c);
                    break;
                case ')':
                    if (!st.isEmpty()) {
                        st.pop();
                        count++;
                    }
                    break;
                default:
                    break;
            }
            i++;
        }
        return count;
    }
}