921. Minimum Add to Make Parentheses Valid #

Problem #

Given a string S of ‘(’ and ‘)’ parentheses, we add the minimum number of parentheses ( ‘(’ or ‘)’, and in any positions ) so that the resulting parentheses string is valid.

Formally, a parentheses string is valid if and only if:

It is the empty string, or
It can be written as AB (A concatenated with B), where A and B are valid strings, or
It can be written as (A), where A is a valid string. Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

Example 1:


Input: "())"
Output: 1

Example 2:


Input: "((("
Output: 3

Example 3:


Input: "()"
Output: 0

Example 4:


Input: "()))(("
Output: 4

Note:

S.length <= 1000
S only consists of ‘(’ and ‘)’ characters.

Problem Summary #

Given a parentheses string, if parentheses can be added at any position in this string, determine the minimum number of additions needed to make the entire string perfectly matched.

Solution Approach #

This is also a stack problem. Use a stack to perform parentheses matching. In the end, the number of parentheses left in the stack is the minimum number that needs to be added.

Code #


package leetcode

func minAddToMakeValid(S string) int {
	if len(S) == 0 {
		return 0
	}
	stack := make([]rune, 0)
	for _, v := range S {
		if v == '(' {
			stack = append(stack, v)
		} else if (v == ')') && len(stack) > 0 && stack[len(stack)-1] == '(' {
			stack = stack[:len(stack)-1]
		} else {
			stack = append(stack, v)
		}
	}
	return len(stack)
}