Graphical LeetCode - 921. Minimal Addition to Make Parentheses Effective (Difficulty: Moderate)

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1. The topic

Bracketed strings are valid only if one of the following is true:

it is an empty string, or

It can be written AB ( A and B concatenated), where Aand B are both valid strings, or

It can be written (A)where A is a valid string.

Given a bracketed string s, move it N times and you can insert a bracket anywhere in the string.

For example, if s = "()))", you can insert an opening bracket as "(()))"or closing bracket as "())))".

Returns the minimum number of parentheses that must be added to make the resulting string s valid .

2. Examples

2.1> Example 1:

【Input】s = "())"
【Output】1

2.2> Example 2:

【Input】s = "((("
【Output】3

hint:

1 <= s.length <= 1000
sContains only '('and ')' characters.

Three, problem-solving ideas

这道题的题目描述真的挺让人费解的。其实题目的意思就是，我们如果想要配对好所有的括号，需要在原有字符串s的基础上，添加多少个括号（可能是左括号、也可能是右括号）。那么针对于这种配对类型类型的题目，第一个想法就是使用堆栈来实现。当然，对于括号配对的特殊性，即：左括号 + 右括号 。我们也可以根据这个规律去计算。如下是两种解题算法的详细解释。

3.1> 思路1：利用栈特性去计算

我们可以通过对字符串s进行每个字符的遍历，放到堆栈中。当发现栈顶字符是‘(’，待入栈的字符是‘)’，则符合括号匹配的情况。那么，此时我们只需将栈顶字符出栈即可。而针对于其他情况，我们都是将遍历的字符入栈即可。那么字符串s遍历完毕之后，我们来调用size()方法计算存储的字符长度，返回的长度就是这道题的结果。具体逻辑如下图所示：

针对该思路的代码实现，请参见：代码实现 4.1> 利用栈特性去计算

3.2> 思路2：找规律去匹配

The [hint] of the passing question sonly contains '(' and ')' characters. Therefore, for the matching of two characters, there are four cases as shown in the figure below. Then, only [case 1] will match successfully , and other cases will fail to match. Then we create two variables: leftCount(number of left parentheses) and rightCount(number of right parentheses).

If the traversed character is a left parenthesis , executeleftCount++

If the traversed character is a closing bracket and leftCount is not 0 , executeleftCount--

If the traversed character is a closing bracket and leftCount is equal to 0 , executerightCount++

For the code implementation of this idea, please refer to: Code Implementation 4.2> Find the rules to match

Fourth, code implementation

4.1> Implementation 1: Use the stack feature to calculate

class Solution {
    public int minAddToMakeValid(String s) {
        Deque<Character> deque = new ArrayDeque();
        for (char sc : s.toCharArray()) {
            if (deque.size() != 0 && (deque.peekLast()).equals('(') && sc == ')') deque.removeLast();
            else deque.addLast(sc);
        }
        return deque.size();
    }
}
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4.2> Realization 2: Find rules to match

class Solution {
    public int minAddToMakeValid(String s) {
        int leftCount = 0, rightCount = 0;
        for (char item : s.toCharArray()) {
            if (item == '(') {
                leftCount++;
            } else {
                if (leftCount == 0) rightCount++;
                else leftCount--;
            }               
        }
        return leftCount + rightCount;
    }
}
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