LeetCode 1963. Minimum Number of Swaps to Make the String Balanced

原题链接在这里：https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/

题目：

You are given a 0-indexed string s of even length n. The string consists of exactly n / 2 opening brackets '[' and n / 2 closing brackets ']'.

A string is called balanced if and only if:

• It is the empty string, or
• It can be written as AB, where both A and B are balanced strings, or
• It can be written as [C], where C is a balanced string.

You may swap the brackets at any two indices any number of times.

Return the minimum number of swaps to make s balanced.

Example 1:

Input: s = "][]["
Output: 1
Explanation: You can make the string balanced by swapping index 0 with index 3.
The resulting string is "[[]]".

Example 2:

Input: s = "]]][[["
Output: 2
Explanation: You can do the following to make the string balanced:
- Swap index 0 with index 4. s = "[]][][".
- Swap index 1 with index 5. s = "[[][]]".
The resulting string is "[[][]]".

Example 3:

Input: s = "[]"
Output: 0
Explanation: The string is already balanced. 

Constraints:

• n == s.length
• 2 <= n <= 106
• n is even.
• s[i] is either '[' or ']'.
• The number of opening brackets '[' equals n / 2, and the number of closing brackets ']' equals n / 2.

题解：

Disregard all the solid pairs. 

Find out the number of unsolid pairs count.

The minimum swap number is the ceiling of count/2.

Time Complexity: O(n). n = s.length().

Space: O(1).

AC Java:

class Solution {
    public int minSwaps(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        
        int count = 0;
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(c == '['){
                count++;
            }else if(count > 0){
                count--;
            }
        }
        
        return (count + 1) / 2;
    }
}

AC C++:

class Solution {
public:
    int minSwaps(string s) {
        int count = 0;
        for(char c : s){
            if(c == '['){
                count++;
            }else if(count > 0){
                count--;
            }
        }

        return (count + 1) / 2;
    }
};

AC Python:

class Solution:
    def minSwaps(self, s: str) -> int:
        count = 0
        for c in s:
            if c == '[':
                count+=1
            elif count > 0:
                count-=1
        
        return (count + 1) // 2
        

类似Minimum Add to Make Parentheses Valid, Minimum Remove to Make Valid Parentheses.